SECTION 13.3
A RC LENGTH AND C URVAT URE
877
48.
34.
35– 40
35.
y
49.
36.
y
50.
37.
y
51.
38.
y
39.
y
40.
y
s
52.
s
53.
s
41.
42.
54.
43.
44.
45.
55.
46.
47.
56.
Arc Length and Curvature
13.3
t
1
t
y
t
s
z
t
t
z
t
0
x
FIGURE 1
y
2
y
s
t
z
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878
CHAPTER 13
VECT OR FU NCTIONS
y
3
t
t
s
t
s
t
t
t
v
Figure 2 shows the arc of the helix
whose length is computed in Example 1.
z
EXAMPLE 1
SOLUTION
s
s
(1, 0, 2π)
(1, 0, 0)
x
y
y
y
s
s
FIGURE 2
4
5
t
z
s(t)
C
6
(t)
(a)
y
z
0
x
FIGURE 3
y
7
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SECTION 13.3
A RC LE NGTH AND C URVAT URE
879
EXAMPLE 2
SOLUTION
s
y
y
s
s
s
s
s
s
Curvature
TEC Visual 13.3A shows animated unit
tangent vectors, like those in Figure 4, for
a variety of plane curves and space curves.
z
0
x
C
y
FIGURE 4
8
Definition
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880
CHAPTER 13
VECT OR FU NCTIONS
9
v
EXAMPLE 3
SOLUTION
10 Theorem
PROOF
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SECTION 13.3
A RC LE NGTH AND C URVAT URE
881
EXAMPLE 4
SOLUTION
s
s
s
s
s
11
EXAMPLE 5
SOLUTION
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882
CHAPTER 13
VECT OR FU NCTIONS
l
l
l
y
2
y=≈
y=k(x)
FIGURE 5
y=≈
0
1
x
The Normal and Binormal Vectors
We can think of the normal vector as indicating
the direction in which the curve is turning at
each point.
(t)
(t)
(t)
FIGURE 6
Figure 7 illustrates Example 6 by showing the
vectors , , and at two locations on the
helix. In general, the vectors , , and , starting at the various points on a curve, form a set
of orthogonal vectors, called the
frame,
that moves along the curve as varies. This
frame plays an important role in the
branch of mathematics known as differential
geometry and in its applications to the motion
of spacecraft.
EXAMPLE 6
SOLUTION
s
s
z
s
y
s
z
x
FIGURE 7
s
s
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SECTION 13.3
A RC LE NGTH AND C URVAT URE
883
TEC Visual 13.3B shows how the TNB frame
moves along several curves.
v
Figure 8 shows the helix and the osculating
plane in Example 7.
EXAMPLE 7
SOLUTION
z
z
z=_x+π2
z
P
x
y
s
FIGURE 8
s
z
s
z
EXAMPLE 8
y
SOLUTION
y=≈
1
2
0
1
x
FIGURE 9
TEC Visual 13.3C shows how the osculating
circle changes as a point moves along a curve.
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884
CHAPTER 13
13.3
VECT OR FU NCTIONS
Exercises
17–20
1–6
1.
2.
3.
17.
18.
s
4.
19.
5.
20.
s
6.
21–23
21.
7–9
22.
7.
23.
8.
9.
24.
; 10.
25.
z
; 26.
z
11.
z
27–29
27.
12.
z
28.
29.
30–31
l
13–14
30.
31.
13.
32.
14.
33.
15.
z
y
16.
P
C
1
Q
0
;
CAS
1
x
1.
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SECTION 13.3
885
47– 48
; 34–35
47.
34.
CAS
A RC LE NGTH AND C URVAT URE
48.
35.
49–50
36–37
49.
36.
50.
s
37.
z
z
; 51.
38–39
38.
39.
y
; 52.
y
a
a
b
b
x
53.
x
z
z
CAS
54.
z
CAS
40.
55.
CAS
z
41.
42.
56.
t
57.
58.
43– 45
43.
44.
45.
59.
46.
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886
CHAPTER 13
VECT OR FUNCTION S
60.
63.
1.
64.
z
2.
65.
3.
66.
61.
62.
;
Motion in Space: Velocity and Acceleration
13.4
z
ª(t)
P
(t)
C
(t+h)- (t)
h
Q
1
(t+h)
O
x
FIGURE 1
y
2
l
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SECTION 13.4
MOTI ON IN SPACE: VELOCITY A ND AC CELER ATION
887
EXAMPLE 1
y
SOLUTION
(1)
(1, 1)
(1)
x
0
s
FIGURE 2
s
TEC Visual 13.4 shows animated velocity
and acceleration vectors for objects moving along
various curves.
Figure 3 shows the path of the particle in
Example 2 with the velocity and acceleration
vectors when
.
z
s
EXAMPLE 2
SOLUTION
(1)
(1)
s
1
x
FIGURE 3
y
v
EXAMPLE 3
SOLUTION
y
y
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888
CHAPTER 13
VECT OR FU NCTIONS
The expression for
that we obtained in
Example 3 was used to plot the path of the
particle in Figure 4 for
.
y
y
6
z 4
2
(1, 0, 0)
0
0
5
y
10
0
15
20
20
x
FIGURE 4
y
y
EXAMPLE 4
The angular speed of the object moving with
position is
, where is the
angle shown in Figure 5.
y
0
SOLUTION
P
¨
x
FIGURE 5
y
v
EXAMPLE 5
¸
a
0
d
x
SOLUTION
t
FIGURE 6
t
t
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SECTION 13.4
MOTI ON IN SPACE: VELOCITY A ND AC CELER ATION
889
t
t
t
t
3
v
v
v
v
If you eliminate from Equations 4, you will
see that is a quadratic function of . So the
path of the projectile is part of a parabola.
v
4
t
v
t
v
v
v
v
t
v
v
t
v
t
t
EXAMPLE 6
SOLUTION
t
v
s
s
s
s
s
s
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890
CHAPTER 13
VECT OR FU NCTIONS
s
s
s
s
Tangential and Normal Components of Acceleration
v
v
v
v
5
6
v
v
v
v
v
7
v
8
v
v
FIGURE 7
v
v
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SECTION 13.4
MOTI ON IN SPACE: VELOCITY A ND AC CELER ATION
891
v
v
v
v
vv
v
vv
v
9
v
v
10
EXAMPLE 7
SOLUTION
s
s
s
s
Kepler’s Laws of Planetary Motion
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892
CHAPTER 13
VECT OR FU NCTIONS
Kepler’s Laws
1.
2.
3.
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SECTION 13.4
MOTI ON IN SPACE: VELOCITY A ND AC CELER ATION
893
z
11
¨
y
x
FIGURE 8
12
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894
13.4
CHAPTER 13
VECT OR FU NCTIONS
Exercises
1.
9–14
9.
11.
10.
s
12.
13.
z
14.
15–16
15.
16.
2.
17–18
;
17.
18.
y
19.
(2.4)
2
(2)
1
(1.5)
0
1
20.
21.
x
2
22.
3–8
23.
3.
4.
24.
s
5.
25.
6.
26.
7.
27.
8.
;
1.
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SECTION 13.4
MOTI ON IN SPACE: VELOCITY A ND AC CELER ATION
895
28.
37– 42
29.
37.
38.
39.
40.
41.
30.
31.
s
42.
43.
y
32.
0
x
; 33.
44.
45.
46.
34.
35.
36.
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896
CHAPTER 13
VECTOR FUNCTIONS
AP PLIED PROJECT
KEPLER’S LAWS
Kepler’s Laws
1.
2.
3.
1.
y
(t)
A(t)
0
(t¸)
x
2.
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CHAPTER 13
R E VIEW
897
3.
4.
13
Review
Concept Check
1.
6.
2.
3.
7.
4.
8.
5.
9.
True-False Quiz
Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement.
1.
2.
7.
8.
9.
3.
4.
10.
11.
5.
12.
13.
6.
14.
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898
CHAPTER 13
VECT OR FU NCTIONS
Exercises
1.
y
s
2.
C
1
l
(3)
(3.2)
3.
z
0
; 4.
s
z
x
5.
x
17.
18.
6.
z
1
z
19.
7.
z
8.
20.
9.
21.
10.
11.
12.
13.
; 14.
15.
z
16.
;
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CHAPTER 13
899
R E VIEW
22.
y
s
s
s
s
vt
s
s
s
s
24.
;
v
t
v
y
1
0
y
y=F(x)
1
œ2
x
y=x
y=0
0
1
t
v
x
v
23.
t
¨
mg
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Problems Plus
y
1.
v
v
0
_R
t
v
t
v t
R x
v
y
0
x
D
v
2.
v
FIGURE FOR PROBLEM 1
y
¸
a
x
¨
FIGURE FOR PROBLEM 2
3.
¨ ¨
FIGURE FOR PROBLEM 3
4.
y
; 5.
v
y
v
t
v
6.
7.
900
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14
Partial Derivatives
Graphs of functions of two variables are
surfaces that can take a variety of
shapes, including that of a saddle or
mountain pass. At this location in
southern Utah (Phipps Arch) you can
see a point that is a minimum in one
direction but a maximum in another
direction. Such surfaces are discussed
in Section 14.7.
Photo by Stan Wagon, Macalester College
901
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902
CHAPTER 14
14.1
PARTIAL DERIVAT IVES
Functions of Several Variables
■
■
■
■
Functions of Two Variables
f
Definition
z
z
z
y
f(x, y)
(x, y)
0
D
FIGURE 1
(a, b)
z
x
0
z
f(a, b)
EXAMPLE 1
s
SOLUTION
s
s
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SECTION 14.1
x+y+1=0
FUNCTIONS O F SEV ERAL VARIAB LES
903
y
x=1
_1
0
x
_1
FIGURE 2
f(x, y)=
œ„„„„„„„
x+y+1
x-1
EXAMPLE 2
y
v
v
v
x=¥
0
x
TABLE 1
/
v
f(x, y)=x (¥-x)
°
FIGURE 3
T
The New Wind-Chill Index
A new wind-chill index was introduced in
November of 2001 and is more accurate than
the old index for measuring how cold it feels
when it’s windy. The new index is based on a
model of how fast a human face loses heat. It
was developed through clinical trials in which
volunteers were exposed to a variety of temperatures and wind speeds in a refrigerated wind
tunnel.
EXAMPLE 3
1
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904
CHAPTER 14
PARTIAL DERIVAT IVES
TABLE 2
2
t
EXAMPLE 4
SOLUTION
y
s
t
≈+¥=9
t
_3
3
z
x
s
z
z
z
s
FIGURE 4
9-≈-¥
g(x, y)=œ„„„„„„„„„
z
z
Graphs
z
S
{ x, y, f (x, y)}
Definition
z
f(x, y)
0
x
D
(x, y, 0)
y
z
z
FIGURE 5
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SECTION 14.1
z
FUNCTIONS O F SEV ERAL VARIAB LES
905
EXAMPLE 5
(0, 0, 6)
z
SOLUTION
z
z
z
(0, 3, 0)
(2, 0, 0)
y
x
FIGURE 6
z
z
0
(3, 0, 0)
v
(0, 0, 3)
z
t
EXAMPLE 6
z
SOLUTION
z
(0, 3, 0)
s
s
z
z
t
y
x
NOTE
FIGURE 7
t
g(x, y)=œ„„„„„„„„„
9-≈-¥
s
z
s
EXAMPLE 7
SOLUTION
300
P
200
100
0
300
FIGURE 8
v
200
100
K
0 0
100
L
200
300
EXAMPLE 8
SOLUTION
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906
CHAPTER 14
PARTIAL DERIVAT IVES
z
z
FIGURE 9
x
h(x, y)=4≈+¥
y
z
z
x
y
x
f(x, y)=(≈+3¥)e _≈_¥
f(x, y)=(≈+3¥)e _≈_¥
z
z
x
y
x
f(x, y)=
FIGURE 10
x+
y
y
f(x, y)=
x
xy
y
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SECTION 14.1
FUNCTIONS O F SEV ERAL VARIAB LES
907
Level Curves
Definition
z
z
45
LONESOME MTN.
0
B
y
x
A
k=45
f(x, y)=20
FIGURE 11
k=40
k=35
k=30
k=25
k=20
FIGURE 12
TEC Visual 14.1A animates Figure 11 by
showing level curves being lifted up to graphs
of functions.
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908
CHAPTER 14
PARTIAL DERIVAT IVES
FIGURE 13
From
4th Edition, 1989.
© 1989 Pearson Education, Inc.
y
EXAMPLE 9
50
5
4
z
SOLUTION
3
2
1
0
1
80
70
60
50
2
3
80
70
60
4
5
x
FIGURE 14
EXAMPLE 10
SOLUTION
y
0
x
FIGURE 15
f(x, y)=6-3x-2y
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SECTION 14.1
v
FUNCTIONS O F SEV ERAL VARIAB LES
909
EXAMPLE 11
t
s
SOLUTION
s
s
t
y
k=3
k=2
k=1
k=0
(3, 0)
0
x
FIGURE 16
g(x, y)=œ„„„„„„„„„
9-≈-¥
EXAMPLE 12
SOLUTION
s
s
y
z
TEC Visual 14.1B demonstrates the
connection between surfaces and their
contour maps.
x
x
FIGURE 17
h(x, y)=4≈+¥+1
y
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910
CHAPTER 14
PARTIAL DERIVAT IVES
K
EXAMPLE 13
SOLUTION
L
FIGURE 18
z
y
z
x
x
y
f(x, y)=_xye_≈_¥
f(x, y)=_xye_≈_¥
z
y
x
y
x
FIGURE 19
f(x, y)=
_3y
≈+¥+1
f(x, y)=
_3y
≈+¥+1
Functions of Three or More Variables
z
z
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SECTION 14.1
FUNCTIONS O F SEV ERAL VARIAB LES
911
EXAMPLE 14
z
z
z
z
SOLUTION
z
z
z
z
z
z
z
≈+¥+z@=9
z
EXAMPLE 15
≈+¥+z@=4
z
SOLUTION
s
z
z
z
z
y
x
z
≈+¥+z@=1
FIGURE 20
3
1.
2.
3.
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912
14.1
CHAPTER 14
PARTIAL DERIVAT IVES
Exercises
1.
v
v
v
v
5.
v
w
w
w
2.
6.
v
TABLE 3
v
v
v
T
7.
°
v
v
v
TABLE 4
√
t
3.
4.
8.
;
1.
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SECTION 14.1
913
FUNCTIONS O F SEV ERAL VARIAB LES
z
z
z
z
y
x
9.
t
z
z
t
t
t
s
10.
y
x
x
y
z
11.
12.
s
z
t
z
s
sz
z
z
zs
t
y
x
z
x
t
y
x
y
33.
13–22
s
13.
15.
17.
s
18.
s
19.
s
14.
s
16.
s
y
s
s
1
0
20.
21.
z
22.
z
s
70 60 50 40
1
30
x
20
10
z
z
34.
23–31
23.
24.
25.
26.
27.
28.
29.
30.
31.
s
s
32.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
914
CHAPTER 14
PARTIAL DERIVAT IVES
35.
39– 42
y
39.
40.
y
_8
_6
x
_4
8
41.
y
y
42.
5
4
3
2
36.
2
1
t
0
y
0
0
x
1
2
3
x
_3
_2
_1
0
1
3
4 5
x
x
43–50
43.
45.
y
44.
s
46.
47.
49.
x
48.
s
50.
51– 52
51.
37.
52.
s
53.
38.
z
54.
y
x
s
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.1
; 55– 58
FUNCTIONS O F SEV ERAL VARIAB LES
915
59–64
59. z
60. z
55.
61. z
62. z
56.
63. z
57.
64. z
58.
z
z
z
y
x
y
y
x
z
x
z
z
x
y
y
y
x
y
y
x
x
y
y
x
y
x
x
y
x
x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
916
CHAPTER 14
PARTIAL DERIVAT IVES
; 76.
65– 68
65.
z
66.
z
67.
z
68.
z
70.
z
z
z
z
; 77.
z
t
69–70
69.
z
; 78.
t
t
t
t
s
s
t
t
t
s
s
; 71–72
s
t
ts
t
71.
72.
; 79.
; 73–74
73.
74.
; 75.
14.2
Limits and Continuity
t
t
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.2
TABLE 1
x
TABLE 2
y
x
L IMITS AND CONTINUITY
917
t
y
t
l
l
l
1
Definition
l
s
l
l
l
l
s
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
918
CHAPTER 14
PARTIAL DERIVAT IVES
z
z
y
L+∑
L
L-∑
(x, y)
∂
D
L+∑
L
L-∑
f
(
x
)
(a, b)
0
S
0
0
x
FIGURE 1
y
D∂
(a, b)
FIGURE 2
z
z
l
l
l
y
b
0
a
x
FIGURE 3
l
l
l
l
l
v
l
EXAMPLE 1
l
SOLUTION
y
f=_1
l
f=1
l
x
l
l
FIGURE 4
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.2
EXAMPLE 2
L IMITS AND CONTINUITY
919
l
SOLUTION
y
y=x
f=0
l
l
l
l
1
f= 2
x
f=0
l
l
FIGURE 5
z
y
TEC In Visual 14.2 a rotating line on the
surface in Figure 6 shows different limits at
the origin from different directions.
x
FIGURE 6
f(x, y)=
xy
≈+¥
v
EXAMPLE 3
l
SOLUTION
l
Figure 7 shows the graph of the function in
Example 3. Notice the ridge above the
parabola
.
l
0.5
z 0
_0.5
2
0
x
_2
2
_2
0 y
l
l
FIGURE 7
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
920
CHAPTER 14
PARTIAL DERIVAT IVES
l
2
l
l
EXAMPLE 4
l
l
l
SOLUTION
s
s
s
3
s
s
s
Another way to do Example 4 is to use the
Squeeze Theorem instead of Definition 1. From
2 it follows that
l
and so the first inequality in 3 shows that the
given limit is 0.
l
Continuity
l
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SECTION 14.2
L IMITS AND CONTINUITY
921
Definition
4
l
t
t
v
EXAMPLE 5
t
l
SOLUTION
l
EXAMPLE 6
SOLUTION
EXAMPLE 7
t
t
t
l
t
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
922
CHAPTER 14
PARTIAL DERIVAT IVES
EXAMPLE 8
Figure 8 shows the graph of the continuous
function in Example 8.
z
y
x
l
l
FIGURE 8
t
t
2
z 0
_2
_2
t
EXAMPLE 9
_2
_1
y
0
1
1
0 x
_1
SOLUTION
t
2 2
t
FIGURE 9
h(x, y)=
(y/x)
x=0
Functions of Three or More Variables
z
z l
z
z
s
z
z
s
z
z
z
z
z l
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.2
L IMITS AND CONTINUITY
923
z
5
l
z
s
l
Exercises
14.2
1.
l
11.
2.
13.
15.
17.
3– 4
19.
l
3.
20.
4.
21.
5–22
5.
7.
9.
;
22.
l
l
l
6.
8.
10.
12.
l
l
14.
s
16.
l
l
18.
s
l
l
l
l
z
z l
z
z
z l
z
z
z
z l
z
z l
z
l
; 23–24
l
23.
l
24.
l
l
1.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
924
CHAPTER 14
PARTIAL DERIVAT IVES
t
25–26
39– 41
l
25. t
l
s
39.
26. t
40.
; 27–28
41.
27.
l
l
l
28.
; 42.
29–38
29.
30.
31.
32.
l
s
l
; 43.
33.
34.
35.
z
36.
z
z
s
44.
z
l
l
37.
45.
38.
46.
14.3
Partial Derivatives
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.3
PARTI AL DER IVATIVES
925
TABLE 1
T
H
°
t
t
t
t
t
t
l
l
t
t
t
t
t
t
t
t
l
l
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
926
CHAPTER 14
PARTIAL DERIVAT IVES
t
f
x
t
1
t
t
t
2
t
l
l
f
3
t
a, b
y
a, b
l
4
l
l
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.3
Notations for Partial Derivatives
927
PARTI AL DER IVATIVES
z
z
z
t
Rule for Finding Partial Derivatives of z
f x, y
1.
2.
EXAMPLE 1
SOLUTION
Interpretations of Partial Derivatives
z
T¡
S
P(a, b, c)
0
x
z
C¡
(a, b, 0)
FIGURE 1
g
f
T™
C™
y
t
t
(a, b)
C¡
C™
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
928
CHAPTER 14
PARTIAL DERIVAT IVES
z
z
z=4-≈-2¥
z
z
z
z
C¡
EXAMPLE 2
y=1
(1, 1, 1)
x
SOLUTION
y
(1, 1)
2
FIGURE 2
z
z
z
z=4-≈-2¥
z
C™
x=1
(1, 1, 1)
x
2
y
(1, 1)
FIGURE 3
4
4
3
3
z 2
z 2
1
1
0
0
y
1
2
1
0
x
0
0
y
1
2
1
0
x
FIGURE 4
4
4
3
3
z 2
z 2
1
1
0
FIGURE 5
0
y
1
2
1
0
x
0
0
y
1
2
1
0
x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.3
v
929
PARTI AL DER IVATIVES
EXAMPLE 3
SOLUTION
Some computer algebra systems can plot
surfaces defined by implicit equations in three
variables. Figure 6 shows such a plot of the
surface defined by the equation in Example 4.
v
z
EXAMPLE 4
z
z
z
z
z
SOLUTION
z
z
z
z
z
z
z
z
FIGURE 6
z
z
z
Functions of More Than Two Variables
z
z
z
l
z
w
w
z
z
z
w
z
l
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930
CHAPTER 14
PARTIAL DERIVAT IVES
EXAMPLE 5
SOLUTION
z
z
z
z
z
z
z
z
Higher Derivatives
z
z
z
z
z
EXAMPLE 6
SOLUTION
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.3
PARTI AL DER IVATIVES
931
20
z 0
_20
Figure 7 shows the graph of the function
in Example 6 and the graphs of its first- and
second-order partial derivatives for
,
. Notice that these graphs are consistent with our interpretations of and as
slopes of tangent lines to traces of the graph of .
For instance, the graph of decreases if we start
at
and move in the positive -direction.
This is reflected in the negative values of . You
should compare the graphs of
and with the
graph of to see the relationships.
_40
_2
_1
y
0
_2
_1
1 0 x
2 2
1
f
40
z
40
20
z 20
0
_20
_2
_1
y
0
1
_2
_1
1 0 x
2 2
0
_2
_1
y
0
fx
_1
y
0
1
_2
_1
1 0 x
2 2
_20
fxx
_40
_2
1
20
z 0
_20
_20
_2
_1
1 0 x
2 2
40
20
z 0
z 0
_2
fy
40
20
1
_2
_1
1 0 x
2 2
_1
y
0
1
fxy fyx
_2
_1
1 0 x
2 2
_40
_2
_1
y
0
fyy
FIGURE 7
Clairaut
Clairaut’s Theorem
Alexis Clairaut was a child prodigy in mathematics: he read l’Hospital’s textbook on
calculus when he was ten and presented a
paper on geometry to the French Academy of
Sciences when he was 13. At the age of 18,
Clairaut published Recherches sur les courbes à
double courbure, which was the first systematic
treatise on three-dimensional analytic geometry
and included the calculus of space curves.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
932
CHAPTER 14
PARTIAL DERIVAT IVES
v
EXAMPLE 7
z
z
z
z
SOLUTION
z
z
z
z
z
z
z
Partial Differential Equations
EXAMPLE 8
SOLUTION
u(x, t)
x
FIGURE 8
EXAMPLE 9
SOLUTION
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.3
5
PARTI AL DER IVATIVES
933
z
z
z
0.103
0.040
0.002
-0.019
-0.037
g
Courtesy Roger Watson
FIGURE 9
g
-0.051
-0.066
-0.109
Nano Teslas
per meter
zz
zz
0.000117
0.000037
0.000002
-0.000017
-0.000036
FIGURE 10
g
Courtesy Roger Watson
-0.000064
-0.000119
-0.000290
Nano Teslas
per m / m
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
934
CHAPTER 14
PARTIAL DERIVAT IVES
The Cobb-Douglas Production Function
6
7
8
9
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.3
14.3
935
PARTI AL DER IVATIVES
Exercises
1.
v
vl
v
4.
v
v
2.
v
t
3.
v
v
°
/
T
v
v
v
v
l
;
CAS
1.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
936
CHAPTER 14
PARTIAL DERIVAT IVES
5 –8
10.
y
z
_4
1
x
3
_2
0
6
8
10
12
14
16
4
2
2
y
1
5.
3
x
18
11.
6.
7.
s
12.
8.
; 13–14
9.
13.
8
15– 40
4
z 0
_4
_8
_3 _2 _1
14.
a
0
y
1
2
3
2
0
_2
x
15.
16.
17.
18.
19. z
20. z
21.
22.
23.
24. w
s
v
25. t
4
v
_4
1
y
29.
b
0
y
28.
2
3
2
0
x
_2
31.
z
4
39.
z 0
40.
_4
c
0
y
1
2
3
2
z
z
0
x
_2
z
32.
34. w
z
z
37.
8
y
30.
33. w
35.
_8
_3 _2 _1
26.
v
27.
z 0
_3 _2 _1
v
v
z
s
z
z
z
z
36.
z
38.
z
z
s
41– 44
41.
s
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.3
42.
43.
z
44.
z
z
71.
z
s
z
z
72.
45.
47–50
z
47.
48.
z
51–52
z
z
52.
z
z
t
s
z
z
z
z
t
s
z
z
z
y
z
z
74.
t
z
t
z
y
53–58
53.
55. w
sz
z
x
z
50.
z
51.
z
73.
46.
z
937
z
45– 46
49.
70.
PARTI AL DER IVATIVES
10 8
54.
s
57. z
4
2
P
56. v
v
6
x
58. v
75.
59–62
76.
59.
60.
61.
62.
s
63–70
63.
s
77.
z
zz
64.
65.
z
z
78.
z
66. t
t
67.
68. z
69. w
sv
t
v w
w
z
79.
z
w
w
t
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
938
CHAPTER 14
PARTIAL DERIVAT IVES
80.
88.
81.
z
z
z
z
z
z
89.
82.
90.
v
v
v
83.
v
91.
v
v
84.
v
92.
85.
93.
86.
; 94.
95.
z
z
96.
87.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.4
939
TANG ENT PL ANE S AND LINEAR APPROXIMA TIONS
99.
;
s
100.
101.
97.
;
98.
CAS
14.4
Tangent Planes and Linear Approximations
Tangent Planes
z
z
z
T¡
C¡
P
T™
C™
0
y
x
z
FIGURE 1
T
T¡
T.
T™
z
1
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
940
CHAPTER 14
PARTIAL DERIVAT IVES
z
z
z
z
2
Note the similarity between the equation of a
tangent plane and the equation of a tangent line:
z
z
v
z
z
z
EXAMPLE 1
SOLUTION
z
z
TEC Visual 14.4 shows an animation
of Figures 2 and 3.
40
40
20
20
20
0
z 0
z 0
_20
_20
40
z
_20
_4
_2
y
FIGURE 2
0
2
4 4
2
0
_2
_4
x
_2
y
z=2≈+¥
0
2
2
0
_2
x
0
y
1
2
2
1
0
x
(1, 1, 3)
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SECTION 14.4
TANGENT PL ANE S AND LINEA R APPR OXIMATIONS
941
FIGURE 3
(1, 1)
f(x, y)=2≈+¥
Linear Approximations
z
z
z
y
3
4
x
z
FIGURE 4
xy
≈+¥
f(0, 0)=0
f(x, y)=
(x, y)≠(0, 0),
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
942
CHAPTER 14
PARTIAL DERIVAT IVES
l
5
This is Equation 2.5.5.
l
z
z
z
6
z
7
Definition
z
z
z
l
8
Theorem 8 is proved in Appendix F.
Figure 5 shows the graphs of the function
and its linearization in Example 2.
v
Theorem
EXAMPLE 2
SOLUTION
6
z
l
4
2
0
1
x
0 1
0y
_1
FIGURE 5
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SECTION 14.4
TANGENT PL ANE S AND LINEA R APPR OXIMATIONS
943
EXAMPLE 3
T
H
°
SOLUTION
y
Differentials
y=ƒ
Îy
dx=Îx
0
a
dy
a+Îx
y=f(a)+fª(a)(x-a)
9
x
z
FIGURE 6
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
944
CHAPTER 14
PARTIAL DERIVAT IVES
z
z
z
10
z
z
z
z
z
z
z
z
z
z
z
z=f(x, y)
{ a+Îx, b+Îy, f (a+Îx, b+Îy)}
dz
Îz
{a, b, f(a, b)}
f(a, b)
0
y
f(a, b)
x
(a+Îx, b+Îy, 0)
(a, b, 0)
Îy=dy
z-f(a, b)=ffx (a, b)(x-a)+ff y (a, b)(y-b)
FIGURE 7
In Example 4, z is close to z because
the tangent plane is a good approximation
to the surface z
near
. (See Figure 8.)
v
EXAMPLE 4
z
z
z
SOLUTION
60
40
z 20
z
0
_20
z
5
4
FIGURE 8
3
x
2
1
0
z
z
0
4 2y
z
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SECTION 14.4
TANGENT PL ANE S AND LINEA R APPR OXIMATIONS
945
z
z
z
z
z
EXAMPLE 5
SOLUTION
Functions of Three or More Variables
z
z
z
z
w
w
z
w
z
z
z
w
z
w
w
w
w
z
z
EXAMPLE 6
z
SOLUTION
z
z
z
z
z
z
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946
CHAPTER 14
PARTIAL DERIVAT IVES
z
z
z
14.4
Exercises
1–6
15.
16.
1. z
2. z
3. z
17–18
s
18. s
17.
4. z
5. z
6. z
19.
; 7–8
; 20.
7. z
21.
8. z
CAS
z
9–10
s
z
s
22.
v
v
v
9.
s
10.
s
s
11–16
v
t
11.
12.
13.
14.
;
s
CAS
1.
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SECTION 14.4
TANGENT PL ANE S AND LINEA R APPR OXIMATIONS
947
23.
v
37.
24.
t
v
v
t
v
T
/
°
T
v
38.
25–30
25. z
26.
27.
28.
29.
30.
31.
39.
v
vw
z
z
z
z
32.
s
z
z
z
40.
z
33.
41.
w
34.
w
w
35.
42.
36.
v
v
v
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
948
CHAPTER 14
PARTIAL DE RIVAT IVES
43– 44
46.
43.
44.
45.
l
14.5
The Chain Rule
t
t
1
z
z
z
z
2
t
z
The Chain Rule (Case 1)
t
z
z
PROOF
z
z
z
l
l
l
z
l
t
t
l
t
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.5
l
z
THE C HAIN R ULE
l
949
l
z
l
l
l
l
l
l
l
z
Notice the similarity to the definition of the
differential:
z
z
z
z
z
z
z
EXAMPLE 1
z
SOLUTION
z
z
z
z
y
(0, 1)
z
C
z
z
x
z
z
v
FIGURE 1
x=
2t, y=
EXAMPLE 2
t
SOLUTION
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
950
CHAPTER 14
PARTIAL DERIVAT IVES
t
z
z
z
z
z
z
z
z
z
z
3
z
The Chain Rule (Case 2)
t
z
EXAMPLE 3
z
z
z
z
z
z
z
z
SOLUTION
z
z
x
x
s
s
x
x
t
t
FIGURE 2
z
z
z
z
z
z
z
z
y
y
s
s
y
y
t
t
z
z
z
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SECTION 14.5
THE C HAIN R ULE
951
z
z
z
z
z
4
v
w
z
The Chain Rule (General Version)
EXAMPLE 4
v
w
v z
v
z
z
v
SOLUTION
z
v
v
v
v
w
FIGURE 3
w
w
w
w
v
v
u
x
s
FIGURE 4
w
w
z
w
r
s
EXAMPLE 5
v
v
z
v
v
z
z
SOLUTION
y
t
z
z
w
r
w
z
t
r
s
z
t
z
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
952
CHAPTER 14
PARTIAL DERIVAT IVES
t
EXAMPLE 6
t
t
t
t
SOLUTION
t
t
t
t
z
EXAMPLE 7
z
z
SOLUTION
z
z
z
z
z
z
5
z
z
z
z
z
z
x
x
r
z
z
z
z
z
z
z
z
y
s r
s
z
z
FIGURE 5
z
z
z
z
z
z
z
z
z
z
Implicit Differentiation
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SECTION 14.5
THE C HAIN R ULE
953
6
EXAMPLE 8
SOLUTION
The solution to Example 8 should be
compared to the one in Example 2 in
Section 2.6.
z
z
z
z
z
z
z
z
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
954
CHAPTER 14
PARTIAL DERIVAT IVES
z
7
z
z
z
z
z
z
EXAMPLE 9
z
z
z
z
SOLUTION
z
z
z
z
z
z
z
z
z
z
13.
w
z
z
t
2. z
t
t
s
z
4. z
6. w
z
z
1. z
5. w
z
Exercises
1– 6
3. z
z
z
The solution to Example 9 should be
compared to the one in Example 4 in
Section 14.3.
14.5
z
14.
z
v
v
v
z
s
z
v
z
v
v
7–12
z
z
7. z
15.
8. z
t
v
t
v
v
tv
t
9. z
10. z
s
11. z
12. z
v
16.
v
t
t
t
1.
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SECTION 14.5
THE C HAIN R ULE
955
17–20
17.
18.
v w
z
z
v w
z
v w
v w
19. w
20.
v w
w
v
37.
v
w
21–26
21. z
z
z
z
v
22.
s
v
23. w
z
w
z
v
w
v
T
16
D
z
w
s
24.
s
v
20
14
5
8
15
10
w
12
10
10
25.
vw
v
v
v
w
w
w
v
20
30
40
t
10
20
30
40 t
38.
w
39.
26.
w
w
w
27–30
27.
28.
29.
30.
31–34
z
31.
33.
z
z
40.
z
32.
34.
z
z
z
z
z
41.
35.
s
42.
36.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
956
CHAPTER 14
PARTIAL DERIVAT IVES
50.
43.
51.
z
52.
z
z
z
53.
44.
z
z
z
v
z
v
54.
v
z
z
z
t
z
v
z
z
z
z
z
z
z
z
55.
t
n
45– 48
45.
z
z
z
z
z
z
z
46.
56.
47.
z
48.
z
z
z
57.
z
z
z
z
58.
z
t
z
z
z
z
z
49–54
z
z
49.
59.
t
z
z
z
v
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SECTION 14.6
DIRECTIONAL DER IVATIV ES AND THE GRAD IENT V ECTOR
957
Directional Derivatives and the Gradient Vector
14.6
Directional Derivatives
z
FIGURE 1
l
1
l
y
z
z
¨
(x¸, y¸)
¨
z
¨
0
FIGURE 2
x
=ka, bl=k
z
z
¨,
z
z
¨l
T
P(x¸, y¸, z¸)
Q(x, y, z)
TEC Visual 14.6A animates Figure 3 by
S
rotating and therefore .
C
Pª (x ¸, y¸, 0)
ha
u
h
hb
FIGURE 3
y
Qª (x, y, 0)
x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
958
CHAPTER 14
PARTIAL DERIVAT IVES
z
B
B
z
z
z
l
2
z
Definition
l
EXAMPLE 1
SOLUTION
s
FIGURE 4
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SECTION 14.6
3
DIRE CTIONAL DER IVATI VES AND THE GRAD IENT VECTO R
959
Theorem
t
PROOF
t
4
t
t
t
l
l
t
t
5
t
6
EXAMPLE 2
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
960
CHAPTER 14
PARTIAL DERIVAT IVES
The directional derivative
in
Example 2 represents the rate of change of z in
the direction of . This is the slope of the tangent line to the curve of intersection of the
surface z
and the vertical
plane through
in the direction of
shown in Figure 5.
SOLUTION
s
z
s
s
0
x
(1, 2, 0)
y
s
s
s
The Gradient Vector
π
6
FIGURE 5
7
8
Definition
EXAMPLE 3
9
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SECTION 14.6
The gradient vector
in Example 4 is
shown in Figure 6 with initial point
.
Also shown is the vector that gives the direction of the directional derivative. Both of these
vectors are superimposed on a contour plot of
the graph of .
y
v
DIRE CTIONAL DER IVATI VES AND THE GRAD IENT VECTO R
961
EXAMPLE 4
SOLUTION
s
±f(2, _1)
(2, _1)
x
s
s
s
FIGURE 6
s
s
s
Functions of Three Variables
z
z
10 Definition
z
11
z
z
l
l
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
962
CHAPTER 14
PARTIAL DERIVAT IVES
z
z
12
z
z
z
z
13
z
z
EXAMPLE 5
z
z
z
z
14
v
z
z
z
z
z
SOLUTION
z
z
z z
s
s
s
z
z
z
z
z
s
s
s
s
Maximizing the Directional Derivative
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.6
963
DIRE CTIONAL DER IVATI VES AND THE GRAD IENT VECTO R
15 Theorem
TEC Visual 14.6B provides visual
confirmation of Theorem 15.
PROOF
y
EXAMPLE 6
Q
2
1
±f( 2, 0)
0
3 x
P
1
SOLUTION
FIGURE 7
At
the function in Example 6 increases
fastest in the direction of the gradient vector
. Notice from Figure 7 that
this vector appears to be perpendicular to the
level curve through
. Figure 8 shows the
graph of and the gradient vector.
20
15
z 10
s
5
0
0
FIGURE 8
1
x
2
3 0
1
y
2
z
EXAMPLE 7
z
z
z
SOLUTION
z
z
z
z
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
964
CHAPTER 14
PARTIAL DERIVAT IVES
s
s
s
Tangent Planes to Level Surfaces
z
z
z
z
z
z
16
z
z
17
z
z
z
z
z
±F (x ¸, y¸, z¸)
g
P
z
18
ª(t¸)
z
0
x
S
C
z
y
z
z
z
FIGURE 9
19
z
z
z
z
z
z
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SECTION 14.6
965
DIRE CTIONAL DER IVATI VES AND THE GRAD IENT VECTO R
z
z
20
z
z
z
z
z
z
z
z
z
z
z
z
z
v
z
EXAMPLE 8
z
SOLUTION
z
z
Figure 10 shows the ellipsoid, tangent plane,
and normal line in Example 8.
z
z
z
z
z
z
z
z
z
y
x
z
FIGURE 10
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
966
CHAPTER 14
PARTIAL DERIVAT IVES
Significance of the Gradient Vector
z
z
z
y
±f(x¸, y¸)
P(x¸, y¸)
300
200
f(x, y)=k
0
x
100
FIGURE 11
FIGURE 12
y
_9
_6
_3
x
FIGURE 13
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.6
14.6
967
DIRE CTIONAL DER IVATI VES AND THE GRAD IENT VECTO R
Exercises
1.
7–10
s
7.
1008
1004
1000
996
992
988
984
980
976
972
1012
1012
1016
s
8.
1020
1024
9.
z
10.
z
z
z
z
11–17
11.
K
12.
13. t
1008
14. t
2.
15.
z
16.
z
z
z
s z
17.
Reprinted by permission of the Commonwealth of Australia.
18.
y
(2, 2)
±f (2, 2)
0
x
s
19.
20.
3.
v
s
z
z
z
21–26
s
21.
22.
4–6
23.
4.
24.
z
5.
25.
z
6.
26.
;
z
s
z
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
968
CHAPTER 14
PARTIAL DERIVAT IVES
27.
35.
28.
36.
29.
30.
z
z
A
31.
B
Reproduced with the permission of Natural Resources Canada 2009,
courtesy of the Centre of Topographic Information.
37.
v
32.
z
v
z
z
v
v
v
z
v
v
v
v
v
38.
y
33.
z
_5
z
_1
34.
z
(4, 6)
_3
0
1
3
5
z
x
39.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.6
40.
969
DIRE CTIONAL DER IVATI VES AND THE GRAD IENT VECTO R
56.
z
z
z
57.
z
58.
41– 46
41.
59.
z
42.
z
43.
z
z
60.
z
44.
z
z
45.
z
61.
z
z
46.
z
z
z
s
s
sz
s
z
62.
z
; 47– 48
63.
z
47.
z
48.
z
z
z
64.
z
49.
;
65.
t
50.
t
z
t
z
51.
z
z
z
z
z
z
zz
s
66.
52.
z
z
;
53.
z
z
z
67.
z
54.
68.
z
z
z
55.
z
l
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
970
CHAPTER 14
PARTIAL DERIVAT IVES
Maximum and Minimum Values
14.7
z
y
x
FIGURE 1
Notice that the conclusion of Theorem 2 can
be stated in the notation of gradient vectors
as
.
1
Definition
2
Theorem
PROOF
t
t
t
z
t
z
z
EXAMPLE 1
(1, 3, 4)
0
x
y
FIGURE 2
z=≈+¥-2x-6y+14
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SECTION 14.7
MAXIMUM A ND MINIMUM VAL UES
971
EXAMPLE 2
SOLUTION
z
x
y
z
FIGURE 3
z
z=¥-≈
Photo by Stan Wagon, Macalester College
3
Second Derivatives Test
NOTE 1
NOTE 2
NOTE 3
v
EXAMPLE 3
SOLUTION
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
972
CHAPTER 14
PARTIAL DERIVAT IVES
z
x
y
FIGURE 4
z=x$+y$-4xy+1
y
A contour map of the function in Example 3 is
shown in Figure 5. The level curves near
and
are oval in shape and indicate
that as we move away from
or
in any direction the values of are increasing.
The level curves near
, on the other hand,
resemble hyperbolas. They reveal that as we
move away from the origin (where the value of
is ), the values of decrease in some directions
but increase in other directions. Thus the contour
map suggests the presence of the minima and
saddle point that we found in Example 3.
0.5
0.9
1
1.1
1.5
2
_0.5
0
x
3
FIGURE 5
TEC In Module 14.7 you can use contour maps
to estimate the locations of critical points.
EXAMPLE 4
SOLUTION
4
5
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SECTION 14.7
MAXIMUM A ND MINIMUM VAL UES
973
6
7
t
_3
2.7
FIGURE 6
s
z
z
TEC Visual 14.7 shows several families
of surfaces. The surface in Figures 7 and 8
is a member of one of these families.
x
FIGURE 7
y
x
y
FIGURE 8
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
974
CHAPTER 14
PARTIAL DERIVAT IVES
y
2
The five critical points of the function in
Example 4 are shown in red in the contour
map of in Figure 9.
1
_1.48
_0.8
3
7
_3
3
x
_1
FIGURE 9
v
EXAMPLE 5
z
z
SOLUTION
s
s
z
z
z
z
s
s
Example 5 could also be solved using
vectors. Compare with the methods of
Section 12.5.
s
z
v
s
EXAMPLE 6
z
SOLUTION
z
z
y
FIGURE 10
x
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.7
z
MAXIMUM A ND MINIMUM VAL UES
975
z
z
z
Absolute Maximum and Minimum Values
FIGURE 11
8
Extreme Value Theorem for Functions of Two Variables
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976
CHAPTER 14
PARTIAL DERIVAT IVES
9
1.
2.
3.
EXAMPLE 7
SOLUTION
y
(0, 2)
L£
(2, 2)
L¢
(3, 2)
L™
L¡
(0, 0)
(3, 0)
x
FIGURE 12
9
0
D
L¡
30
L™
2
FIGURE 13
f(x, y)=≈-2xy+2y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.7
MAXIMUM A ND MINIMUM VAL UES
977
PROOF OF THEOREM 3, PART ( )
10
Exercises
14.7
1.
3.
y
t
2.
t
t
3.2
t
t
t
t
t
t
t
t
3.7
4
_1
0
3– 4
;
1
1
2
3.7
3.2
4.2
5
1
x
6
_1
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
978
CHAPTER 14
PARTIAL DERIVAT IVES
4.
22.
y
23.
1.5
_2.9
_2.7
_2.5
24.
1
1.5
1.7
1.9
_1
1
x
; 25–28
_1
25.
26.
27.
5–18
28.
29–36
5.
6.
29.
7.
8.
30.
9.
10.
31.
11.
12.
32.
13.
33.
14.
15.
34.
16.
17.
35.
18.
36.
19.
; 37.
20.
s
s
; 21–24
21.
; 38.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.7
MAXIMUM A ND MINIMUM VAL UES
979
53.
54.
39.
z
40.
41.
z
z
42.
z
55.
43.
44.
45.
46.
y
(x i, yi )
47.
di
(⁄, ›)
z
mx i+b
48.
0
49.
x
50.
51.
52.
56.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
980
CHAPTER 14
PARTIAL DERIVATIVES
AP PLIED PROJECT
DESIGNING A DUMPSTER
1.
2.
■
■
■
■
3.
4.
DI SCOVERY PROJECT
QUADRATIC APPROXIMATIONS AND CRITICAL POINTS
z
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.8
981
LAGRA NGE MULTIPLIERS
2.
;
3.
;
4.
5.
;
14.8
Lagrange Multipliers
z
z
z
g(x, y)=k
t
z
y
t
f(x, y)=11
f(x, y)=10
f(x, y)=9
f(x, y)=8
f(x, y)=7
0
t
t
t
x
t
FIGURE 1
TEC Visual 14.8 animates Figure 1 for both
level curves and level surfaces.
z
z
t
t
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
982
CHAPTER 14
PARTIAL DERIVAT IVES
z
z
t
z
z
z
z
z
z
z
z
z
z z
z
z
z
t
t
z
t
Method of Lagrange Multipliers
z
t
t
z
t
z
1
Lagrange multipliers are named after the
French-Italian mathematician Joseph-Louis
Lagrange (1736–1813). See page 210 for a
biographical sketch of Lagrange.
t
z
z
t
z
z
z
z
In deriving Lagrange’s method we assumed
that t
. In each of our examples you
can check that t
at all points where
t
z
. See Exercise 23 for what can
go wrong if t
.
t
z
t
z
z
z
t
t
t
z
tz
t
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.8
LAGRA NGE MULTIPLIERS
983
t
t
t
t
v
t
t
EXAMPLE 1
z
SOLUTION
z
t
t
t
z
z
z
z
z
t
t
tz
z
z
z
2
z
z
3
z
z
4
z
5
z
z
Another method for solving the system of equations (2–5) is to solve each of Equations 2, 3,
and 4 for and then to equate the resulting
expressions.
6
z
z
7
z
z
8
z
z
z
z
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
984
CHAPTER 14
PARTIAL DERIVAT IVES
z
z
z
z
z
z
z
z
z
z
v
z
z
z
z
In geometric terms, Example 2 asks for
the highest and lowest points on the curve
in Figure 2 that lie on the paraboloid
z
and directly above the constraint circle
.
z
z
EXAMPLE 2
SOLUTION
t
z=≈+2¥
t
t
t
t
t
9
C
10
11
x
≈+¥=1
y
FIGURE 2
The geometry behind the use of Lagrange
multipliers in Example 2 is shown in Figure 3.
The extreme values of
correspond to the level curves that touch the
circle
.
y
EXAMPLE 3
≈+2¥=2
SOLUTION
0
x
≈+2¥=1
FIGURE 3
z
EXAMPLE 4
z
SOLUTION
s
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.8
985
LAGRA NGE MULTIPLIERS
z
z
z
t
z
z
t t
12
13
z
14
z
z
15
z
Figure 4 shows the sphere and the nearest point
in Example 4. Can you see how to find the
coordinates of without using calculus?
z
z
s
x
P
s
y
(3, 1, _1)
z
FIGURE 4
s
s
h=c
±f
±g
C
P
s
s
s
s
s
s
s
s
Two Constraints
±h
t
g=k
FIGURE 5
s
s
z
z
z
z
t
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
986
CHAPTER 14
PARTIAL DERIVAT IVES
t
t
t
z
z
z
z
t
z
16
t
z
z
z
z
t
t
tz
z
t
z
z
z
The cylinder
intersects the
plane
z
in an ellipse (Figure 6).
Example 5 asks for the maximum value of
when
z is restricted to lie on the ellipse.
v
z
z
z
SOLUTION
t
z
4
3
17
2
18
z 1
19
t
z
z
z
z
20
0
21
_1
_2
z
EXAMPLE 5
_1
0
y
1
FIGURE 6
s
z
s
s
s
s
s
s
s
s
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 14.8
987
Exercises
14.8
1.
LAGRA NGE MULTIPLIERS
16.
t
y
g(x, y)=8
60
z
z
t
70
z
17.
z
18.
z
z
z
z
z
z
19–21
40
50
19.
20.
0
x
30
21.
20
10
22.
; 2.
s
s
;
23.
3–14
t
3.
4.
5.
CAS
6.
7.
z
z
8.
z
z
9.
z
10.
z
11.
z
z
z
12.
z
z
z
13.
z
z
z
24.
z
z
z
z
z
25.
z
14.
15–18
15.
;
z
z
z
CAS
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
988
CHAPTER 14
PARTIAL DERIVAT IVES
26.
CAS
45 – 46
27.
28.
s
45.
z
46.
z
z
z
z
z
z
z
z
47.
z
s
29– 41
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
s
41.
48.
42.
s
43.
z
z
44.
;
s
s
s
z
z
APPLIED PROJECT
ROCKET SCIENCE
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APPLIED PROJECT
ROCK ET SC IENCE
989
1.
Courtesy of Orbital Sciences Corporation
v
2.
v
v
3.
v
4.
v
5.
6.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
990
CHAPTER 14
PARTIAL DERIVATIVES
AP PLIED PROJECT
HYDRO-TURBINE OPTIMIZATION
1.
2.
3.
4.
5.
6.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 14
991
R E VIEW
Review
14
Concept Check
1.
11.
12.
2.
z
z
z
z
z
13.
3.
l
4.
14.
5.
15.
6.
16.
7.
z
17.
z
18.
8.
19.
9.
10.
z
z
z
t
z
z
True-False Quiz
Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement.
5.
l
l
l
6.
1.
l
7.
2.
8.
3.
4.
l
z
z
z
9.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
992
CHAPTER 14
PARTIAL DERIVAT IVES
10.
s
11.
s
12.
Exercises
1–2
1.
s
2.
s
s
3– 4
3.
4.
x
y
5–6
s
5.
6.
7.
z
x
2
2
y
12.
13–17
8.
y
1
2
13.
14. t
15.
16.
17.
1.5
v w
v
v
v
z
z
z
v sw
18.
4
x
9–10
9.
l
10.
l
11.
;
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 14
19–22
993
R E VIEW
40.
19.
20. z
21.
z
23.
z
24.
z
22. v
z
z
z
41.
z
v
z
v
z
z
z
z
z
z
v
z
42.
z
v
v
z
z
z
v
z
25–29
43.
25. z
z
z
44.
26. z
27.
z
28.
z
29.
z
z
45– 46
z
; 30.
45.
z
46.
31.
z
s
z
z
z
s
47.
32.
48.
33.
s
z
z
z
z
s
34.
49.
35.
z
z
36.
v
v
37.
z
t
z
v
t
t
t
z
38.
w
v
39.
v
v
z
z
z
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994
CHAPTER 14
PARTIAL DERIVAT IVES
50.
60.
z
z
61.
z
62.
z
z
z
51–54
z
z
51.
z
63.
z
52.
53.
64.
54.
55–56
65.
55.
56.
¨
; 57.
; 58.
59–62
66.
z
z
59.
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Problems Plus
1.
2.
3.
w
x
¨
¨
x
w-2x
4.
z
z
z
z
z
5.
z
6.
t
z
t
t
z
t
t
t
t
t
t
t
t
995
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y
x x+y y=1000
4
x y+y x=100
2
0
2
4
x
7.
8.
z
996
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15
Multiple Integrals
FPO
New Art to
come
Geologists study how mountain ranges were formed
and estimate the work required to lift them from sea
level. In Section 15.8 you are asked to use a triple
integral to compute the work done in the formation
of Mount Fuji in Japan.
© S.R. Lee Photo Traveller / Shutterstock
997
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998
CHAPTER 15
MUL TIPLE INTEGR ALS
Double Integrals over Rectangles
15.1
Review of the Definite Integral
1
l
y
2
l
x
y
Îx
f(x i )
0
FIGURE 1
z
a
x
b
0
a
x¡
⁄
¤
x™
‹
xi-1
x£
xi
xi
xn-1
b
x
xn
Volumes and Double Integrals
z=f(x, y)
c
R
d
z
y
z
z
FIGURE 2
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SECTION 15.1
D OUBLE INTEG RALS OVER R ECTANGLES
999
y
R ij
d
(xi, yj)
(x ij , y ij )
yj
yj-1
Îy
›
c
(x£™, y£™)
0
FIGURE 3
a
⁄
¤
x i-1 x i
b
x
Îx
3
z
z
a
x
f(x *ij , y*ij )
0
c
b
0
d
y
y
x
R ij
FIGURE 4
FIGURE 5
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1000
CHAPTER 15
M ULTIP LE INT EGRAL S
The meaning of the double limit in Equation 4 is
that we can make the double sum as close as we
like to the number [for any choice of
in ] by taking and sufficiently large.
4
5
l
Definition
Notice the similarity between Definition 5
and the definition of a single integral in
Equation 2.
yy
Although we have defined the double integral by
dividing into equal-sized subrectangles, we
could have used subrectangles
of unequal
size. But then we would have to ensure that all
of their dimensions approach in the limiting
process.
l
yy
yy
6
l
z
yy
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SECTION 15.1
y
(1, 2)
2
R¡™
1
v
(2, 2)
DOUB L E INT EGRALS OVE R REC TANGL ES
1001
EXAMPLE 1
z
R™™
(2, 1)
(1, 1)
R¡¡
0
SOLUTION
R™¡
1
x
2
FIGURE 6
z
16
z=16-≈-2¥
2
2
y
x
FIGURE 7
FIGURE 8
z=16-≈-2¥
m=n=4, VÅ41.5
v
m=n=8, VÅ44.875
m=n=16, VÅ46.46875
EXAMPLE 2
yy s
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1002
CHAPTER 15
z
M ULTIP LE INT EGRAL S
SOLUTION
(0, 0, 1)
z
S
x
(1, 0, 0)
(0, 2, 0)
s
s
z
z
z
y
yy s
FIGURE 9
The Midpoint Rule
Midpoint Rule for Double Integrals
yy
v
y
2
3
2
1
EXAMPLE 3
xx
SOLUTION
R¡™
R™™
R¡¡
R™¡
0
1
2
(2, 2)
yy
x
FIGURE 10
yy
NOTE
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SECTION 15.1
DOUB L E INT EGRALS OVE R REC TANGL ES
1003
Average Value
y
yy
yy
z
FIGURE 11
EXAMPLE 4
12
40 36
44
12
16
32
28
16
24
40
20
36
32
12
28
24
0
4
8 12 16
32
28
20
24
8
FIGURE 12
SOLUTION
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1004
CHAPTER 15
M ULTIP LE INT EGRAL S
yy
y
276
12
40 36
44
20
12
16
32
28
16
24
40
36
32
12
28
24
0
FIGURE 13
0
4
16 20
8 12
32
28
24
8
388 x
yy
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SECTION 15.1
DOUB L E INT EGRALS OVE R REC TANGL ES
1005
Properties of Double Integrals
7
yy
8
yy
Double integrals behave this way because the
double sums that define them behave this way.
yy
t
yy t
yy
t
9
yy
yy t
Exercises
15.1
1.
z
y
x
2.
xx
6.
3.
xx
4.
z
5.
xx
7.
s
1.
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1006
CHAPTER 15
M ULTIP LE INT EGRAL S
28
24
20
16
32 4444
24
32
8.
40
3236
16
xx
44
48
28
56
52
y
2
32
36
40
44
20
24
28
48
5256
1
11–13
0
1
2
x
11.
12.
13.
9.
14.
xx
xx
xx
xx
xx
s
15.
yy
y
4
2
s
xx
16.
s
17.
0
yy
4 x
18.
10.
15.2
2
yy
Iterated Integrals
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SECTION 15.2
ITE RATED INTEG RALS
1007
x
x
y
y
1
y y
2
y y
y y
3
y y
y y
EXAMPLE 1
y y
yy
SOLUTION
y
y y
y y
y
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1008
CHAPTER 15
M ULTIP LE INT EGRAL S
y y
y y
y
4
Theorem 4 is named after the Italian mathematician Guido Fubini (1879–1943), who proved a
very general version of this theorem in 1907. But
the version for continuous functions was known
to the French mathematician Augustin-Louis
Cauchy almost a century earlier.
Fubini’s Theorem
yy
y y
y y
z
C
x
x
a
0
z
xx
A(x)
y
b
y
FIGURE 1
TEC Visual 15.2 illustrates Fubini’s
Theorem by showing an animation of
Figures 1 and 2.
z
y
z
0
x
FIGURE 2
yy
c
y
d
y
yy
y
yy
y y
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SECTION 15.2
v
Notice the negative answer in Example 2;
nothing is wrong with that. The function is
not a positive function, so its integral doesn’t
represent a volume. From Figure 3 we see that
is always negative on , so the value of the
integral is the negative of the volume that lies
above the graph of and below .
R
0
z
_12
SOLUTION 1
yy
0.5
1
y
y y
SOLUTION 2
yy
1.5
2 2
y
y
z=x-3¥
0
1009
xx
EXAMPLE 2
_4
_8
ITE RATED INTEG RALS
1
x
yy
0
FIGURE 3
y
v
xx
EXAMPLE 3
SOLUTION 1
yy
y y
y
y
SOLUTION 2
yy
For a function that takes on both positive and
negative values, xx
is a difference
of volumes:
, where is the volume
above and below the graph of , and is the
volume below and above the graph. The fact
that the integral in Example 3 is means that
these two volumes and are equal. (See
Figure 4.)
1
z 0
_1
v
v
y
z=y
0
1
(xy)
y
2
3 2
yy
y
1
x
FIGURE 4
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1010
CHAPTER 15
M ULTIP LE INT EGRAL S
v
v
y
In Example 2, Solutions 1 and 2 are equally
straightforward, but in Example 3 the first solution is much easier than the second one. Therefore, when we evaluate double integrals, it’s
wise to choose the order of integration that
gives simpler integrals.
yy
v
EXAMPLE 4
z
z
SOLUTION
16
12
z 8
yy
4
0
0
1
y
2 2
1
x
0
y y
y
y
FIGURE 5
t
yy
y y
x
5
y y
y y
t
y
t
y
t
t
y
t
y
t
yy t
y
t
y
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SECTION 15.2
ITE RATED INTEG RALS
1011
EXAMPLE 5
yy
y
y
z
The function
in
Example 5 is positive on , so the integral represents the volume of the solid that lies above
and below the graph of shown in Figure 6.
0
FIGURE 6
15.2
Exercises
x
1–2
y
x
x
1.
18.
yy
19.
yy
2.
3–14
3.
yy
4.
yy
20.
yy
5.
yy
6.
y y
21.
yy
7.
y y
8.
y y
22.
9.
yy
10.
y y
yy
11.
yy
12.
y y
13.
yy
14.
y y
v
v
v
23–24
s
s
23.
yy
24.
yy
15–22
15.
16.
17.
yy
25.
z
yy
26.
z
yy
;
CAS
1.
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1012
CHAPTER 15
MULTIPLE INTEGRALS
27.
35–36
z
35.
s
36.
28.
z
z
37–38
29.
z
z
30.
z
37.
yy
38.
yy
31.
z
z
CAS
; 32.
CAS
CAS
z
33.
39.
yy
z
xx
40.
34.
z
yy
y y
t
z
t
15.3
t
Double Integrals over General Regions
1
y
y
R
D
0
FIGURE 1
D
x
0
x
FIGURE 2
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.3
1013
DOUBL E INTEGRALS OVER GENERAL REGIONS
z
f
g
yy
2
0
y
D
x
yy
xx
FIGURE 3
xx
z
z
xx
F
g
0
y
D
x
xx
xx
FIGURE 4
t
t
y
t
y
y=g™(x)
y
y=g™(x)
y=g™(x)
D
D
D
y=g¡(x)
0
t
a
y=g¡(x)
y=g¡(x)
b
x
0
a
x
b
0
a
b
x
FIGURE 5
xx
y
y=g™(x)
d
yy
D
FIGURE 6
yy
t
c
0
yy
y=g¡(x)
a
x
b
x
y
t
y
y
t
t
t
t
t
t
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1014
CHAPTER 15
M ULTIP LE INT EGRAL S
3
t
yy
y
y y
t
t
t
d
x=h¡(y)
D
x=h™( y)
c
t
0
t
x
y
4
d
x=h¡( y)
D
x=h™(y)
0
c
x
v
y
y=1+≈
(_1, 2)
D
_1
FIGURE 8
yy
5
FIGURE 7
(1, 2)
EXAMPLE 1
yy
xx
SOLUTION
y=2≈
1
x
yy
y y
y
y
y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.3
DOUBL E INTEGRALS OVER GENERAL REGIONS
1015
NOTE
t
t
y
(2, 4)
z
EXAMPLE 2
y=2x
SOLUTION 1
y=≈
D
0
1
z
x
2
yy
FIGURE 9
y y
y
4
(2, 4)
x= y
x=œy
D
x
0
FIGURE 10
SOLUTION 2
Figure 11 shows the solid whose volume
is calculated in Example 2. It lies above the
-plane, below the paraboloid z
,
and between the plane
and the
parabolic cylinder
.
s
z
y=≈
yy
z=≈+¥
yy
s
s
x
y=2x
y
FIGURE 11
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1016
CHAPTER 15
M ULTIP LE INT EGRAL S
v
xx
EXAMPLE 3
SOLUTION
y
y
(5, 4)
y=œ„„„„
2x+6
y=x-1
0
_3
x=y+1
x
0
(_1, _2)
y=_œ„„„„
2x+6
(5, 4)
¥
x= 2 -3
x
_2
(_1, _2)
FIGURE 12
yy
y y
y
z
(0, 0, 2)
x+2y+z=2
x=2y
T
y
(0, 1, 0)
0
yy
1
y y
y y
s
s
s
”1, 2 , 0’
x
y
1
z
EXAMPLE 4
FIGURE 13
z
SOLUTION
x+2y=2
y=1-x/2)
z
z
D
”1,
z
z
1
’
2
y=x/ 2
0
1
x
z
z
z
FIGURE 14
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.3
DOUBL E INTEGRALS OVER GENERAL REGIONS
1017
yy
yy
y
y
y
v
y=1
D
x x
EXAMPLE 5
SOLUTION
x
x
y=x
0
1
yy
x
yy
FIGURE 15
y
1
x=0
0
D
x=y
y y
x
yy
y y
FIGURE 16
y
y
Properties of Double Integrals
6
7
yy
t
yy
yy
yy t
yy
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1018
CHAPTER 15
M ULTIP LE INT EGRAL S
t
yy
8
y
x
D
D¡
yy t
x
x
D™
0
x
yy
9
yy
yy
FIGURE 17
y
y
D™
D
0
D¡
x
0
D=D¡
FIGURE 18
10
x
D™, D¡
D™
yy
z
xx
z=1
0
x
y
D
11
yy
FIGURE 19
D
g
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.3
DOUBL E INTEGRALS OVER GENERAL REGIONS
1019
xx
EXAMPLE 6
SOLUTION
yy
Exercises
15.3
1–6
14.
1.
yy
2.
yy
3.
yy
4.
yy
5.
yy
6.
yy
s
15–16
v
s
v
7–10
7.
yy
8.
yy
9.
yy
10.
yy
yy
w v
15.
yy
16.
yy
17–22
17.
yy
18.
yy
19.
yy
20.
yy
21.
yy
22.
yy
11.
12.
s
13–14
13.
yy
;
CAS
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1020
CHAPTER 15
M ULTIP LE INT EGRAL S
23–32
40.
23.
z
24.
z
25.
z
41.
z
42.
z
z
z
43– 48
26.
z
z
43.
y y
44.
yy
45.
y y
46.
y y
47.
y y
48.
yy
27.
z
28.
z
z
z
29.
z
z
s
z
30.
z
49–54
z
31.
z
49.
y y
50.
y y
51.
y y
52.
yy
53.
y y
54.
y y
56.
yy
z
32.
z
; 33.
xx
; 34.
s
s
s
s
s
55– 56
z
z
55.
yy
35–36
y
35.
1
z
D
z
_1
36.
z
z
0
_1
y
1
(1, 1)
1
x=y-Á
y=(x+1)@
x
_1
0
x
_1
37–38
37.
CAS
yy
38.
y y
57–58
57.
yy
58.
yy
39– 42
39.
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.4
59–60
64.
yy s
65.
yy
66.
yy
67.
yy
59.
60.
61.
62.
yy
yy
yy
63–67
CAS
1021
D OUBL E INTEGRALS IN POLAR COORDINATES
s
68.
z
z
63.
yy
15.4
s
Double Integrals in Polar Coordinates
xx
y
y
≈+¥=4
≈+¥=1
R
0
x
R
0
FIGURE 1
R=s(r, ¨) | 0¯r¯1, 0¯¨¯2πd
≈+¥=1
x
R=s(r, ¨ ) | 1¯r¯2, 0¯¨¯πd
y
P (r, ¨ ) =P (x, y)
r
y
¨
O
x
x
FIGURE 2
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1022
CHAPTER 15
M ULTIP LE INT EGRAL S
xx
¨=¨ j
¨=¨ j-1
r=b
¨=∫
R ij
(ri , ¨j )
R
Ψ
r=a
∫
O
r=ri
¨=å
r=ri-1
å
O
FIGURE 3
FIGURE 4
xx
1
t
t
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SECTION 15.4
DOUBL E INTEGRALS IN POLAR COORDINATES
y y
yy
1023
t
l
l
y y
t
t
y y
2
Change to Polar Coordinates in a Double Integral
yy
y y
|
dA
d¨
r
dr
EXAMPLE 1
r d¨
xx
SOLUTION
O
FIGURE 5
yy
y y
y y
y
Here we use the trigonometric identity
y
y
See Section 7.2 for advice on integrating
trigonometric functions.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1024
CHAPTER 15
M ULTIP LE INT EGRAL S
v
z
EXAMPLE 2
z
z
SOLUTION
(0, 0, 1)
z
0
D
x
yy
y
y
FIGURE 6
y y
y
yy
y y
s
s
x
r=h™(¨)
¨=∫
D
∫
O
å
3
¨=å
r=h¡(¨)
yy
y y
FIGURE 7
D=s(r, ¨) | 寨¯∫, h¡(¨)¯r¯h™(¨)d
yy
y y
y
π
¨= 4
v
π
¨=_ 4
EXAMPLE 3
SOLUTION
FIGURE 8
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SECTION 15.4
yy
y
DOUBL E INTEGRALS IN POLAR COORDINATES
1025
y
y
y
y
v
z
EXAMPLE 4
SOLUTION
z
y
(x-1)@+¥=1
( r=2
¨)
D
0
1
2
x
x
y
FIGURE 10
FIGURE 9
yy
y
y
y
y
y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1026
CHAPTER 15
15.4
M ULTIP LE INT EGRAL S
Exercises
1– 4
14.
xx
y
4
1.
y
2.
1
xx
15–18
y=1-≈
15.
16.
0
4
x
_1
0
x
1
17.
18.
y
3.
4.
1
y
6
19–27
3
0
_1
1
0
x
x
19.
z
20.
s
z
21.
z
z
22.
5– 6
5.
y
y
6.
23.
24.
xx
26.
9.
xx
28.
10.
xx
11.
xx
xx
13.
xx
s
z
z
27.
12.
z
z
xx
8.
z
z
25.
7–14
7.
y y
z
z
29–32
s
s
29.
y y
31.
yy
s
s
30.
y y
32.
y y
s
s
s
1.
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SECTION 15.5
33–34
33.
34.
APPL IC ATIONS OF D OUBLE INTEGRALS
1027
40.
yy
xx
l
xx
y y
yy
s
y y
35.
yy
36.
l
y
y
y
s
37.
yy
s
s
y
38.
s
39.
y y
s
15.5
s
y y
s
y y
s
s
41.
y
y
s
Applications of Double Integrals
Density and Mass
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1028
CHAPTER 15
M ULTIP LE INT EGRAL S
y
(x, y)
D
0
x
FIGURE 1
y
(xij , yij )
R ij
1
0
yy
l
x
FIGURE 2
yy
2
y
EXAMPLE 1
y=1
1
D
(1, 1)
SOLUTION
yy
y=1-x
0
FIGURE 3
yy
x
y
y
Moments and Centers of Mass
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SECTION 15.5
l
yy
l
yy
3
4
(x, y)
APPL IC ATIONS OF DOUBLE INTEGRALS
1029
D
5
FIGURE 4
yy
yy
yy
v
EXAMPLE 2
SOLUTION
y
(0, 2)
yy
y=2-2x
y y
3 11
” 8 , 16 ’
D
0
(1, 0)
x
y
FIGURE 5
yy
y y
y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1030
CHAPTER 15
M ULTIP LE INT EGRAL S
yy
y y
y
v
y
a
D
_a
SOLUTION
≈+¥=a@
s
3a
”0, 2π ’
0
EXAMPLE 3
s
a
x
s
FIGURE 6
yy
yy
y y
y
yy
Compare the location of the center of mass in
Example 3 with Example 4 in Section 8.3,
where we found that the center of mass of
a lamina with the same shape but uniform
density is located at the point
.
s
y
yy
y
y
Moment of Inertia
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SECTION 15.5
6
7
8
v
APPL IC ATIONS OF DOUBLE INTEGRALS
l
yy
l
yy
1031
yy
l
EXAMPLE 4
SOLUTION
yy
y
y y
y
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1032
CHAPTER 15
M ULTIP LE INT EGRAL S
9
10
v
EXAMPLE 5
SOLUTION
Probability
x
y
yy
y y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.5
APPL IC ATIONS OF DOUBLE INTEGRALS
1033
z
z=f(x, y)
c
a
FIGURE 7
g
a
u
g D=[a, b]x[c, d]
u
x
d
b
y
D
yy
yy
y y
EXAMPLE 6
SOLUTION
y y
y y
y
y
y y
y
y y
y
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1034
CHAPTER 15
M ULTIP LE INT EGRAL S
EXAMPLE 7
SOLUTION
y
20
x+y=20
FIGURE 8
y y
y
D
0
yy
20 x
y
y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.5
APPL IC ATIONS OF DOUBLE INTEGRALS
1035
Expected Values
y
yy
11
yy
s
EXAMPLE 8
SOLUTION
s
s
1500
1000
500
0
5.95
3.95
y
4
6
6.05
4.05
x
FIGURE 9
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1036
CHAPTER 15
M ULTIP LE INT EGRAL S
y y
y y
15.5
Exercises
1.
13.
2.
s
s
s
14.
15.
3–10
3.
4.
16.
5.
6.
17.
7.
18.
8.
19.
9.
10.
20.
s
11.
21–24
12.
21.
22.
CAS
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.6
SU RFACE AREA
1037
23.
24.
CAS
CAS
25–26
31.
25.
26.
32.
27.
28.
33.
29.
30.
15.6
Surface Area
In Section 16.6 we will deal with areas of more
general surfaces, called parametric surfaces, and
so this section need not be covered if that later
section will be covered.
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1038
CHAPTER 15
M ULTIP LE INT EGRAL S
z
ÎTij
Pij
ÎS ij
S
Îy
0
R ij
D
x
y
(x i , yj )
Îx
1
ÎA
l
FIGURE 1
z
Pij
ÎTij
0
Îx
Îy
y
x
FIGURE 2
s
l
s
l
z
2
yy s
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.6
z
3
y
SU RFACE AREA
1039
z
z
EXAMPLE 1
(1, 1)
SOLUTION
y=x
T
(0, 0)
x
(1, 0)
yy s
FIGURE 3
z
y
y y
s
s
z
EXAMPLE 2
y
T
x
s
z
z
SOLUTION
FIGURE 4
z
z
9
z
yy s
yy s
D
x
3
FIGURE 5
y
y y
s
y
y
s
s
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1040
CHAPTER 15
15.6
M ULTIP LE INT EGRAL S
Exercises
1–12
16.
1.
z
z
CAS
2.
z
3.
z
4.
z
5.
CAS
17.
CAS
18.
z
z
6.
z
z
CAS
19.
z
7.
z
CAS
20.
z
8.
z
9.
z
21.
10.
z
s
z
11.
z
22.
z
z
12.
z
z
z
l
13–14
23.
13.
z
14.
z
z
z
24.
z
z
z
15.
z
CAS
CAS
x
y
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.7
TRIPLE I NTEGR AL S
1041
Triple Integrals
15.7
z
1
z
z
z
B
z
x
z
y
Bijk
z
z
2
Îz
z
Îy
Îx
z
3
Definition
yyy
x
z
z
l
y
FIGURE 1
z
yyy
4
z
z
l
Fubini’s Theorem for Triple Integrals
yyy
z
yy y
z
z
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1042
CHAPTER 15
M ULTIP LE INT EGRAL S
z
yyy
v
yyy
z
xxx
EXAMPLE 1
z
z
z
z
z
SOLUTION
z
yyy
yy y
z
z
yyy
z
z
z
z
yy
y
z
z
z
z
z
z
z
z
yyy
z
z=u™(x, y)
E
z=u¡(x, y)
0
x
D
y
z
5
z
FIGURE 2
z
g
z
6
yyy
z
yy y
z
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.7
z
TRIPLE I NTEGR AL S
1043
z=u™(x, y)
t
z
E
t
z
z=u¡(x, y)
x
a
b
0
y=g¡(x)
D
y=g™(x)
y
yyy
7
y y y
t
z
z
t
z
FIGURE 3
g
D
g
z
z=u™(x, y)
E
z
z=u¡(x, y)
x=h¡(y)
0
x
z
yyy
8
y y
z
y
z
z
D
x=h™(y)
EXAMPLE 2
FIGURE 4
xxx
z
z
g
z
SOLUTION
z
z
z
(0, 0, 1)
z
z
z
z=1-x-y
E
(0, 1, 0)
0
(1, 0, 0)
x
y
z=0
z
9
z
FIGURE 5
yyy z
y
1
y=1-x
y=0
z z
z
z
y y
D
0
yy y
z
1
x
y
FIGURE 6
z
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1044
CHAPTER 15
M ULTIP LE INT EGRAL S
z
z
z
z
D
0
yyy
10
y
E
x
yy y
z
z
z
z
x=u¡(y, z)
x=u™(y, z)
z
FIGURE 7
z
z
g
z
z
z
z
z
y=u™(x, z)
D
yyy
11
yy y
z
z
z
z
E
0
y=u¡(x, z)
x
y
v
FIGURE 8
g
xxx
EXAMPLE 3
z
s
z
SOLUTION
z
z
z
y
y=≈+z@
TEC Visual 15.7 illustrates how solid regions
(including the one in Figure 9) project onto
coordinate planes.
y=4
D¡
E
0
4
x
y
0
FIGURE 9
g
z
x
FIGURE 10
g
s
y=≈
P
z
z
s
z
s
s
z
yyy s
y
z
y y y
s
s
s
z
z
s
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.7
TRIPLE I NTEGR AL S
1045
z
z
≈+z@=4
z
D£
z
0
_2
z
x
2
yyy s
yy y
z
z
z
s
z
yy
z
z s
z
FIGURE 11
xz
y y
| The most difficult step in evaluating a triple
integral is setting up an expression for the region
of integration (such as Equation 9 in Example 2).
Remember that the limits of integration in the
inner integral contain at most two variables, the
limits of integration in the middle integral contain at most one variable, and the limits of integration in the outer integral must be constants.
s
z s
s
z
z
z
yyy s
yy
z
z
z s
z
y y
y
y
y
1
y=≈
D¡
0
1
z
1
x xx
EXAMPLE 4
x
z
z
z
SOLUTION
z=y
y yy
D™
0
1
z
yyy
z
z
y
z
z
z
1
z=≈
s
D£
0
1
FIGURE 12
x
E
z
z
z
z
z
s
z=y
x
0
1
1
y=≈
x=1
FIGURE 13
E
y
z
z
z
yyy
s
z
z
z
y yy
s
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1046
CHAPTER 15
M ULTIP LE INT EGRAL S
Applications of Triple Integrals
xxx
x
xx
z
z
z
xxx
z
z
z
z
yyy
12
z
yyy
z
yy y
yy
z
z
EXAMPLE 5
z
z
SOLUTION
z
z
z
z
(0, 0, 2)
y
x+2y+z=2
x=2y
T
(0, 1, 0)
0
1
x+2y=2
y=1- x/2)
y
”1, 21 ’
D
y=x/2
1
”1, 2 , 0’
0
x
1
x
FIGURE 15
FIGURE 14
yyy
y y
y
z
y y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.7
TRIPLE I NTEGR AL S
z
z
yyy
13
14
z
1047
yyy
z
z
yyy
z
yyy z
z
z
z
z
15
16
yyy
z
z
z
yyy
z
z
yyy
z
z
z
z
yyy
z
yyy
z
y y y
z
y y y
z
z
z
z
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1048
CHAPTER 15
M ULTIP LE INT EGRAL S
z
v
z z
z=x
E
EXAMPLE 6
SOLUTION
z
0
y
1
x
z
z
y
z
z
yyy
x=¥
D
0
x=1
x
y y y
z
y y
y
FIGURE 16
y
z
z
z
yyy
y y y
z
y y
y
yyy z
y y y
z
z
z
z
y y
z
y
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.7
1049
TRIPLE I NTEGR AL S
Exercises
15.7
1.
19–22
z
19.
xxx
2.
z
z
z
20.
z
z
z
21.
z
3–8
3.
yy y
5.
yy y
z
7.
y yy
z
8.
y yy
z
z
z
z
s
z
4.
y y y
6.
y yy
z
22.
z
z z
s
z
z
z
z
23.
z
z
CAS
z
24.
9–18
9.
z
xxx
z
10.
xxx
z
z
25–26
z
12.
xxx
z
13.
xxx
z
25.
s
16.
xxx
17.
xxx
18.
26.
xxx
xxx
xxx
CAS
z
z
xxx
15.
z
CAS
z
z
14.
s
z
z
11.
xxx
z
z
z
xxx
xxx
z
s
z
z
z
z
z
z
27–28
z
27.
yy y
z
xxx
29–32
z
z
28.
z
yy y
z
z
z
z
29.
z
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1050
CHAPTER 15
30.
z
M ULTIP LE INT EGRAL S
38.
31.
z
xxx
z
z
z
32.
z
z
39– 42
33.
39.
yy y
z
s
z
40.
z
z
z
z
z
41.
z
z
z
z
42.
z
1
z=1-y
z
z
43– 46
y=œx
43.
0
1
y
44.
x
34.
yy
y
z
45.
z
46.
z
z=1-≈
z
z
z
s
z
47– 48
1
z
47.
0
x
1
y=1-x
1
48.
y
z
z
CAS
s
z
s
z
z
49.
35–36
z
z
z
35.
y yy
36.
y yy
z
z
z
z
z
z
CAS
xxx
50.
z
37–38
37.
z
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.8
TRIPLE I NTEGR AL S IN CYLINDRIC AL COO RDINATES
1051
51.
z
z
z
z
53.
54.
52.
z
z
z
z
z
z
z
z
z
z
z
55.
53–54
yyy
z
yyy
D I SCO VE RY PROJ ECT
z
z
CAS
VOLUMES OF HYPERSPHERES
1.
2.
3.
z
x
w
x
4.
15.8
Triple Integrals in Cylindrical Coordinates
y
P(r, ¨)=P(x, y)
r
y
¨
O
x
x
FIGURE 1
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1052
CHAPTER 15
M ULTIP LE INT EGRAL S
z
Cylindrical Coordinates
P (r, ¨, z)
z
z
z
O
r
¨
x
y
(r, ¨, 0)
1
z
z
2
z
z
FIGURE 2
EXAMPLE 1
SOLUTION
z
”2,
2π
3 , 1’
1
2
0
2π
3
x
s
s
y
z
FIGURE 3
s
s
s
z
z
0
(c, 0, 0)
x
(0, c, 0)
s
s
y
z
z
FIGURE 4
r=c
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.8
v
z
1053
TRIPLE I NTEGR AL S IN CYLINDRIC AL COO RDINATES
z
EXAMPLE 2
z
SOLUTION
z
z
0
y
x
z
z
FIGURE 5
z=r
z
Evaluating Triple Integrals with Cylindrical Coordinates
z
z
z
z=u™(x, y)
z=u¡(x, y)
r=h¡(¨) 0
D
yyy
3
z
yy y
4
yyy
z
y y
y
r=h ™(¨)
x
FIGURE 6
¨=b
¨=a
y
z
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1054
CHAPTER 15
M ULTIP LE INT EGRAL S
z
z
dz
d¨
r
r d¨
z
z
dr
z
v
FIGURE 7
z
EXAMPLE 3
z
dV=r dz dr d¨
SOLUTION
z
z
z=4
(0, 0, 4)
z
z
z
z
s
z
(0, 0, 1)
0
y yy
s
z
y y
y
(1, 0, 0)
x
yyy
z=1-r @
y
y
FIGURE 8
y y
EXAMPLE 4
y
s
s
z
s
SOLUTION
s
z
z
s
s
s
z
z
z
z
z
z=2
y y
2
z=œ„„„„
≈+¥
x
2
2
y
s
s
y
s
z
yyy
y y y
y
z
y
FIGURE 9
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.8
1055
TRIPLE I NTEGR AL S IN CYLINDRIC AL COO RDINATES
Exercises
15.8
1–2
xxx
18.
z
1.
z
s
3.
s
4.
z
z
xxx
20.
3– 4
z
xxx
19.
s
2.
z
z
xxx
21.
5–6
z
z
5.
6.
22.
z
7–8
7. z
8.
23.
z
z
s
z
24.
9–10
9.
z
10.
z
z
z
z
z
25.
z
z
11–12
11.
z
12.
26.
z
;
13.
27.
z
; 14.
z
28.
z
15–16
15.
y
y y
z
xxx
z
;
z
z
16.
yy y
z
17–28
17.
z
s
z
29–30
29.
y y
s
30.
y y
s
s
y
s
y
z z
s
z
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1056
CHAPTER 15
M ULTIP LE INT EGRAL S
31.
© S.R. Lee Photo Traveller / Shutterstock
t
L A B O R AT O R Y P R O J E C T THE INTERSECTION OF THREE CYLINDERS
1.
z
z
2.
CAS
3.
4.
5.
CAS
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.9
TRIPLE INTEG RAL S IN SPHERICAL COOR DINA TES
1057
Triple Integrals in Spherical Coordinates
15.9
Spherical Coordinates
z
P ( ∏, ¨, ˙)
˙
∏
z
O
¨
x
y
FIGURE 1
z
z
z
z
z
c
0
0
c
y
x
0
0
y
x
y
y
x
x
0<c<π/2
FIGURE 2 ∏=c
FIGURE 3 ¨=c
c
π/2<c<π
FIGURE 4 ˙=c
z
Q
z
˙
z
P(x, y, z)
P(∏, ¨, ˙)
∏
˙
O
x
x
r
¨
y
y
z
1
P ª(x, y, 0)
FIGURE 5
2
z
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1058
CHAPTER 15
M ULTIP LE INT EGRAL S
v
SOLUTION
z
π
3
O
s
(2, π/4, π/3)
2
s
π
4
x
EXAMPLE 1
y
s
s
z
FIGURE 6
s
v
s
s
EXAMPLE 2
SOLUTION
s
z
s
|
WARNING There is not universal
agreement on the notation for spherical coordinates. Most books on physics reverse the
meanings of and and use in place of .
TEC In Module 15.9 you can investigate
families of surfaces in cylindrical and spherical coordinates.
z
s
Evaluating Triple Integrals with Spherical Coordinates
z
∏i
˙k
0
x
ri=∏ i
˙ k Ψ
Î˙
Î∏
∏ i Î˙
y
˙k
ri Ψ=∏ i
˙ k Ψ
FIGURE 7
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SECTION 15.9
TRIPLE INTEG RAL S IN SPHERICAL COOR DINA TES
1059
z
yyy
z
z
l
l
3
yyy
z
y y y
z
z
˙ d¨
˙
∏
˙ d∏ d¨ d˙
x
d∏
∏ d˙
0
FIGURE 8
dV=∏@
∏
y
d¨
t
t
t
t
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1060
CHAPTER 15
M ULTIP LE INT EGRAL S
v
xxx
EXAMPLE 3
z
z
z
SOLUTION
z
yyy
y y y
z
y
y
y
NOTE
y y
v
s
s
y
s
z
EXAMPLE 4
z
s
z
z
(0, 0, 1)
π
4
x
FIGURE 9
Figure 10 gives another look (this time drawn
by Maple) at the solid of Example 4.
z
s
z
≈+¥+z@=z
z=œ„„„„
≈+¥
y
SOLUTION
s
FIGURE 10
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.9
yyy
1061
TRIPLE INTEG RAL S IN SPHERICAL COOR DINA TES
y y y
y
y
y
TEC Visual 15.9 shows an animation of
Figure 11.
z
z
x
∏
FIGURE 11
15.9
˙
x
y
˙
¨
z
˙
x
y
π/4
¨
y
¨
2π.
Exercises
9–10
1–2
9.
1.
z
z
10.
z
z
2.
11–14
11.
3– 4
s
3.
4.
s
s
12.
s
13.
14.
5–6
5.
15.
6.
7–8
7.
;
z
z
z
s
16.
8.
CAS
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1062
CHAPTER 15
M ULTIP LE INT EGRAL S
32.
17–18
17.
y y y
18.
y y y
33.
34.
19–20
z
z
19.
z
20.
35–38
3
35.
z
2
z
y
x
x
1
2
21–34
21.
xxx
22.
xxx
CAS
z
z
z
CAS
z
xxx
xxx
26.
z
z
38.
39– 41
z
25.
xxx
37.
z
xxx
24.
36.
z
xxx
23.
y
s
39.
y y
40.
y y
s
41.
y y
s
z
z
z
y
s
s
s
s
z
s
y
y
s
s
s
s
z
z
z
z
z
z
27.
42.
28.
29.
30.
z
s
z
31.
z
; 43.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPLIED PROJECT
44.
ROLL ER DERBY
1063
46.
y y y
z
s
z
z
z
47.
z
z
CAS
45.
APPLIED PROJECT
ROLLER DERBY
h
t
v
v
v
å
t
v
1.
t
v
2.
v
t
t
s
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1064
CHAPTER 15
MULTIPLE INTEGRALS
3.
t
4.
5.
l
l
6.
15.10 Change of Variables in Multiple Integrals
y
1
t
t
y
t
t
y
2
t
y
yy
yy
v
v
v
3
t
v
v
v
v
t
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.10
CHAN GE OF VARIA BLES IN MULTIPL E INTEGR ALS
v
1065
v
v
√
y
T
S
(u¡, √¡)
R
T –!
(x¡, y¡)
u
0
x
0
FIGURE 1
v
v
v
v
EXAMPLE 1
v
v
v
v
SOLUTION
√
v
S£
(0, 1)
S¢
(1, 1)
S
0
v
S™
S¡ (1, 0)
v
u
v
v
4
T
y
v
(0, 2)
¥
x= 4 -1
¥
x=1- 4
5
R
(_1, 0)
v
0
(1, 0)
v
x
FIGURE 2
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1066
CHAPTER 15
M ULTIP LE INT EGRAL S
v
v
√
v
y
u=u ¸
(u¸, √)
Î√
T
Îu
(u¸, √ ¸)
R
(x¸, y¸)
√=√¸
(u, √¸)
0
u
0
x
FIGURE 3
v
t
v
v
v
v
v
v
v
v
(u¸, √¸+Î√)
t
v
tv
v
v
v
v
v
R
(u¸, √¸)
v
v
v
(u¸+Î u, √¸)
v
FIGURE 4
Î√
v
v
v
l
v
√
(u¸, √¸)
v
Îu
v
u
v
FIGURE 5
6
v
v
v
v
v
v
v
v
v
v
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.10
CHAN GE OF VARIA BLES IN MULTIPL E INTEGR ALS
1067
v
v
v
7
The Jacobian is named after the German mathematician Carl Gustav Jacob Jacobi (1804–1851).
Although the French mathematician Cauchy first
used these special determinants involving partial derivatives, Jacobi developed them into a
method for evaluating multiple integrals.
v
v
v
v
t
Definition
v
v
v
v
v
v
v
8
v
v
v
v
√
y
Sij
S
Î√
Îu
T
(x i , y j)
(u i , √ j )
FIGURE 6
0
R ij
R
u
0
x
yy
t
v
v
v
v
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1068
CHAPTER 15
M ULTIP LE INT EGRAL S
v
yy
9
t
v
v
v
v
Change of Variables in a Double Integral
v
yy
yy
v
v
v
v
v
v
v
v
v
¨
¨=∫
∫
r=a
r=b
S
å
t
¨=å
0
a
b
r
T
y
r=b
¨=∫
R
∫
0
r=a
yy
¨=å
å
x
yy
y y
FIGURE 7
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 15.10
1069
CHAN GE OF VARIA BLES IN MULTIPL E INTEGR ALS
v
EXAMPLE 2
xx
v
SOLUTION
v
v
v
v
v
v
yy
yy
v
yy
y
v
yy
v
v
v
v
v
y
v
v
v
v
v
v
v
v
v
NOTE
v
xx
EXAMPLE 3
SOLUTION
v
10
v
v
v
11
v
v
v
v
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1070
CHAPTER 15
M ULTIP LE INT EGRAL S
√
v
√=2
(_2, 2)
(2, 2)
S
u=_√
u=√
(_1, 1)
(1, 1)
0
T
v
√=1
v
u
v
v
v
T –!
v
y
v
v
v
x-y=1
0
_1
1
R
2
yy
x
yy
x-y=2
v
yy
_2
v
v
y
v
v
y
FIGURE 8
v
v
v
v
v
v
v
v v
Triple Integrals
vw
t
z
v w
v w
v w
z
v
v
w
z
12
13
v w
z
yyy
z
yyy
v w
v
w
z
z
v
w
v w z
v w
z
v w
v w
EXAMPLE 4
SOLUTION
z
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SECTION 15.10
CHAN GE OF VARIA BLES IN MULTIPL E INTEGR ALS
1071
z
z
yyy
z
yyy
15.10 Exercises
11–14
1–6
1.
v
2.
v
v
v
v
v
11.
3.
12.
4.
5.
v
6.
v
v w
w
13.
w
z
w
z
v
14.
7–10
7.
v
15–20
v
v
v
15.
8.
xx
v
v
v
v
v
16.
xx
9.
v
v
10.
;
v
v
17.
v
xx
v
v
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1072
18.
CHAPTER 15
xx
s
19.
M ULTIP LE INT EGRAL S
s
s
v
s
v
; 20. xx
23.
yy
24.
xx
25.
yy
26.
xx
27.
xx
v
xxx
z
v z
w
z
22.
23–27
xx
v
21.
v
28.
yy
y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 15
REVIEW
1073
Review
15
Concept Check
1.
6.
xx
z
xx
7.
xxx
xxx
xx
xx
2.
xx
xx
xxx
z
xxx
z
xxx
z
z
z
8.
z
3.
4.
9.
5.
t
10.
v
v
True-False Quiz
Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement.
1.
y y
2.
yy
3.
yy
4.
y y
yy
y
yy
s
7.
yy
s
6.
yy s
s
y
8.
xxx
z
z
9.
y yy
5.
yy
y
z
z
z
s
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1074
CHAPTER 15
M ULTIP LE INT EGRAL S
Exercises
1.
12.
xx
y y y
y
3
2
2
1
4
3
5
7
6
8
9
13–14
10
13.
14.
15.
xx
16.
xx
1
3 x
2
17.
yy
s
s
2.
3–8
3.
yy
15–28
1
0
y y
y y
4.
5.
y y
7.
y y y
6.
s
8.
z
y
4
9.
2
_4
_2
0
yy
y yy
4
R
R
4 x
19.
xx
20.
xx
21.
xx
z z
y
10.
2
yy
y y
xx
9–10
18.
_4
0
22.
xx
23.
xxx
4 x
24.
xxx
25.
xxx
s
z
z
11.
y y
;
z
z
CAS
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 15
26.
27.
xxx
z
z
CAS
z
xxx
z
xxx
zs
z
1075
z
41.
z
28.
40.
REVIEW
y y
z
s
s
42.
y y
29–34
29.
z
30.
s
y
s
s
s
z
z
; 43.
xx
z
CAS
31.
44.
z
32.
z
z
z
45.
33.
z
z
34.
z
z
s
46.
35.
47.
y y y
36.
z
z
z
48.
yy y
37.
49.
z
v
z
yy
z
38.
z
z
39.
z
z
50.
z
v
s
s
z
w
sz
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1076
CHAPTER 15
51.
M ULTIP LE INT EGRAL S
xx
l
52.
yy
yy
54.
yy
l
yyy
z
53.
l
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Problems Plus
1.
yy
2.
yy
x
3.
4.
z
yyy
5.
yy
l
yy
6.
v
v
s
s
v
7.
y yy
y yy
z
z
z
z
1077
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
8.
y
9.
z
z
10.
z
s
11.
yyy
12.
l
z
y
z
s
13.
z
z
1078
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16
Vector Calculus
Parametric surfaces, which are studied in
Section 16.6, are frequently used by
programmers creating animated films. In
this scene from Antz, Princess Bala is
about to try to rescue Z, who is trapped
in a dewdrop. A parametric surface
represents the dewdrop and a family of
such surfaces depicts its motion. One of
the programmers for this film was heard
to say, “I wish I had paid more attention
in calculus class when we were studying
parametric surfaces. It would sure have
helped me today.”
© Dreamworks / Photofest
1079
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1080
16.1
CHAPTER 16
V E CTOR CALC ULUS
Vector Fields
Nova Scotia
Adapted from ONERA photograph, Werle, 1974
FIGURE 1
FIGURE 2
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SECTION 16.1
y
1
Definition
2
Definition
VECTOR F I ELDS
1081
(x, y)
(x, y)
x
0
FIGURE 3
R@
z
z
z
(x, y, z)
z
(x, y, z)
0
z
z
z
y
x
z
FIGURE 4
z
z
R#
v
EXAMPLE 1
y
(2, 2)
(0, 3)
0
(1, 0)
SOLUTION
x
FIGURE 5
(x, y)=_y +x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1082
CHAPTER 16
V E CTOR CALC ULUS
s
s
5
_5
s
6
5
_6
5
_5
_5
6
5
_6
FIGURE 6
_5
FIGURE 7
(x, y)=k_y, xl
(x, y)=ky,
v
FIGURE 8
xl
(x, y)=k (1+¥), (1+≈)l
z
EXAMPLE 2
z
SOLUTION
z
0
y
FIGURE 9
x
(x, y, z)=z
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SECTION 16.1
1083
VECTOR F I ELDS
z
1
z
0
z
_1
1
5
0
z3
_1
_1
0
y
_1
0
1
x
1
_1
FIGURE 10
(x, y, z)=y +z +x
TEC In Visual 16.1 you can rotate the
vector fields in Figures 10–12 as well as
additional fields.
1
0
y
1
1
FIGURE 11
(x, y, z)=y -2 +x
0
x
_1
_1
y0
1
1
_1
0 x
FIGURE 12
y
x
z
(x, y, z)= - +
z
z
4
z
z
EXAMPLE 3
z
z
0
x
y
EXAMPLE 4
FIGURE 13
u
z
z
3
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1084
CHAPTER 16
V E CTOR CALC ULUS
z
z
s
z
y
x
z
z
z
z
z
EXAMPLE 5
FIGURE 14
z
z
4
Gradient Fields
∇
∇
4
z
v
_4
4
z
z
z
z
EXAMPLE 6
SOLUTION
_4
FIGURE 15
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.1
VECTOR F I ELDS
1085
∇
z
z
s
z
z
z
z
z
z
z
16.1
Exercises
1–10
13.
14.
1.
2.
3.
4.
5.
3
s
6.
_3
3
3
_3
3
s
7.
z
8.
z
9.
z
10.
z
_3
_3
3
3
_3
3
_3
3
11–14
11.
_3
12.
CAS
_3
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1086
CHAPTER 16
V E CTOR CALC ULUS
29–32
15–18
15.
z
17.
z
18.
z
16.
z
z
29.
30.
31.
32.
4
z
1
_4
1
s
4
4
_4
4
z 0
z 0
_1
_1
_1
y
0
_1 0
1
y
_1
1 0x
1
1
0
_1
x
1
1
_4
4
4
_4
z 0
z 0
_4
4
_4
4
_1
_1
_1 0
1
y
1
0
_1
x
_1
y
0
_4
_1
1 0x
1
_4
33.
CAS
19.
34.
CAS
20.
35.
21–24
21.
CAS
22.
23.
z
24.
z
s
z
z
25–26
∇
25.
26.
s
36.
27–28
27.
28.
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SECTION 16.2
LINE I NTEGR ALS
1087
Line Integrals
16.2
1
y
P i (x i , y i )
Pi-1
Pi
C
Pn
P™
P¡
P¸
0
x
ti
a
t i-1
ti
b t
FIGURE 1
2
Definition
y
3
y
l
y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1088
CHAPTER 16
V E CTOR CALC ULUS
The arc length function is discussed in
Section 13.3.
z
0
C
y
y
f(x, y)
y
(x, y)
x
x
FIGURE 2
x
EXAMPLE 1
y
SOLUTION
≈+¥=1
(y˘0)
0
_1
x
1
y
FIGURE 3
y
y
y
s
y
C¢
C∞
C™
C£
C¡
0
FIGURE 4
x
y
y
y
y
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SECTION 16.2
SOLUTION
C™
C¡
(1, 2)
(1, 1)
y
x
(0, 0)
y
y
s
s
FIGURE 5
C=C¡
1089
x
EXAMPLE 2
y
LINE I NTEGR ALS
C™
y
y
y
y
y
s
y
x
l
4
v
y
y
y
EXAMPLE 3
SOLUTION
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1090
CHAPTER 16
V E CTOR CALC ULUS
y
y
y
y
y
y
1
_1
0
1
x
FIGURE 6
5
y
l
6
y
l
x
7
y
y
y
y
y
y
y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.2
LINE I NTEGR ALS
1091
8
v
y
(0, 2)
C¡
0
C™
x
EXAMPLE 4
SOLUTION
4
x
x=4-¥
(_5, _3)
FIGURE 7
y
y
y
y
y
y
y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1092
CHAPTER 16
V E CTOR CALC ULUS
B
C
A
a
b
t
y
B
A
y
y
y
_C
FIGURE 8
y
y
Line Integrals in Space
z
z
z
y
9
y
z
z
l
y
z
z
z
y
z
y
y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.2
LINE I NTEGR ALS
1093
z
y
z
z
z
l
y
y
10
z
x
EXAMPLE 5
z
z
z
v
z
z
z
z
z
z
z
SOLUTION
6
y
4
z
y
2
C
0
_1
z
y
z
s
s
s
s
_1
y
0
0
1 1
EXAMPLE 6
x
x
y
z
z
FIGURE 9
SOLUTION
z
(3, 4, 5)
C¡
(2, 0, 0)
x
FIGURE 10
0
C™
z
y
(3, 4, 0)
y
z
z
y
y
z
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1094
CHAPTER 16
V E CTOR CALC ULUS
y
z
y
y
z
z
z
Line Integrals of Vector Fields
x
l
z
(x i , y i , z i )
Pi-1
0
z
(t i )
Pi
P i (x i , y i , z i )
x
Pn
y
z
z
P¸
FIGURE 11
z
11
z
z
12
z
y
z
z
y
z
y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.2
1095
LINE I NTEGR ALS
x
13 Definition
y
y
y
z
z
z
z
Figure 12 shows the force field and the curve in
Example 7. The work done is negative because
the field impedes movement along the curve.
y
EXAMPLE 7
SOLUTION
1
0
1
y
x
y
y
FIGURE 12
x
NOTE
Figure 13 shows the twisted cubic in
Example 8 and some typical vectors acting at
three points on .
x
y
y
2
1.5
{ (1)}
z 1
0.5
EXAMPLE 8
x
z
z
z
(1, 1, 1)
{ (3/4) }
0
0
y1 2
2
z
C
SOLUTION
{ (1 / 2)}
1
x
0
FIGURE 13
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1096
CHAPTER 16
V E CTOR CALC ULUS
y
y
y
y
y
y
z
y
y
z
z
y
z
z
z
16.2
9.
x
z
z
1.
x
2.
x
10.
x
3.
x
11.
x
4.
x
5.
x
12.
x
s
s
13.
14.
7.
x
15.
x
8.
x
16.
x
CAS
z
z
z
z
x
x
x
;
z
Exercises
1–16
6.
z
z
x
x
z
z
z
z
z
s
z
z
z
z
z
z
1.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.2
17.
x
x
24.
x
25.
x
26.
x
z
z
LINE I NTEGR ALS
1097
z
z
z
z
z
y
3
2
CAS
27–28
1
_3
_2
_1 0
_1
2
1
27.
3x
_2
28.
_3
s
s
18.
x
29.
y
;
s
C¡
C™
z
x
z
;
CAS
x
19–22
x
30.
x
31.
z
z
32.
19.
CAS
20.
z
21.
z
22.
z
z
z
z
33.
34.
23–26
23.
x
35.
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1098
CHAPTER 16
V E CTOR CALC ULUS
45.
z
36.
z
46.
37.
47.
y
y
38.
48.
z
z
z
y
z
z
y
z
z
y
49.
y
z
50.
39.
y
51.
40.
41.
z
y
z z
C
42.
z
z
C
1
43.
0
1
x
52.
44.
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SECTION 16.3
THE FUNDAM ENTAL T HEOREM FOR LINE INTEG RAL S
1099
I
y
The Fundamental Theorem for Line Integrals
16.3
y
1
∇
2
Theorem
∇
y
B(x™, y™)
A(x¡, y¡)
0
C
y
x
NOTE
∇
z
A(x¡, y¡, z¡)
C
0
x
y
B(x™, y™, z™)
y
z
z
y
z
z
FIGURE 1
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1100
CHAPTER 16
V E CTOR CALC ULUS
PROOF OF THEOREM 2
y
y
z
z
y
EXAMPLE 1
SOLUTION
∇
z
y
s
z
y
s
s
s
Independence of Path
x
x
y
y
∇
x
x
x
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SECTION 16.3
THE FUNDAM ENTAL THEOR EM FO R LINE INTEG RAL S
1101
x
C
y
FIGURE 2
y
y
y
x
C™
B
A
y
y
C¡
x
FIGURE 3
3
y
y
y
y
x
Theorem
x
x
x
4
Theorem
x
∇
PROOF
y
x
y
(x¡, y)
C¡
(a, b)
0
FIGURE 4
C™
(x, y)
y
D
y
y
y
x
y
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1102
CHAPTER 16
V E CTOR CALC ULUS
y
y
(x, y)
C™
C¡
(a, b)
0
(x, y¡)
D
y
y
y
y
y
∇
x
FIGURE 5
∇
5
Theorem
FIGURE 6
FIGURE 7
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SECTION 16.3
6
v
10
_10
THE FUNDAM ENTAL THEOR EM FO R LINE INTEG RAL S
1103
Theorem
EXAMPLE 2
10
SOLUTION
C
_10
FIGURE 8
Figures 8 and 9 show the vector fields in
Examples 2 and 3, respectively. The vectors in
Figure 8 that start on the closed curve all
appear to point in roughly the same direction as
. So it looks as if x
and therefore
is not conservative. The calculation in Example
2 confirms this impression. Some of the vectors
near the curves and in Figure 9 point in
approximately the same direction as the curves,
whereas others point in the opposite direction.
So it appears plausible that line integrals around
all closed paths are . Example 3 shows that
is indeed conservative.
v
EXAMPLE 3
SOLUTION
2
C™
C¡
_2
2
_2
∇
EXAMPLE 4
x
FIGURE 9
∇
SOLUTION
∇
7
8
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1104
CHAPTER 16
V E CTOR CALC ULUS
t
9
t
t
10
t
t
y
v
y
EXAMPLE 5
∇
z
z
z
SOLUTION
11
z
12
z
13
z
z
14
t
z
z
z
t
z
t
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.3
THE FUNDAM ENTAL THEOR EM FO R LINE INTEG RAL S
t
t
z
z
1105
z
z
z
z
z
z
z
z
z
z
z
∇
Conservation of Energy
y
y
y
y
y
15
16
∇
z
∇
z
y
z
y
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1106
16.3
CHAPTER 16
V E CTOR CALC ULUS
Exercises
1.
9.
x
10.
y
C
20
30
40
50
60
11.
x
10
y
0
2.
x
3
x
2
x
1
y
0
12–18
1
2
x
∇
x
3–10
3
12.
3.
13.
4.
5.
14.
6.
7.
15.
z
z
z
z
8.
CAS
1.
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SECTION 16.3
16.
z
17.
z
18.
z
z
s
z
z
THE FUNDAM ENTAL THEOR EM FO R LINE INTEG RAL S
1107
28.
z
z
z
z
z
y
z
y
29.
z
z
z
19–20
19.
x
20.
x
30.
x
z
z z
31–34
31.
32.
33.
21.
34.
35.
22.
x
x
x
36.
23–24
s
23.
24.
z
25–26
25.
y
26.
x
CAS
y
x
27.
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1108
16.4
CHAPTER 16
V E CTOR CALC ULUS
Green’s Theorem
y
D
C
0
x
y
FIGURE 1
y
C
D
D
C
0
x
0
x
FIGURE 2
Green’s Theorem
Recall that the left side of this equation
is another way of writing x
, where
.
y
NOTE
y
g
y
1
y
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SECTION 16.4
GR EEN’S THEOR EM
1109
PROOF OF GREEN’S THEOREM FOR THE CASE IN WHICH D IS A SIMPLE REGION
George Green
Green’s Theorem is named after the selftaught English scientist George Green
(1793–1841). He worked full-time in his father’s
bakery from the age of nine and taught himself
mathematics from library books. In 1828 he
published privately An Essay on the Application
of Mathematical Analysis to the Theories of
Electricity and Magnetism, but only 100 copies
were printed and most of those went to his
friends. This pamphlet contained a theorem
that is equivalent to what we know as Green’s
Theorem, but it didn’t become widely known
at that time. Finally, at age 40, Green entered
Cambridge University as an undergraduate
but died four years after graduation. In 1846
William Thomson (Lord Kelvin) located a copy
of Green’s essay, realized its significance, and
had it reprinted. Green was the first person to
try to formulate a mathematical theory of electricity and magnetism. His work was the basis
for the subsequent electromagnetic theories of
Thomson, Stokes, Rayleigh, and Maxwell.
y
2
3
y
yy
y
yy
t
t
t
yy
4
t
y y
y
t
t
t
t
y=g™(x)
C£
C¢
y=g¡(x)
0
t
D
a
C™
y
C¡
b
y
x
t
t
FIGURE 3
y
y
y
y
y
y
y
y
t
y
t
y
y
y
t
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1110
CHAPTER 16
V E CTOR CALC ULUS
y
x
EXAMPLE 1
y
yy
SOLUTION
y=1-x
(0, 1)
C
D
(0, 0)
(1, 0)
y
x
y y
FIGURE 4
y
v
y
x
EXAMPLE 2
s
SOLUTION
Instead of using polar coordinates, we could
simply use the fact that is a disk of radius 3
and write
y
s
s
yy
y y
y
y
y
xx
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SECTION 16.4
y
5
y
GR EEN’S THEOR EM
1111
y
EXAMPLE 3
SOLUTION
y
y
y
P
0
3
10
4
P
P
■
FIGURE 5
■
u
Extended Versions of Green’s Theorem
C¡
D¡
D™
C£
FIGURE 6
_C£
y
C™
y
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1112
CHAPTER 16
V E CTOR CALC ULUS
C
y
FIGURE 7
v
y
EXAMPLE 4
x
SOLUTION
≈+¥=4
C
D
0
≈+¥=1
y
x
FIGURE 8
yy
y y
y
y
C™
D
C¡
FIGURE 9
Dª
y
Dªª
y
FIGURE 10
y
v
EXAMPLE 5
y
y
x
SOLUTION
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SECTION 16.4
GR EEN’S THEOR EM
1113
y
C
Cª
D
x
y
y
FIGURE 11
y
y
y
y
y
y
y
y
y
SKETCH OF PROOF OF THEOREM 16.3.6
y
y
yy
x
x
16.4
Exercises
1– 4
1.
x
;
CAS
2.
x
3.
x
1.
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1114
CHAPTER 16
4.
V E CTOR CALC ULUS
x
19.
; 20.
5–10
5.
x
6.
x
7.
x
8.
x
9.
10.
21.
y
s
x
x
22.
x
11–14
y
11.
12.
23.
24.
13.
14.
CAS
y
s
25.
y
15–16
y
26.
15.
x
27.
16.
17.
28.
18.
s
29.
x
x
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SECTION 16.5
CURL AND DIVERGENCE
1115
30.
t
31.
yy
16.5
yy
v
v
v
v
v
v
Curl and Divergence
Curl
1
z
z
∇
∇
z
∇
∇
z
z
z
∇
z
z
2
z
∇
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1116
CHAPTER 16
V E CTOR CALC ULUS
EXAMPLE 1
z
z
z
SOLUTION
z
z
z
z
CAS Most computer algebra systems have com-
mands that compute the curl and divergence of
vector fields. If you have access to a CAS, use
these commands to check the answers to the
examples and exercises in this section.
z
z
z
z
z
z
z
3
Theorem
PROOF
Notice the similarity to what we know
from Section 12.4:
for every
three-dimensional vector .
z
z
z
z
z
z
∇
Compare this with Exercise 29 in
Section 16.3.
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SECTION 16.5
v
z
EXAMPLE 2
CURL AND DIVERGENCE
z
1117
z
SOLUTION
z
4
v
Theorem
EXAMPLE 3
z
z
z
z
SOLUTION
z
z
z
z
z
z
5
z
6
z
7
8
z
z
z
z
z
z
z
z
z
z
z
t
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1118
CHAPTER 16
V E CTOR CALC ULUS
t
z
t
z
z
z
z
t
z
z
z
z
z
z
z
z
(x, y, z)
z
z
(x, y, z)
FIGURE 1
Divergence
z
9
z
z
10
EXAMPLE 4
z
z
z
SOLUTION
z
z
z
z
z
11 Theorem
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.5
CURL AND DIVERGENCE
1119
PROOF
Note the analogy with the scalar triple
product:
.
z
z
z
v
z
z
z
z
EXAMPLE 5
z
z
z
SOLUTION
z
The reason for this interpretation of
will
be explained at the end of Section 16.9 as a
consequence of the Divergence Theorem.
z
z
z
z
z
z
z
z
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1120
CHAPTER 16
V E CTOR CALC ULUS
Vector Forms of Green’s Theorem
y
y
z
y
12
yy
y
(t)
D
(t)
(t)
C
0
x
FIGURE 2
y
y
y
y
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SECTION 16.5
y
13
1121
yy
Exercises
16.5
12.
1–8
1.
z
2.
z
z
3.
z
z
4.
z
5.
z
6.
z
7.
z
z
z
z
z
z
z
z
s
z
z
z
z
z
8.
z
13–18
z
z
z
z
9–11
z
9.
CURL AND DIVERGENCE
z
10.
y
y
∇
13.
z
z
z
z
14.
z
z
z
z
15.
z
16.
z
17.
z
18.
z
z
z
z
z
z
z
z
z
z
z
z
z
z
19.
z
20.
0
11.
x
0
z
x
z
z
21.
y
t
z
z
t
22.
0
x
z
z
t
z
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1122
CHAPTER 16
V E CTOR CALC ULUS
23–29
x
z
z
z
z
z
z
t
t
t
t
36.
z
z
xx
z
37.
23.
24.
z
25.
26.
z
27.
t
28.
29.
z
30–32
z
30.
B
31.
d
v
P
32.
¨
0
33.
y
yy
y
t
t
yy
t
t
x
t
38.
t
t
34.
yy
t
t
y
t
t
t
35.
t
t
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.6
t
39.
t
16.6
1123
PAR AMETR IC SURFACES AND THEIR AREAS
z
t
z
z
x
z
Parametric Surfaces and Their Areas
Parametric Surfaces
v
v
v
1
v
v
v
z
v
z
v
z
v
2
v
z
z
v
v
v
v
v
√
z
S
D
(u, √)
0
u
0
FIGURE 1
x
(u, √)
y
EXAMPLE 1
v
v
SOLUTION
v
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1124
CHAPTER 16
z
V E CTOR CALC ULUS
z
(0, 0, 2)
z
z
0
v
x
v
y
v
(2, 0, 0)
v
v
FIGURE 2
z
z
v
(0, 3, 2)
v
0
v
v
x
v
y
z
√
FIGURE 3
(u¸, √¸)
√=√¸
TEC Visual 16.6 shows animated versions
of Figures 4 and 5, with moving grid curves, for
several parametric surfaces.
D
C¡
u=u ¸
0
C™
0
u
y
x
FIGURE 4
v
v
v
v
v
√
EXAMPLE 2
v
u
v
v
v
v
SOLUTION
v
v
x
FIGURE 5
y
v
v
v
v
z
v
v
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.6
1125
PAR AMETR IC SURFACES AND THEIR AREAS
v
z
z
EXAMPLE 3
SOLUTION
√
v
A
P¸
v
v
A
FIGURE 6
A
v
v
v
v
z
z
z
v
¨
2π
v
D
v
z
z
v
EXAMPLE 4
z
˙=c
SOLUTION
¨=k
k
0
c
˙
π
z
z
˙=c
0
x
y
¨=k
NOTE
v
FIGURE 7
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1126
CHAPTER 16
V E CTOR CALC ULUS
One of the uses of parametric surfaces is in
computer graphics. Figure 8 shows the result of
trying to graph the sphere
z
by solving the equation for z and graphing the
top and bottom hemispheres separately. Part
of the sphere appears to be missing because
of the rectangular grid system used by the
computer. The much better picture in Figure 9
was produced by a computer using the
parametric equations found in Example 4.
FIGURE 8
FIGURE 9
EXAMPLE 5
z
SOLUTION
z
z
z
z
v
z
EXAMPLE 6
SOLUTION
z
TEC In Module 16.6 you can investigate
several families of parametric surfaces.
z
z
z
EXAMPLE 7
z
s
SOLUTION 1
s
z
s
SOLUTION 2
z
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SECTION 16.6
For some purposes the parametric representations in Solutions 1 and 2 are equally good,
but Solution 2 might be preferable in certain
situations. If we are interested only in the part
of the cone that lies below the plane z
,
for instance, all we have to do in Solution 2 is
change the parameter domain to
1127
PAR AMETR IC SURFACES AND THEIR AREAS
s
z
Surfaces of Revolution
z
z
0
y
z
3
y=ƒ
ƒ
x
x
¨
z
(x, y, z)
EXAMPLE 8
ƒ
SOLUTION
FIGURE 10
z
z
y
x
z
Tangent Planes
FIGURE 11
v
v
v
v
z
v
v
v
v
4
v
v
v
v
z
v
v
z
√
(u ¸, √¸)
√=√¸
D
0
FIGURE 12
v
P¸
√
u
C¡
u=u ¸
0
u
x
C™
y
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1128
CHAPTER 16
V E CTOR CALC ULUS
v
v
v
v
v
5
z
v
v
v
v
v
v
Figure 13 shows the self-intersecting
surface in Example 9 and its tangent plane
at
.
z
EXAMPLE 9
v z
v
SOLUTION
(1, 1, 3)
z
y
z
v
x
v
v
v
v
FIGURE 13
v
v
v
v
v
z
z
Surface Area
v
√
z
R ij
Î√
Pij
Îu
(u i , √ j )
FIGURE 14
0
u
Sij
0
x
y
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SECTION 16.6
PAR AMETR IC SURFACES AND THEIR AREAS
1129
v
v
v
v
v
v
Sij
v
Pij
v
v
v
v
v
v
v
Î√
v
√
Îu
u
xx
v
6
v
Definition
v
FIGURE 15
v
v
v
z
v
v
yy
v
z
z
v
v
v
v
EXAMPLE 10
SOLUTION
z
z
z
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1130
CHAPTER 16
V E CTOR CALC ULUS
s
s
s
yy
y y
y
y
Surface Area of the Graph of a Function
z
z
7
z
8
Notice the similarity between the surface area
formula in Equation 9 and the arc length formula
z
9
z
z
from Section 8.1.
v
EXAMPLE 11
z
z
SOLUTION
z
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SECTION 16.6
PAR AMETR IC SURFACES AND THEIR AREAS
1131
z
9
z
z
yy s
yy s
D
x
3
FIGURE 16
y
y y
y
s
y
s
s
z
s
s
s
yy
y y
s
y
s
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1132
CHAPTER 16
V E CTOR CALC ULUS
Exercises
16.6
1–2
z
1.
v
2.
v
v
v
v
v
z
v
v
x
x
3–6
3.
v
4.
v
v
v
z
z
v
v
y
y
v
5.
6.
x
; 7–12
v
7.
v
v
y
x
y
z
z
v
v
8.
v
v
v
v
9.
v
v
v
10.
v
11.
y
x
v
v
v
x
v
v
v
z
v
z
y
v
v
12.
v
19–26
v
19.
13–18
20.
v
13.
v
v
v
14.
v
v
v
15.
v
v
21.
v
22.
v
16.
v
23.
z
v
17.
18.
;
z
z
v
v
z
s
24.
v
v
z
z
v
z
z
z
v
z
v
z
25.
z
z
CAS
1.
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SECTION 16.6
26.
z
38.
1133
PAR AMETR IC SURFACES AND THEIR AREAS
v
v
v
39–50
CAS
27–28
39.
27.
28.
z
40.
3
v
v
v
v
z
z 0
_3
_3
z
42.
_1
_1
0
x
0 5
y
41.
0
y
0
0
1 1
s
z
_1
x
43.
; 29.
; 30.
z
44.
z
45.
z
46.
z
z
47.
; 31.
z
z
z
48.
v
; 32.
v
v
v
v
49.
v
z
v
v
50.
z
z
51.
z
52–53
33–36
33.
v
v
z
34.
v
35.
v
36.
v
v
z
53.
z
v
z
v
52.
v
v
v
v
v
CAS
54.
z
CAS
37–38
55.
37.
v
v
v
v
z
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1134
CHAPTER 16
VE CTOR CALCULUS
61.
CAS
z
z
z
62.
CAS
z
56.
v
v
CAS
v
v
57.
z
v
z
z
x
58.
y
v
v z
v
63.
;
z
64.
z
CAS
59.
v
v z
z
;
v
;
z
60.
(x, y, z)
v
v z
0
;
z
16.7
x
z
å
¨
y
(b, 0, 0)
Surface Integrals
z
Parametric Surfaces
v
v
v
z
v
v
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SECTION 16.7
√
R ij
D
Îu
SUR FACE I NTEGR ALS
1135
v
Î√
0
u
z
S
yy
1
P ij
z
l
Sij
0
x
y
v
v
FIGURE 1
z
z
v
v
v
v
v
We assume that the surface is covered only
once as
v ranges throughout . The value
of the surface integral does not depend on the
parametrization that is used.
yy
2
yy
z
y
v
y
z
yy
v
yy
v
v
v
EXAMPLE 1
z
v
z
z
v
z
xx
SOLUTION
z
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1136
CHAPTER 16
V E CTOR CALC ULUS
yy
yy
y y
Here we use the identities
y
y
y
y
Instead, we could use Formulas 64 and 67 in
the Table of Integrals.
z
yy
z
z
z
yy
yy
z
z
z
yy z
z
Graphs
z
t
t
z
t
3
t
t
z
t
z
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SECTION 16.7
yy
4
yy
z
SUR FACE I NTEGR ALS
z
t
1137
z
z
z
z
z
yy
z
yy
z
z z
z
xx
EXAMPLE 2
z
SOLUTION
z
y
x
yy
FIGURE 2
z
z
yy
yy
y
z
s
s
y
s
s
s
yy
z
S£ (z=1+x )
v
EXAMPLE 3
yy
z
xx
yy
z
z
z
z
z
y
S¡ (≈+¥=1)
0
SOLUTION
z
x
z
z
S™
FIGURE 3
z
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1138
CHAPTER 16
V E CTOR CALC ULUS
Therefore
r
i
sin
0
rz
and
r
1
r
y y
2
0
1
2
1 cos
2
1
0
1 3
2 2
Since
2
lies in the plane z
sin 2
sin j
1
rz
0
y
cos i
is
yy z
1
k
0
1
scos 2
rz
Thus the surface integral over
yy z
j
cos
0
z z
y
2
2 cos
1
2
1
0
1
4
2 sin
1
2
1
2
cos
cos 2
3
2
2
0
sin 2
0, we have
yy z
yy 0
2
0
2
The top surface 3 lies above the unit disk and is part of the plane z 1
. So,
taking t ,
1
in Formula 4 and converting to polar coordinates, we have
yy z
yy
3
1
y y
2
0
z
1
1
0
1
y y
0
0
s2
y
1
2
2
2
0
s2
2
s1
cos
s2
1
2
1
3
2
z
1
2
0
cos
cos
2
sin
3
0
s2
Therefore
yy z
yy z
yy z
1
3
2
2
0
s2
yy z
3
3
2
s2
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SECTION 16.7
1139
SUR FACE I NTEGR ALS
Oriented Surfaces
P
To define surface integrals of vector fields, we need to rule out nonorientable surfaces such
as the Möbius strip shown in Figure 4. [It is named after the German geometer August
Möbius (1790 –1868).] You can construct one for yourself by taking a long rectangular
strip of paper, giving it a half-twist, and taping the short edges together as in Figure 5.
If an ant were to crawl along the Möbius strip starting at a point , it would end up on
the “other side” of the strip (that is, with its upper side pointing in the opposite direction).
Then, if the ant continued to crawl in the same direction, it would end up back at the
same point without ever having crossed an edge. (If you have constructed a Möbius strip,
try drawing a pencil line down the middle.) Therefore a Möbius strip really has only
one side. You can graph the Möbius strip using the parametric equations in Exercise 32 in
Section 16.6.
FIGURE 4
A Möbius strip
TEC Visual 16.7 shows a Möbius strip
with a normal vector that can be moved along
the surface.
B
C
A
D
B
D
A
C
FIGURE 5
Constructing a Möbius strip
z
From now on we consider only orientable (two-sided) surfaces. We start with a surface
that has a tangent plane at every point , , z on (except at any boundary point). There
are two unit normal vectors n1 and n 2
n1 at , , z . (See Figure 6.)
If it is possible to choose a unit normal vector n at every such point , , z so that n
varies continuously over , then is called an oriented surface and the given choice of n
provides with an orientation. There are two possible orientations for any orientable surface (see Figure 7).
n¡
n™
0
x
FIGURE 6
n
n
y
n
n
n
n
FIGURE 7
The two orientations
of an orientable surface
n
n
n
n
For a surface z t , given as the graph of t, we use Equation 3 to associate with
the surface a natural orientation given by the unit normal vector
t
5
t
i
n
t
1
2
j
k
t
2
Since the k-component is positive, this gives the
orientation of the surface.
If is a smooth orientable surface given in parametric form by a vector function
r , v , then it is automatically supplied with the orientation of the unit normal vector
6
n
and the opposite orientation is given by
r
r
rv
rv
n. For instance, in Example 4 in Section 16.6 we
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1140
CHAPTER 16
V E CTOR CALC ULUS
found the parametric representation
r
for the sphere
,
2
r
2
sin
z2
2
r
cos i
sin 2
So the orientation induced by r
r
r
cos
k
. Then in Example 10 in Section 16.6 we found that
sin
2
cos i
r
r
r
sin j
2
and
n
sin
,
sin 2
2
r
2
sin j
sin
cos
k
sin
is defined by the unit normal vector
cos i
sin
sin j
cos
k
1
r
,
Observe that n points in the same direction as the position vector, that is, outward from the
sphere (see Figure 8). The opposite (inward) orientation would have been obtained (see
Figure 9) if we had reversed the order of the parameters because r
r
r
r.
z
z
0
y
y
x
x
FIGURE 8
FIGURE 9
Positive orientation
Negative orientation
For a closed surface, that is, a surface that is the boundary of a solid region , the
convention is that the positive orientation is the one for which the normal vectors point
from , and inward-pointing normals give the negative orientation (see Figures 8
and 9).
Surface Integrals of Vector Fields
z
Sij
F=∏v
n
S
0
x
FIGURE 10
y
Suppose that is an oriented surface with unit normal vector n, and imagine a fluid with
density
, , z and velocity field v , , z flowing through . (Think of as an imaginary surface that doesn’t impede the fluid flow, like a fishing net across a stream.) Then the
rate of flow (mass per unit time) per unit area is v. If we divide into small patches ,
as in Figure 10 (compare with Figure 1), then is nearly planar and so we can approximate the mass of fluid per unit time crossing
in the direction of the normal n by the
quantity
v n
where , v, and n are evaluated at some point on . (Recall that the component of the vector v in the direction of the unit vector n is v n.) By summing these quantities and taking the limit we get, according to Definition 1, the surface integral of the function v n
over :
7
yy
v n
yy
, ,z v , ,z
n , ,z
and this is interpreted physically as the rate of flow through .
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SECTION 16.7
If we write F
becomes
v, then F is also a vector field on
yy F
3
1141
SUR FACE I NTEGR ALS
and the integral in Equation 7
n
A surface integral of this form occurs frequently in physics, even when F is not v, and is
called the
(or
) of F over .
8 Definition If F is a continuous vector field defined on an oriented surface
with unit normal vector n, then the surface integral of F over is
yy F
yy F
S
n
This integral is also called the flux of F across .
In words, Definition 8 says that the surface integral of a vector field over is equal to
the surface integral of its normal component over (as previously defined).
If is given by a vector function r , v , then n is given by Equation 6, and from Definition 8 and Equation 2 we have
yy F
S
r
r
yy F
rv
rv
F r ,v
where
Compare Equation 9 to the similar expression
for evaluating line integrals of vector fields in
Definition 16.2.13:
y
F
y
r
Fr
r
r
rv
rv
r
rv
r
rv
is the parameter domain. Thus we have
yy F
9
yy F
S
r
EXAMPLE 4 Find the flux of the vector field F , , z
Figure 11 shows the vector field F in Example 4
at points on the unit sphere.
z
sphere
2
2
z2
zi
j
k across the unit
1.
SOLUTION As in Example 1, we use the parametric representation
r ,
sin
cos i
Then
sin
Fr ,
sin j
cos
i
cos
sin
k
sin j
0
0
sin
2
cos k
and, from Example 10 in Section 16.6,
y
x
FIGURE 11
r
sin 2
r
cos i
sin 2
sin j
sin
cos
sin 3
sin 2
cos
k
Therefore
Fr
,
r
r
cos
sin 2
sin 2
cos
cos
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1142
CHAPTER 16
V E CTOR CALC ULUS
and, by Formula 9, the flux is
yy F
S
yy F
r
y y
2
0
2 sin 2
0
2 y sin2
0
y
0
r
0
cos
y
cos
2
0
y
sin 3
sin 3
cos
2
0
sin 2
y
cos
0
y
sin3
2
since y cos
2
sin 2
sin2
0
0
0
4
3
by the same calculation as in Example 1.
If, for instance, the vector field in Example 4 is a velocity field describing the flow of a
fluid with density 1, then the answer, 4 3, represents the rate of flow through the unit
sphere in units of mass per unit time.
In the case of a surface given by a graph z t , , we can think of and as parameters and use Equation 3 to write
F
r
r
i
j
t
k
t
i
j
k
Thus Formula 9 becomes
yy F
10
t
S
t
This formula assumes the upward orientation of ; for a downward orientation we multiply by 1. Similar formulas can be worked out if is given by
, z or
,z.
(See Exercises 37 and 38.)
i
v EXAMPLE 5 Evaluate xx F S, where F , , z
boundary of the solid region enclosed by the paraboloid z
plane z 0.
z
x
FIGURE 12
z k and
2
2
is the
and the
consists of a parabolic top surface 1 and a circular bottom surface 2. (See
SOLUTION
Figure 12.) Since is a closed surface, we use the convention of positive (outward)
orientation. This means that 1 is oriented upward and we can use Equation 10 with
2
being the projection of 1 onto the -plane, namely, the disk 2
1. Since
S¡
S™
j
1
, ,z
y
on
1
and
, ,z
t
, ,z
2
t
z
1
2
2
2
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SECTION 16.7
SUR FACE I NTEGR ALS
1143
we have
yy F
t
S
t
1
yy
2
yy
1
y y
1
y y
1
2
0
0
2
0
y
2
2
4
2
1
4
2
1
3
cos
2
2
4 2 cos sin
1
4 3 cos
0
0
The disk
2
2
sin
1
4
sin
2
0
is oriented downward, so its unit normal vector is n
yy F
yy F
S
2
yy
k
2
S
yy F
yy F
S
1
S
2
k and we have
yy 0
z
since z 0 on 2 . Finally, we compute, by definition, xx F
face integrals of F over the pieces 1 and 2 :
yy F
2
2
0
S as the sum of the sur-
0
2
Although we motivated the surface integral of a vector field using the example of fluid
flow, this concept also arises in other physical situations. For instance, if E is an electric
field (see Example 5 in Section 16.1), then the surface integral
yy E
S
is called the electric flux of E through the surface . One of the important laws of electrostatics is Gauss’s Law, which says that the net charge enclosed by a closed surface is
11
0
yy E
S
where 0 is a constant (called the permittivity of free space) that depends on the units used.
(In the SI system, 0 8.8542 10 12 C 2 N m2.) Therefore, if the vector field F in
Example 4 represents an electric field, we can conclude that the charge enclosed by is
4
0.
3
Another application of surface integrals occurs in the study of heat flow. Suppose the
temperature at a point , , z in a body is
, , z . Then the heat flow is defined as the
vector field
F
∇
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1144
CHAPTER 16
V E CTOR CALC ULUS
where
is an experimentally determined constant called the conductivity of the substance. The rate of heat flow across the surface in the body is then given by the surface
integral
yy F
yy ∇
S
S
v EXAMPLE 6 The temperature in a metal ball is proportional to the square of the
distance from the center of the ball. Find the rate of heat flow across a sphere of
radius with center at the center of the ball.
SOLUTION Taking the center of the ball to be at the origin, we have
2
, ,z
where
2
z2
is the proportionality constant. Then the heat flow is
F , ,z
2 i
2 j
2z k
where is the conductivity of the metal. Instead of using the usual parametrization of
the sphere as in Example 4, we observe that the outward unit normal to the sphere
2
2
2
z2
at the point , , z is
1
n
and so
2
yy F
2
S
z2
2
2
, so F n
yy F
n
2
16.7
j
2
F n
But on we have
flow across is
i
zk
2
2
2
2
z2
. Therefore the rate of heat
yy
4
2
8
3
Exercises
1. Let
be the boundary surface of the box enclosed by the
planes
0,
2,
0,
4, z 0, and z 6. Approxz
imate xx 0.1
by using a Riemann sum as in Definition 1, taking the patches to be the rectangles that are the
faces of the box and the points * to be the centers of the
rectangles.
2
consists of the cylinder 2
1, 1 z 1,
together with its top and bottom disks. Suppose you know that
is a continuous function with
2. A surface
1, 0, 0
2
0,
1, 0
3
0, 0,
1
4
Estimate the value of xx
, ,z
by using a Riemann sum,
taking the patches to be four quarter-cylinders and the top
and bottom disks.
CAS Computer algebra system required
2
z 2 50, z 0, and
be the hemisphere 2
suppose is a continuous function with 3, 4, 5
7,
3, 4, 5
8,
3, 4, 5
9, and
3, 4, 5
12.
By dividing into four patches, estimate the value of
, ,z
.
xx
3. Let
2
ts 2
, ,z
function of one variable such that t 2
xx , , z , where is the sphere
4. Suppose that
z 2 , where t is a
5. Evaluate
2
2
z 2 4.
5–20 Evaluate the surface integral.
5.
xx
z
,
is the parallelogram with parametric equations
1 2
2, 0 v
v, z
v, 0
v,
1
1. Homework Hints available at stewartcalculus.com
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.7
6.
7.
xx
,
8.
9.
10.
11.
12.
xx
2
14.
15.
16.
17.
18.
23. F , , z
1
2
xx
z ,
is the part of the plane 2
octant
v2
1
,
is the triangular region with vertices 1, 0, 0 , 0,
and 0, 0, 4
3 2
3 2
xx
z ,
is the surface
,0
2
z
20.
xx
z 2, 0
1,
1
2
j
is the surface z
z k,
1, with upward orientation
,
1,
1
i
j 5 k,
is the boundary of the region
enclosed by the cylinder 2 z 2 1 and the planes
0
and
2
2
2
z that lies inside the
2
i
z
j
s1
z 2 k,
2
,0
is the boundary of the solid
2
32. F , , z
z
i
j
k,
is the surface of the tetrahedron with vertices 0, 0, 0 ,
1, 0, 0 , 0, 1, 0 , and 0, 0, 1
2
z 2 4 that lies
1 and above the -plane
CAS
2
2
half-cylinder 0
2
z
2
4, z
0
,
z
is the part of the cylinder 2 z 2 1 that lies between the
planes
0 and
3 in the first octant
2
,
z
2
is the part of the cylinder 2
9 between the planes
z 0 and z 2, together with its top and bottom disks
S for the given vector
21–32 Evaluate the surface integral xx F
field F and the oriented surface . In other words, find the flux of F
across . For closed surfaces, use the positive (outward) orientation.
z
i 3z j
k,
is the parallelogram of Exercise 5 with upward orientation
33. Evaluate xx
where
CAS
2
2
z2
correct to four decimal places,
,0
1, 0
1.
is the surface z
34. Find the exact value of xx
z
,0
1, 0
CAS
35. Find the value of xx
CAS
36. Find the flux of
2
21. F , , z
1, 0
31. F , , z
z ,
is the boundary of the region enclosed by the cylinder
2
z 2 9 and the planes
0 and
5
2
4
2
i 2 j 3z k,
is the cube with vertices 1,
1
2
2
i
0
1
xx
xx
j z k,
consists of the paraboloid
and the disk 2 z 2 1,
2
2
0, oriented in the
27. F , , z
that lies between the
1, 0
,
is the part of the paraboloid
cylinder 2 z 2 4
2
z
z ,
is the hemisphere
25,
30. F , , z
2z 2, 0
,
is the part of the sphere
2
inside the cylinder 2
4 in the first octant,
29. F , , z
2
xx
19.
zi
j
k,
2
z2
is the hemisphere 2
direction of the positive -axis
2, 0 ,
1, 0
z2
26. F , , z
28. F , , z
2
3
2 2
z
,
is the part of the cone z 2
planes z 1 and z 3
xx
i zj
k,
2
is the part of the sphere 2
with orientation toward the origin
,
xx
xx
25. F , , z
4 that lies in the first
z
xx
xx
is the part of the
z k,
2
that lies above the square
1, and has upward orientation
2
i
j z 3 k,
2 between the planes
is the part of the cone z s 2
z 1 and z 3 with downward orientation
3 that lies above the
2
zj
24. F , , z
2
z ,
is the part of the plane z
rectangle 0, 3
0, 2
i
4
paraboloid z
0
1, 0
2
is the surface z
13.
2
,
is the surface with vector equation
r ,v
2 v, 2 v 2, 2 v 2 , 2
xx
j
k,
zi
is the helicoid of Exercise 7 with upward orientation
cos v,
is the helicoid with vector equation
cos v, sin v, v , 0
1, 0 v
r ,v
1145
22. F , , z
z ,
is the cone with parametric equations
sin v, z
,0
1, 0 v
xx
SUR FACE I NTEGR ALS
2
2
z
1.
, where
is the surface
2 2
z
correct to four decimal places,
2
where is the part of the paraboloid z 3 2 2
that
lies above the -plane.
F , ,z
sin
z i
2
j
z2
5
k
across the part of the cylinder 4 2 z 2 4 that lies above
the -plane and between the planes
2 and
2 with
upward orientation. Illustrate by using a computer algebra system to draw the cylinder and the vector field on the same
screen.
37. Find a formula for
xx
where is given by
points toward the left.
F
S similar to Formula 10 for the case
, z and n is the unit normal that
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1146
CHAPTER 16
V E CTOR CALC ULUS
38. Find a formula for xx F
S similar to Formula 10 for the case
where is given by
, z and n is the unit normal that
points forward (that is, toward the viewer when the axes are
drawn in the usual way).
39. Find the center of mass of the hemisphere
2
2
z2
2
,
44. Seawater has density 1025 kg m3 and flows in a velocity field
v
i
j, where , , and z are measured in meters and the
components of v in meters per second. Find the rate of flow
2
outward through the hemisphere 2
z 2 9, z 0.
45. Use Gauss’s Law to find the charge contained in the solid
0, if it has constant density.
z
hemisphere
40. Find the mass of a thin funnel in the shape of a cone
s 2
, ,z
z
,1 z
10 z.
2
2
2
z2
2
,z
0, if the electric field is
i
j
E , ,z
4, if its density function is
2z k
46. Use Gauss’s Law to find the charge enclosed by the cube
with vertices
41. (a) Give an integral expression for the moment of inertia
z
1,
1,
1 if the electric field is
E , ,z
i
j
zk
about the z-axis of a thin sheet in the shape of a surface if
the density function is .
(b) Find the moment of inertia about the z-axis of the funnel in
Exercise 40.
47. The temperature at the point
2
be the part of the sphere 2
z 2 25 that lies
above the plane z 4. If has constant density , find
(a) the center of mass and (b) the moment of inertia about
the z-axis.
48. The temperature at a point in a ball with conductivity
, , z in a substance with conductivity
6.5 is
, ,z
2 2 2 z 2. Find the rate of
heat flow inward across the cylindrical surface 2 z 2 6,
0
4.
42. Let
is
inversely proportional to the distance from the center of the
ball. Find the rate of heat flow across a sphere of radius
with center at the center of the ball.
43. A fluid has density 870 kg m3 and flows with velocity
2
2
v zi
j
k , where , , and z are measured in
meters and the components of v in meters per second. Find the
2
rate of flow outward through the cylinder 2
4,
0 z 1.
16.8
Stokes’ Theorem
z
n
n
0
x
FIGURE 1
r r 3 for
some constant , where r
i
j z k. Show that the flux
of F across a sphere with center the origin is independent of
the radius of .
49. Let F be an inverse square field, that is, F r
S
C
y
Stokes’ Theorem can be regarded as a higher-dimensional version of Green’s Theorem.
Whereas Green’s Theorem relates a double integral over a plane region to a line integral
around its plane boundary curve, Stokes’ Theorem relates a surface integral over a surface
to a line integral around the boundary curve of (which is a space curve). Figure 1 shows
an oriented surface with unit normal vector n. The orientation of induces the positive
orientation of the boundary curve shown in the figure. This means that if you walk in
the positive direction around with your head pointing in the direction of n, then the surface will always be on your left.
Stokes’ Theorem Let be an oriented piecewise-smooth surface that is bounded
by a simple, closed, piecewise-smooth boundary curve with positive orientation.
Let F be a vector field whose components have continuous partial derivatives on
an open region in 3 that contains . Then
y
F
r
yy curl F
S
Since
y
F
r
y
F T
and
yy curl F
S
yy curl F
n
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.8
George Stokes
Stokes’ Theorem is named after the Irish mathematical physicist Sir George Stokes (1819–1903).
Stokes was a professor at Cambridge University
(in fact he held the same position as Newton,
Lucasian Professor of Mathematics) and was
especially noted for his studies of fluid flow
and light. What we call Stokes’ Theorem was
actually discovered by the Scottish physicist
Sir William Thomson (1824–1907, known as
Lord Kelvin). Stokes learned of this theorem
in a letter from Thomson in 1850 and asked
students to prove it on an examination at
Cambridge University in 1854. We don’t know
if any of those students was able to do so.
S T OKES ’ THEOREM
1147
Stokes’ Theorem says that the line integral around the boundary curve of of the tangential component of F is equal to the surface integral over of the normal component of the
curl of F.
The positively oriented boundary curve of the oriented surface is often written as
, so Stokes’ Theorem can be expressed as
yy curl F
1
y
S
F
r
There is an analogy among Stokes’ Theorem, Green’s Theorem, and the Fundamental
Theorem of Calculus. As before, there is an integral involving derivatives on the left side
of Equation 1 (recall that curl F is a sort of derivative of F ) and the right side involves the
values of F only on the
of .
In fact, in the special case where the surface is flat and lies in the -plane with
upward orientation, the unit normal is k, the surface integral becomes a double integral,
and Stokes’ Theorem becomes
y
F
r
yy curl F
S
yy
curl F
k
This is precisely the vector form of Green’s Theorem given in Equation 16.5.12. Thus we
see that Green’s Theorem is really a special case of Stokes’ Theorem.
Although Stokes’ Theorem is too difficult for us to prove in its full generality, we can
give a proof when is a graph and F, , and are well behaved.
PROOF OF A SPECIAL CASE OF STOKES’ THEOREM We assume that the equation of is
, where t has continuous second-order partial derivatives and
z t , , ,
is a simple plane region whose boundary curve 1 corresponds to . If the orientation of
is upward, then the positive orientation of corresponds to the positive orientation of
i
j
k, where the partial deriva1. (See Figure 2.) We are also given that F
tives of , , and are continuous.
Since is a graph of a function, we can apply Formula 16.7.10 with F replaced by
curl F. The result is
z
n
z=g(x, y)
S
0
x
C
D
C¡
y
2
yy curl F
S
FIGURE 2
z
z
z
z
where the partial derivatives of , , and
is a parametric representation of
are evaluated at
, ,t ,
, then a parametric representation of
1
z
t
. If
is
,
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1148
CHAPTER 16
V E CTOR CALC ULUS
z
y
z
z
z
z
z
z
z
z
z
z
y
z
z
z
z
z
y
EXAMPLE 1
C
z
z
z
z
z
yy
x
z
z
z
z
S
z
z
z
v
z
x
SOLUTION
y+z=2
z
D 0
x
z
y
z
FIGURE 3
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.8
z
y
yy
S T OKES ’ THEOREM
1149
t
yy
y y
y
z
≈+¥+z@ =4
S
C
v
xx
EXAMPLE 2
z
z
z
SOLUTION
z
z
0
x
z
s
≈+¥=1
y
z
z
s
z
s
FIGURE 4
s
yy
s
y
y
y
s
3
yy
s
s
y
y
y
yy
y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1150
CHAPTER 16
V E CTOR CALC ULUS
x
jC
FIGURE 5
jC
>0
<0
z
y
yy
yy
yy
Imagine a tiny paddle wheel placed in the
fluid at a point , as in Figure 6; the paddle
wheel rotates fastest when its axis is parallel
to
.
l
4
y
l
x
FIGURE 6
y
yy
yy
x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.8
16.8
S T OKES ’ THEOREM
Exercises
1.
10.
yy
yy
z
z
z
x
11.
z
z
z
4
4
z
z
z
;
P
H
;
x
12.
x
2
2
x
y
2
2
xx
2–6
2.
1151
z
z
z
z
z
y
z
;
z
;
3.
z
z
z
z
13–15
z
4.
5.
z
s
z
z
13.
z
z
z
14.
z
z
z
z
z
z
z
z
15.
z
z
z
6.
z
z
z
z
z
16.
z
7.
z
8.
z
9.
x
x
7–10
z
z
z
z
z
17.
sz
z
z
z
z
z
z
z
;
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1152
CHAPTER 16
VE CTOR CALCULUS
18.
20.
x
z
x
z
19.
xx
t
t
x
xx
WRITING PROJECT
t
z
x
t
t
THREE MEN AND TWO THEOREMS
The photograph shows a stained-glass
window at Cambridge University in honor of
George Green.
Courtesy of the Masters and Fellows of Gonville and
Caius College, Cambridge University, England
1.
2.
3.
4.
5.
6.
7.
8.
16.9
The Divergence Theorem
y
yy
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.9
yy
1
The Divergence Theorem is sometimes called
Gauss’s Theorem after the great German mathematician Karl Friedrich Gauss (1777–1855), who
discovered this theorem during his investigation
of electrostatics. In Eastern Europe the Divergence Theorem is known as Ostrogradsky’s
Theorem after the Russian mathematician
Mikhail Ostrogradsky (1801–1862), who published this result in 1826.
yyy
THE DIVERGENCE THEOR EM
1153
z
The Divergence Theorem
yy
yyy
PROOF
z
yyy
yy
yyy
yy
yy
yyy
yyy
z
yy
yy
yy
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1154
CHAPTER 16
V E CTOR CALC ULUS
2
yy
yyy
3
yy
yyy
4
yy
yyy
z
z
yyy
z
S™ {z=u™(x, y)}
S£
yyy
yy y
z
z
z
z
yy
z
E
0
x
5
z
S¡ {z=u¡(x, y)}
D
y
yy
yy
FIGURE 1
yy
6
yy
yy
z
yy
yy
yy
yy
z
yy
yy
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.9
yy
yyy
THE DIVERGENCE THEOR EM
1155
z
Notice that the method of proof of the
Divergence Theorem is very similar to that
of Green’s Theorem.
v
z
EXAMPLE 1
z
z
SOLUTION
z
z
z
yy
The solution in Example 1 should be compared
with the solution in Example 4 in Section 16.7.
v
z
(0, 0, 1)
yyy
yyy
xx
EXAMPLE 2
z
z
y=2-z
z
z
0
SOLUTION
(1, 0, 0)
x
z
(0, 2, 0) y
z=1-≈
z
FIGURE 2
z
z
yy
z
yyy
y y
z
yyy
y
z
z
z
y y
z
z
y
y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1156
CHAPTER 16
V E CTOR CALC ULUS
™
¡
S™
FIGURE 3
_ ¡
S¡
yyy
7
yy
yy
yy
yy
yy
yy
EXAMPLE 3
z
yy
SOLUTION
yy
yy
yy
yy
yyy
yy
yy
yy
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
SECTION 16.9
THE DIVERGENCE THEOR EM
1157
z
z
yy
yyy
yyy
l
y
8
yy
l
P¡
x
P™
FIGURE 4
16.9
=≈ +¥
Exercises
1– 4
1.
7.
z
2.
z
z
z
z
z
z
8.
z
9.
z
z
z
z
z
z
z
10.
3.
z
z
z
z
z
z
z
z
4.
z
z
z
z
11.
5–15
xx
5.
z
z
z
z
z
z
z
z
12.
z
z
z
z
z
z
6.
z
z
z
z
z
13.
z
CAS
z
z
z
z
s
1.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1158
CHAPTER 16
V E CTOR CALC ULUS
14.
CAS
15.
z
z
23.
s
z
24.
yy
z
z
z
CAS
16.
z
z
25–30
z
z
xx
17.
z
z
z
25.
z
z
18.
z
z
z
z
19.
2
yy
26.
27.
yy
29.
yy
t
30.
yy
t
z
z
yy
z
28.
yyy
yy
yyy
t
yyy
t
z
t
t
t
31.
P¡
_2
yy
2
P™
yyy
_2
20.
32.
z
2
z
tz
t
P¡
_2
2
yy
P™
_2
CAS
21–22
z
21.
22.
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SECTION 16.10
1159
SUMMARY
16.10 Summary
y
a
b
(b)
y
C
(a)
y
yy
D
y
S
C
S
yyy
yy
E
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1160
16
CHAPTER 16
V E CTOR CALC ULUS
Review
Concept Check
1.
2.
10.
3.
11.
z
t
12.
z
v
t
z
4.
z
z
13.
5.
x
6.
v
x
t
z
7.
14.
8.
15.
16.
9.
True-False Quiz
Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement.
1.
8.
9.
2.
3.
10.
4.
x
11.
5.
6.
7.
x
x
xx
12.
z
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CHAPTER 16
REVIEW
1161
Exercises
1.
12.
x
z
z
13–14
x
y
13.
C
14.
z
z
z
x
15.
P
x
16.
2–9
2.
x
3.
x
4.
x
5.
x
6.
x
7.
x
8.
x
9.
x
y
s
z
z
x
17.
18.
z
s
z z
z
z
19.
z
z z
z
20.
z
z
z
21.
t
x
10.
22.
z
t
t
t
z
23.
t
t
t
x
z
24.
11–12
z
∇
x
11.
;
z
z z
CAS
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1162
CHAPTER 16
V E CTOR CALC ULUS
25.
37.
z
z
26.
z
z
z
x
v
v
v
v
z
;
(0, 0, 2)
0
CAS
z
(1, 1, 0)
z
x
27–30
28.
xx
xx
38.
z
z
z
z
x
z
z
29.
30.
y
(3, 0, 0)
xx
27.
(0, 3, 0)
z
xx
z
y
z
z
xx
z
z
z
z
x
0
31.
z
z
z
39.
32.
z
z
z
z
z
(0, 2, 2)
x
z
(2, 0, 2)
1
xx
z
1
z
z
1
y
(2, 2, 0)
x
z
35.
z
z
z
z
z
33.
34.
z
xx
z
z
xx
40.
z
xx
36.
41.
z
z
z
z
z
yy
y
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Problems Plus
1.
yy
S
S(a)
a
P
2.
y
3.
y
; 4.
z
z
z
v
v
z
z
z
z
5.
;
1163
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6.
P
$
#
C
%
0
①
②
③
!
@
V
②
③
④
④
⑤
①
x
x
1164
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17
Second-Order
Differential Equations
The motion of a shock absorber in a car
is described by the differential equations
that we solve in Section 17.3.
© Christoff / Shutterstock
1165
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1166
17.1
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
Second-Order Linear Equations
1
2
3
Theorem
PROOF
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SECTION 17.1
1167
t
t
4
S E COND-ORD ER LIN EAR EQUATIONS
Theorem
5
6
7
s
s
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1168
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
CASE I
8
In Figure 1 the graphs of the basic solutions
and t
of the differential
equation in Example 1 are shown in blue and
red, respectively. Some of the other solutions,
linear combinations of and t, are shown
in black.
EXAMPLE 1
SOLUTION
8
5f+g
_1
f
f+5g
f+g
g
g-f
f-g
_5
1
EXAMPLE 2
FIGURE 1
SOLUTION
s
s
s
CASE II
9
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SECTION 17.1
S E COND-ORD ER LIN EAR EQUATIONS
1169
10
v
Figure 2 shows the basic solutions
and t
in
Example 3 and some other members of the
family of solutions. Notice that all of them
approach 0 as l .
EXAMPLE 3
SOLUTION
f-g 8
f
_2
f+g
5f+g
g-f
g
f+5g
2
CASE III
_5
FIGURE 2
s
11
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1170
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
Figure 3 shows the graphs of the solutions
in Example 4,
and
t
, together with some linear
combinations. All solutions approach 0
as l
.
f+g
_3
v
EXAMPLE 4
SOLUTION
s
3
s
g
f-g
f
2
Initial-Value and Boundary-Value Problems
_3
FIGURE 3
EXAMPLE 5
SOLUTION
Figure 4 shows the graph of the solution of the
initial-value problem in Example 5. Compare with
Figure 1.
20
12
13
_2
0
2
FIGURE 4
EXAMPLE 6
SOLUTION
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SECTION 17.1
S E COND-ORD ER LIN EAR EQUATIONS
1171
The solution to Example 6 is graphed in
Figure 5. It appears to be a shifted sine curve
and, indeed, you can verify that another way of
writing the solution is
s
where
5
2π
_2π
_5
FIGURE 5
v
EXAMPLE 7
SOLUTION
Figure 6 shows the graph of the solution of
the boundary-value problem in Example 7.
5
_1
5
_5
FIGURE 6
Summary: Solutions of
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1172
17.1
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
Exercises
1–13
22.
1.
2.
23.
3.
4.
24.
5.
6.
7.
8.
9.
10.
25–32
25.
26.
11.
27.
28.
12.
29.
13.
30.
31.
32.
; 14–16
33.
14.
15.
16.
34.
l
17–24
35.
17.
18.
19.
20.
21.
;
17.2
1.
Nonhomogeneous Linear Equations
1
2
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SECTION 17.2
3
NONHOMOGENEOUS LINEA R EQUATIONS
1173
Theorem
PROOF
The Method of Undetermined Coefficients
v
EXAMPLE 1
SOLUTION
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1174
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
Figure 1 shows four solutions of the differential
equation in Example 1 in terms of the particular
solution and the functions
and t
.
8
yp+2f+3g
yp+3g
_3
yp+2f
3
yp
_5
FIGURE 1
EXAMPLE 2
Figure 2 shows solutions of the differential
equation in Example 2 in terms of and the
functions
and t
.
Notice that all solutions approach as l
and all solutions (except ) resemble sine
functions when is negative.
SOLUTION
4
yp+f+g
yp+g
yp
_4
2
yp+f
_2
FIGURE 2
v
EXAMPLE 3
SOLUTION
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SECTION 17.2
v
NONHOMOGENEOUS LINEA R EQUATIONS
1175
EXAMPLE 4
SOLUTION
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1176
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
In Figure 3 we show the particular solution
of the differential equation in
Example 4. The other solutions are given in
terms of
and t
.
5
yp+2f+g
yp+g
yp+f
_4
1
yp
_2
FIGURE 3
EXAMPLE 5
SOLUTION
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SECTION 17.2
NONHOMOGENEOUS LINEA R EQUATIONS
1177
The graphs of four solutions of the differential
equation in Example 5 are shown in Figure 4.
4
_2π
2π
yp
_4
FIGURE 4
Summary of the Method of Undetermined Coefficients
1.
2.
EXAMPLE 6
SOLUTION
The Method of Variation of Parameters
4
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1178
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
5
6
7
8
9
EXAMPLE 7
SOLUTION
10
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SECTION 17.2
NONHOMOGENEOUS LINEA R EQUATIONS
1179
11
Figure 5 shows four solutions of the
differential equation in Example 7.
2.5
0
π
2
yp
_1
FIGURE 5
17.2
Exercises
1–10
; 11–12
1.
11.
12.
2.
3.
13–18
4.
5.
13.
6.
14.
7.
15.
8.
16.
9.
17.
10.
18.
;
1.
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1180
CHAPTER 17
SE COND- ORDER DIFFE RENTIAL EQ UATIONS
19–22
24.
19.
25.
20.
21.
26.
22.
27.
23–28
28.
23.
17.3
Applications of Second-Order Differential Equations
Vibrating Springs
m
0
m
x
x
FIGURE 1
1
s
m
0
x
x
FIGURE 2
s
s
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SECTION 17.3
v
A PPLICA TIONS OF SECOND-ORDER DIF FERENTIAL EQ UATIO NS
1181
EXAMPLE 1
SOLUTION
2
Damped Vibrations
m
Schwinn Cycling and Fitness
FIGURE 3
3
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1182
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
4
x
s
s
CASE I
0
t
s
x
0
l
l
t
CASE II
FIGURE 4
CASE III
x
x=Ae– (c/2m)t
0
s
t
x=_Ae– (c/2m)t
FIGURE 5
l
l
l
l
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SECTION 17.3
v
A PPLICA TIONS OF SECOND-ORDER DIF FERENTIAL EQ UATIO NS
1183
EXAMPLE 2
SOLUTION
Figure 6 shows the graph of the position function for the overdamped motion in Example 2.
0.03
0
1.5
FIGURE 6
Forced Vibrations
5
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1184
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
s
6
Electric Circuits
R
L
E
C
FIGURE 7
7
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SECTION 17.3
v
A PPLICA TIONS OF SECOND-ORDER DIF FERENTIAL EQ UATIO NS
1185
EXAMPLE 3
SOLUTION
8
s
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1186
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
l
NOTE 1
l
0.2
0
Qp
Q
l
l
1.2
_0.2
FIGURE 8
5
NOTE 2
7
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SECTION 17.3
17.3
A PPLICA TIONS OF SECOND-ORDER DIF FERENTIAL EQ UATIO NS
1187
Exercises
10.
1.
s
2.
11.
3.
12.
4.
;
13.
5.
6.
14.
; 7.
;
; 8.
9.
15.
16.
s
;
17.
;
1.
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1188
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
18.
t
t
¨
17.4
L
Series Solutions
1
v
EXAMPLE 1
SOLUTION
2
3
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SECTION 17.4
By writing out the first few terms of 4 , you can
see that it is the same as 3 . To obtain 4 , we
replaced by
and began the summation
at 0 instead of 2.
SERIES SOLUTIONS
1189
4
5
6
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1190
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
NOTE 1
v
EXAMPLE 2
SOLUTION
7
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SECTION 17.4
SERIES SOLUTIONS
1191
8
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1192
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
NOTE 2
NOTE 3
2
T¸
2
_2
T¡¸
NOTE 4
_8
FIGURE 1
15
fi
_2.5
2.5
›
_15
FIGURE 2
17.4
Exercises
11.
1–11
1.
2.
3.
4.
5.
6.
12.
7.
8.
;
9.
10.
;
1.
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CHAPTER 17
17
REVIEW
1193
Review
Concept Check
1.
2.
4.
5.
3.
True-False Quiz
Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement.
3.
1.
4.
2.
Exercises
1–10
11–14
1.
11.
2.
12.
3.
13.
4.
5.
14.
15–16
15.
6.
7.
8.
16.
17.
18.
9.
19.
10.
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1194
CHAPTER 17
S E COND -ORDE R DIFF ERENT IAL EQ UATIONS
20.
21.
t
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Appendixes
F Proofs of Theorems
G Complex Numbers
H Answers to Odd-Numbered Exercises
A1
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A2
APPENDIX F
F
PROOFS OF THEORE MS
Proofs of Theorems
Section 11.8
Theorem
1.
2.
PROOF OF 1
l
PROOF OF 2
Theorem
1.
2.
3.
PROOF
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APPENDIX F
3
PROO FS OF THEORE MS
A3
Theorem
1.
2.
3.
PROOF
Section 14.3
Clairaut’s Theorem
PROOF
t
t
t
l
t
t
t
l
l
l
l
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A4
APPENDIX F
PROOFS OF THEORE MS
Section 14.4
Theorem
8
PROOF
z
z
z
l
y
l
(a+Îx, b+Îy)
(a, b+Îy)
(a, √)
(a, b)
0
(u, b+Îy)
z
1
R
t
x
t
t
FIGURE 1
t
t
t
v
v
v
v
z
v
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APPENDIX G
COMP LEX N UMBER S
A5
v
l
v l
l
l
l
l
Complex Numbers
G
_4+2i
_2-2i
2+3i
i
0
_i
1
3-2i
FIGURE 1
EXAMPLE 1
z
z
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A6
APPENDIX G
COMPLEX NUMBERS
EXAMPLE 2
SOLUTION
z
z=a+bi
z
i
0
Properties of Conjugates
_i
z
z=a-bi
w
z
w
zw
FIGURE 2
zw
z
z
z
z
z
bi
z=a+bi
s
z
b
0
zz
a
FIGURE 3
zz
z
z
zw
zw
w
ww
w
s
s
s
s
EXAMPLE 3
SOLUTION
s
s
s
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX G
COMP LEX N UMBER S
z
z
z
A7
z
Polar Form
z
a+bi
r
¨
0
b
a
FIGURE 4
z
z
z
s
z
z
z
z
z
EXAMPLE 4
w
z
SOLUTION
z
π
4
s
w
s
s
w
π
_
6
2
FIGURE 5
s
z
1+i
œ2
0
s
s
w
œ„
3-i
z
w
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A8
APPENDIX G
COMPLEX NUMBERS
z
z™
z
zz
z¡
¨™
¨¡
¨¡+¨™
zz
1
z¡z™
FIGURE 6
z
z
z
r
¨
0
z
z
_¨
1
r
z
z
1
z
FIGURE 7
z
z
s
EXAMPLE 5
SOLUTION
s
s
z=1+i
œ2
zw
2 œ„2
s
π
12
0
2
w=œ
s
s
FIGURE 8
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APPENDIX G
COMP LEX N UMBER S
A9
z
z
z
2
De Moivre’s Theorem
zz
z
z
EXAMPLE 6
SOLUTION
s
s
w
z
w
w
z
z
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A10
APPENDIX G
CO MPLEX NUM BERS
w
w
3
z
Roots of a Complex Number
z
w
w
z
z
z
z
EXAMPLE 7
z
SOLUTION
w
œ„
2 i w¡
w™
_ œ2
w¸
0
FIGURE 9
s
w
s
s
w
s
w
s
s
s
œ2
w£
w∞
_œ„2 i
w
w¢
z=_8
w
s
w
s
s
s
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APPENDIX G
CO MPLEX NUM BERS
A11
Complex Exponentials
z
z
4
z
z
5
z
z
z
z
z
z
z
z
6
7
EXAMPLE 8
SOLUTION
We could write the result of Example 8(a) as
This equation relates the five most famous numbers in all of mathematics:
and .
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A12
APPENDIX G
CO MPLEX NUM BERS
Exercises
G
1–14
33–36
33.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13. s
14. s
35.
s
36.
37– 40
37.
38.
39.
40.
41– 46
s
15–17
15.
s
34.
41.
42.
43.
44.
45.
46.
s
16.
17.
47.
18.
w
z
z
w
z
zw
48.
zw
z
z
w
19–24
19.
20.
21.
22.
23. z
24. z
z
t
49.
z
t
t
25–28
25.
26.
27.
28.
29–32
29. z
zw z w
s
50.
s
z
z
x
y
w
s
w
30. z
s
w
31. z
s
w
32. z
s
w
y
y
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APPENDIX H
ANSWERS TO ODD-NUMBERED EXERCISES
A13
Answers to Odd-Numbered Exercises
H
CHAPTER 10
13.
EXERCISES 10.1
1.
N
PAGE 665
3.
t=_2
(2, 6)
2
y
t=0
(1, 1)
1
(1, 1)
x
t=2
(6, 2)
π
t= 6
t=0
(0, 0)
π
t= 3
π
t= 2
(0, 0)
2
15.
y
1
x
y
5.
(7, 5)
t=_1
(3, 2)
t=0
0
(_1, _1)
t=1
17.
y
x
1
(_5, _4)
t=2
x
0
7.
y
(_3, 0)
t=2
19.
(0, _1)
t=1
x
21.
23.
(1, _2)
t=0
25.
(0, _3)
t=_1
(_3, _4)
t=_2
27.
1 t=
(0, 1) t=1
9.
(_1, 0)
t=0
(0, 1) t=0
t=0
(0, _1) t=_1
1
(1, 0) t=1
π
29.
(2, _3) t=4
_4
4
11.
1
_π
31.
33.
_1
1
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A14
APPENDIX H
ANSWERS TO ODD-NUMBERED EXERCISES
19.
37.
21.
23.
41.
25.
43.
45.
27.
31.
37.
39.
29.
x
x
33.
35.
s
s
41.
43. s
s
s
45. s
47.
_1
47.
_3
1.4
_1
_2.1
49.
51.
s
_1.4
49.
EXERCISES 10.2
N
PAGE 675
1.
51.
s
s
55.
3.
7.
9.
5.
20
_10
10
_2
11.
2.1
13.
57.
59.
15.
61.
17.
65.
x
x
s
s
s
s
63.
71.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
EXERCISES 10.3
11.
PAGE 686
N
1.
¨=
π
”2, 3 ’
7π
3
r=3
r=2
π
3
O
A15
ANSWERS TO ODD-NUMBERED EXERCISES
O
_ 3π
”1, _ 3π ’
4
4
¨=
13.
17.
19.
21.
25.
29.
π
2
O
π
”_1, 2 ’
s
5π
3
s
15.
23.
27.
31.
O
(4, 0)
O
3.
(2, 3π/2)
π
(1, π)
33.
O
O
_
35.
2π
3
¨=
5
(2π, 2π)
”2, _ 2π ’
3
2
π
3
π
”4, 6 ’
1
6
s
3
3π
4
37.
O
¨=
3π
5π
6
¨=
π
6
O
s
s
5.
¨=
(2, 0)
”_2, 4 ’
s
39.
π
8
4
s
41.
43.
(3, π/6)
(3, π/4)
7.
r=1
O
O
O
45.
9.
¨=
(3, π)
2π
3
π
¨= 3
47.
(3, 0)
O
¨=
3π
4
π
¨=
4
O
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A16
APPENDIX H
ANSWERS TO ODD-NUMBERED EXERCISES
49
51.
15.
x=1
(2, 0)
1.4
(6, 0)
_2.1
2.1
_1.4
53.
17.
55. s
57.
25.
59.
s
61.
19.
s
33.
s
s
21.
27.
s
s
s
35.
s
23.
29.
31.
37.
39.
63.
s
65.
67.
2.6
69.
41.
43.
45.
3.5
s
s
47.
1
49.
_3.4
1.8
_3
3
_0.75
1.25
_2.5
_2.6
71.
_1
51.
55.
53.
s
EXERCISES 10.5
N
PAGE 700
1.
73.
3.
y
y
2
6
75.
”0, 32’
EXERCISES 10.4
1.
9.
N
5.
11.
y=-32
7.
(3, π/2)
(1, π)
r=2
x
_2
6 x
(2, π/2)
O
¨
(5, 0)
5.
7.
y
(3, 3π/2)
O
13.
(_1/2, 0)
PAGE 692
3.
x=21
(_2, 5)
3
y=1
_4
(_5, _1)
0
x
(_2, _1)
4
x=1
_3
9.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
s
11.
s
13.
y
2
_2œ„2
_3
0
2
œ„
2œ„2
0
x
33.
A17
35.
37.
y
1
2
œ„
2
_œ„
31.
ANSWERS TO ODD-NUMBERED EXERCISES
39.
41.
43.
45.
47.
3 x
_1
2
_œ„
49.
_2
s
15.
51.
s
17.
55.
59.
(1, 3)
61.
63.
EXERCISES 10.6
(1,_3)
s
19.
{0, œ„„
34}
y
y=53 x
(3, 5)
(0, 5)
3.
5.
7.
9.
y
(0, _5)
” 45 , π’
y
(_10, 0)
y=x
(10, 0)
x
y=_1
11.
{10œ„
2, 0}
x
y
y=_x
23.
” 45 , 0’
O
4 3π
”9, 2 ’
(10, 10)
{_10œ„
2, 0}
(4, π/2)
y=_ 53x
{0, _ œ„„
34}
s
”31 , π
2’
” 23, π’
O
y
s
PAGE 708
1.
x
21.
N
” 23, 0’
x
x
0
(2, _2)
(4, _2)
(3-œ„5, _2)
13.
(3+œ„5, _2)
9
3 π
” 2, 2 ’
9
” 4 , π’
25.
27.
29.
y=2/3
x= 2
9
O
” 8 , 0’
3 3π
s
” 2, 2 ’
s
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A18
APPENDIX H
ANSWERS TO ODD-NUMBERED EXERCISES
15.
7.
”4, 2π ’
3
s
2π
3
3
”- 4 , 0’
1
” 4 , π’
s
O
O
s
3
x=_ 8
9.
17.
11.
1
_2
π
6
(1, 0)
(2, π)
2
O
1
y=-2
3π
”1, 2 ’
_3
13.
15.
(2, π)
_2
=
l
3π
’
2
(2, 0)
O
3
y= 2
_1
2
”1, π2 ’
17.
_2
19.
”_3,
1
2
O
19.
r=
=
¨
¨
-0.3
=
=
-0.75
25.
21.
27.
29.
CHAPTER 10 REVIEW
31.
N
True-False Quiz
1.
3.
23.
25.
PAGE 709
27.
29.
5.
7.
s
9.
s
Exercises
1.
(
)
(
)
3.
y
(0, 6), t=_4
(5, 1),
t=1
31.
(1, 1), ¨=0
37.
x
5.
¨=
π
”1, 2 ’
s
39.
33.
35.
s
s
s
s
s
41.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
43.
45.
67.
69.
71.
73.
77.
79.
85.
47.
œ„
(1, 0)
ANSWERS TO ODD-NUMBERED EXERCISES
A19
75.
s
81.
s
83.
(_1, 3)
EXERCISES 11.2
N
PAGE 735
1.
œ„
49.
51.
53.
3.
5.
7.
55.
57.
s
s
s
9.
s
san d
ssn d
_3
11.
PROBLEMS PLUS
1.
N
ssn d
PAGE 712
s
3.
s
san d
11
0
CHAPTER 11
EXERCISES 11.1
10
13.
N
1
PAGE 724
ssn d
1.
3.
5.
7.
19.
15.
25.
37.
49.
21.
51.
9.
11.
15.
23.
35.
47.
57.
65.
13.
17.
25.
37.
49.
59.
27.
39.
29.
41.
51.
61.
31.
43.
53.
63.
33.
45.
55.
san d
0
17.
27.
39.
53.
57.
19.
21.
31.
33.
43.
23.
35.
45.
47.
55.
59.
61.
65.
29.
41.
11
63.
67.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A20
APPENDIX H
ANSWERS TO ODD-NUMBERED EXERCISES
35.
39.
69.
71.
s
73.
77.
45.
79.
85.
EXERCISES 11.7
87.
1.
13.
25.
33.
89.
EXERCISES 11.3
1.
PAGE 744
N
a™
0
1
a¢
7.
17.
29.
37.
N
9.
19.
31.
11.
21.
23.
PAGE 769
3
9.
11.
21.
23.
35.
2
15.
27.
5.
17.
29.
7.
19.
31.
9.
21.
35.
_2
PAGE 755
7.
l
7.
9.
EXERCISES 11.6
N
3.
N
PAGE 775
5.
9.
11.
11.
23.
21.
29.
41.
EXERCISES 11.9
1.
s¡ s£ s∞
37.
1.
5.
19.
J¡
8
PAGE 750
13.
25.
s¸ s™ s¢
_8
3.
N
29.
33.
33.
45.
N
23.
27.
31.
25.
17.
21.
25.
13.
11.
15.
19.
x
4
31.
5.
9.
13.
a∞
19.
EXERCISES 11.5
1.
3.
15.
27.
5.
3.
7.
7.
EXERCISES 11.4
3.
17.
27.
33.
3.
15.
27.
35.
EXERCISES 11.8
1
x 1.3
a£
2
5.
17.
39.
1.
11.
23.
33.
45.
PAGE 764
1.
y
y=
3.
15.
27.
29.
35.
37.
N
13.
25.
15.
13.
31.
35.
PAGE 761
15.
5.
17.
29.
7.
19.
31.
9.
11.
21.
33.
13.
23.
25.
17.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
19.
ANSWERS TO ODD-NUMBERED EXERCISES
A21
29.
21.
31.
s£
0.25
s¡
f
s™
_4
s™
s¢
f
s∞
33.
s∞
s¢
35.
4
37.
s¡
_0.25
s£
39.
s£
23.
s™
s¡
f
T¸=T¡=T™=T£
_1.5
1.5
Tˆ=T˜=T¡¸=T¡¡
f
27.
29.
35.
EXERCISES 11.10
1.
41.
31.
39.
N
33.
6
T£
f
_3
3.
T¡
T™
T™
T£
7.
13.
T∞
T¡
PAGE 789
5.
9.
T¢=T∞=Tß=T¶
_1.5
25.
4
T¢ Tß
f
_6
T¢ T∞ Tß
11.
43.
45.
15.
17.
47.
19.
49.
25.
27.
51.
53.
55.
59.
65.
61.
67.
s
57.
63.
69.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A22
APPENDIX H
EXERCISES 11.11
N
ANSWERS TO ODD-NUMBERED EXERCISES
PAGE 798
9.
3
1.
_1
1.5
T¢=T∞
T¸=T¡
f
_2π
T£ f
2π
_4
11.
_2
Tß
T™=T£
T¢ T∞
T£
T™
5
f
T¢
π
2
0
π
4
T∞
T™
T£
2
f
_2
3.
13.
2
f
T£
4
0
15.
17.
19.
23.
29.
37.
21.
25.
27.
31.
CHAPTER 11 REVIEW
5.
1.1
T£
True-False Quiz
1.
3.
11.
13.
19.
21.
f
0
N
π
π
2
f
_1.1
5.
15.
7.
9.
17.
Exercises
1.
15.
T£
PAGE 802
27.
3.
5.
17.
7.
19.
29.
9.
21.
31.
11.
23.
13.
25.
35.
37.
7.
41.
2
T£
f
_1
s
45.
3
47.
_4
43.
49.
51.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
25.
53.
27.
29.
55.
z
z
z
z
57.
s
31.
33.
35.
39.
37.
z
L¡
C
T£
z
P
f
L™
59.
A
PROBLEMS PLUS
N
B
PAGE 805
1.
3.
5.
s
41.
43.
z
s
EXERCISES 12.2
9.
11.
19.
A23
ANSWERS TO ODD-NUMBERED EXERCISES
1.
13.
l
3.
PAGE 822
l l
5.
s
N
l l
l l
l
+
+
21.
+
CHAPTER 12
EXERCISES 12.1
1.
N
_
PAGE 814
3.
5.
z
y=2-x
z
y=2-x, z=0
0
2
2
x
9.
11.
y
0
x
s
x
15.
y
A (0, 3, 1)
z
z
B(2, 2)
z
17.
s
A(_1, 3)
B(3, 2)
13.
z
s
y
y
0
z
z
z
23.
7.
A(_1, 1)
z
_
- -
+ +
s
7.
9.
11.
13.
15.
19.
21.
_
-
k6, _2l
k_1, 4l
B (2, 3, _1)
0
k5, 2l
x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A24
APPENDIX H
ANSWERS TO ODD-NUMBERED EXERCISES
17.
s
15.
z
19.
k3, 0, 1l
k3, 8, 1l
x
33.
29.
33.
35. s
s
s
s
35.
s
s
s
39.
41.
s
s
s
s
43.
s
45.
25.
s
s
29.
31.
y
k0, 8, 0l
s
s
27.
19.
21.
23.
17.
53.
27.
EXERCISES 12.5
31.
N
PAGE 848
1.
37.
39.
3.
z
s
41.
45.
5.
43.
y
z
s
7.
z
z
0
x
9.
11.
t
47.
1.
9.
3.
19.
17.
27.
29.
33.
31.
47.
49.
53.
s
s
s
s
17.
19.
25.
29.
33.
37.
21.
23.
27.
z
31.
z
z
z
z
35.
z
z
39.
z
z
43.
z
z
3
2
(0, 0, 10)
s
”0, 0, ’
(0, _2, 0)
s
s
s
s
0
39.
43.
s
0
(1, 0, 0)
(0, 2, 0)
y
x
y
(5, 0, 0)
s
s
x
51.
N
45.
51.
55.
57.
s
55.
EXERCISES 12.4
1.
7.
13.
7.
z
41.
25.
41.
5.
21.
s
23.
37.
15.
PAGE 830
s
z
13.
11.
15.
35.
N
z
z
z
EXERCISES 12.3
z
PAGE 838
.
5.
9.
11.
59.
63.
65.
67.
47.
49.
53.
z
z
z
z
61.
z
s
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
69. s
79.
s
71.
EXERCISES 12.6
N
s
73.
15.
s
77.
A25
ANSWERS TO ODD-NUMBERED EXERCISES
z
PAGE 856
y
1.
x
z
3.
5.
z
z
17.
z
(0, 0, 1)
(0, 6, 0)
y
x
y
x
x
y
(1, 0, 0)
19.
z
7.
y
y
x
21.
23.
27.
z
z
29.
y
25.
x
9.
z
y
z
x
z
11.
z
31.
z
z
y
x
y
x
13.
z
33.
z
z
(0, 0, 3)
(0, 4, 3)
y
x
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A26
APPENDIX H
ANSWERS TO ODD-NUMBERED EXERCISES
35.
(2,-1,1) z
z
29.
31.
z
z
y
x
0
37.
39.
4
_4
_4
y
0
z
41.
0
44
x
33.
35.
z
_2
_2
_4
x
2
z 0
z 0
y
y
x
y
0
2 2
0x
z
(0, 1, 2)
_2
z=2
y
(0, 2, 0)
x
(1, 1, 0)
(0, 2, 0)
x
y
(0, 1,-2)
37.
z=œ≈+¥
z
PROBLEMS PLUS
43.
45.
z
PAGE 861
1. s
3.
z
z
47.
N
z
z
5.
51.
CHAPTER 13
z
EXERCISES 13.1
1.
N
PAGE 869
3.
7.
CHAPTER 12 REVIEW
N
5.
9.
y
z
π
PAGE 858
(0, 2, 0)
True-False Quiz
1.
3.
11.
13.
19.
21.
5.
15.
7.
9.
x
1
17.
x
Exercises
1.
z
z
3.
5.
9.
13.
15.
17.
19.
25.
s
11.
s
7.
s
11.
z
x
z
21.
z
z
27.
13.
z
s
23.
z
y=≈
1
y
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
y
15.
35.
1
_2π
2π
_1
x
2π x
_2π
z
z
2
(0, 0, 2)
1.
N
R
C
(4.5)- (4)
(4.5)
1
(4)
19.
(4.5)- (4)
0.5
z
25.
29.
0
y
(4.2)- (4)
0.2
R
C
(4.5)
1
y
x
1
z
z
(4.2)- (4)
P
0
23.
Q
(4.2)
_2
17.
47.
PAGE 876
y
x
21.
27.
_1
x
z
EXERCISES 13.2
y
1
0
41.
43.
45.
_2
_1
z
0
_1
_1
y 0 1
1
2
A27
37.
1
z
z
ANSWERS TO ODD-NUMBERED EXERCISES
Q
(4.2)
x
(4)
P
(4)
31.
0
x
1
1
l
3.
z 0
y
_1
(_3, 2)
0 x
_1
_1
0
y
1
1
ª(_1)
(_1)
0
33.
5.
10
x
y
2
, œ„2 ’
” œ„
2
π
z 0
ª” 4 ’
0
_10
10
0
x
_10
10
0
y
_10
π
”4’
x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A28
APPENDIX H
7.
ANSWERS TO ODD-NUMBERED EXERCISES
y
s
41.
ª(0)
k(t)
(1, 1)
(0)
0
x
9.
11.
13.
s
17.
2π
15.
19.
s
21.
23.
25.
27.
29.
31.
33.
s
z
4π
6π
43.
s
45.
49.
51.
z
s
47.
z
z
35.
37.
39.
41.
47.
49.
EXERCISES 13.3
1.
9.
s
3.
11.
13.
5.
s
15.
7.
s
s
17.
53.
55.
63.
PAGE 884
N
s
z
EXERCISES 13.4
s
z
65.
N
PAGE 894
1.
s
3.
s
19.
21.
27.
s
s
23.
29.
(2)
s
(_2, 2)
(2)
25.
s
31.
35.
s
33.
4
5.
y=x –@
y=k(x)
” π3 ’
(0, 2)
s
” π3 ’
4
_4
3
” 2 , œ„
3’
(3, 0)
_1
37.
(t)
0.6
5
z
0
_5
0
39.
7.
50
y
100
z
s
0
250 x
500
_5
0
(1, 1, 2)
5 t
(1)
(1)
y
x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
s
9.
11. s
13.
A29
ANSWERS TO ODD-NUMBERED EXERCISES
17.
s
19.
s
21.
15.
17.
23.
PROBLEMS PLUS
z
1.
3.
0.6
0.4
0.2
0
200
19.
23.
25.
29.
31.
33.
x
0
_200
0 y
10
_10
v
5.
7.
N
PAGE 900
t
v
v
s
21.
27.
CHAPTER 14
s
EXERCISES 14.1
N
PAGE 912
1.
_12
_4
35.
3.
37.
43.
39.
s
41.
45.
CHAPTER 13 REVIEW
True-False Quiz
1.
3.
9.
11.
N
5.
PAGE 897
7.
5.
13.
Exercises
1.
7.
v
v
v
z
9.
11.
z
z
z
13.
(0, 1, 0)
y
y
(2, 1, 0)
x
0
x
y=2x
3.
5.
11.
13.
s
s
7.
9.
s
15.
y
0
≈+¥=1
x
15.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A30
APPENDIX H
ANSWERS TO ODD-NUMBERED EXERCISES
17.
29. z
y
z
(0, 0, 9)
1
_1
0
1
x
19.
(0, 1, 0)
(3, 0, 0)
_1
x
y
y
y=≈
s
31. z
z
(0, 0, 2)
_1
21.
z
z
0
x
1
(0, 2, 0)
(1, 0, 0)
z
y
x
33.
35.
39.
23. z
37.
41.
z
z
14
(0, 0, 1)
(0, _1, 0)
x
25.
z
z
5
y
x
0
y
x
y
43.
z
45.
y
(0, 0, 10)
s
y
x
2
x
1
0
_1
(2.5, 0, 0)
0
43 2 1
(0, 2, 0)
0
_2
1 234
y
x
47.
27. z
49.
y
z
1 2
y
3
2
1
0
0
x
y
3
0
_2
_1
x
_3
0
x
1
2
3
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
73.
51.
ANSWERS TO ODD-NUMBERED EXERCISES
A31
10
z
5
z=4
0
z
z=3
_5
_10
2
z=2
x 0
z=1
0
_2 2
_2
y
75.
53.
77.
79.
EXERCISES 14.2
55.
1.
5.
13.
19. s
23.
z 0
_2
y
0
2 2
0
_2
x
PAGE 923
N
3.
7.
15.
21.
9.
11.
17.
s
25.
57.
1.0
z 0.5
29.
31.
33.
35.
z
37.
0.0
_4
27.
39.
z
41.
43.
y 0
0
4 4
59.
63.
67.
69.
_4
x
z
61.
65.
2
1
0
_1
_2
y
0
EXERCISES 14.3
2
N
2
0x
_2
PAGE 935
1.
71.
20
3.
0
z
_20
_40
_5
y
0
5
5
0x
_5
v
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A32
APPENDIX H
ANSWERS TO ODD-NUMBERED EXERCISES
5.
7.
9.
11.
35.
s
z
37.
z
z
39.
z
16
41.
16
(1, 2, 8)
x
2
C¡
0
4
y
(1, 2)
2
x
C™
0
4
(1, 2)
20
0
_20
_2
0
x
2 _2
0
2
z
z s
v
wv
z
z
z
z
z
z
45.
z
z
z
z
z
z
z
z
51.
53.
55. w
wvv
57. z
63.
67.
z
71.
13.
z
43.
49.
y
z
z
s
z
47.
(1, 2, 8)
z
z
z
t
v
v
w
v
v
v
z
z
z
65.
69.
73.
z
z
z
83.
87.
y
93.
101.
95.
99.
z
0.2
z 0
_0.2
20
z
_1
0
_20
_2
0
x
_2
2
0 y
y 0
2
EXERCISES 14.4
40
z
1. z
5.
7.
20
0
_2
x
0
2 _2
0
1
2
N
0
1
PAGE 946
3.
z
z
9.
400
y
1
z 0
z 200
15.
17.
19.
21.
10
z
x 0 _10
v
v
tv
v
v
v
v
27.
29.
z
31.
33. w
w z
z
z
z
z
w
z
z
z
v
23.
25.
27.
29.
31.
37.
39.
0
y
_5
y
13.
19.
v
0
5
11.
23.
25. t
_1
0
z
_1
x
21.
2
2
x
0
15.
z
z
z
33.
z
41.
35.
43.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
EXERCISES 14.5
PAGE 954
N
43.
1.
5.
7. z
z
9. z
z
11.
z
z
z
z
y
s
(3, 2)
0
0
s
15.
_1
1
x
55.
w
w
w
w
w
2
1
23.
1.
z
3.
5.
7.
9.
27.
29.
35.
39.
41.
45.
z
z
z
N
PAGE 977
z
11.
13.
17.
37.
s
43.
z
z
51.
33.
z
z
2
59.
EXERCISES 14.7
z
z
z
z
z
z
z
z
15.
21.
23.
s
s
25.
EXERCISES 14.6
1.
7.
9.
11.
17.
23.
27.
29.
31.
33.
35.
41.
PAGE 967
N
3.
z
x
w
21.
25.
y
63.
67.
w
Î
f (3, 2)
2x+3y=12
z 1
z
17.
31.
xy=6
2
13.
w
z
49.
z
19.
A33
z
z
45.
47.
s
3.
ANSWERS TO ODD-NUMBERED EXERCISES
s
z
s
13.
19.
25.
s
5.
z
z
s
27.
z
15.
29.
31.
33.
35.
s
21. s
37.
(_1, 0, 0)
(1, 2, 0)
0
s
s
z
s
_2
_3
39.
z
_1
z
_1
0
x
1
4
2y
_2
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A34
39.
45.
51.
APPENDIX H
s
ANSWERS TO ODD-NUMBERED EXERCISES
s
41.
s
47.
s
49.
9.
11.
43.
s
53.
EXERCISES 14.8
N
13.
PAGE 987
15.
1.
3.
5.
7.
9.
11.
v sw
17.
s
s
19.
21.
s
z
15.
s
s
17.
19.
s
s
s
21.
s
29– 41.
43.
45.
47.
CHAPTER 14 REVIEW
25.
s
s
s
N
s
s
7.
9.
3.
61.
63.
65.
z
1
y
_1
x
1
x
_1
z
z
zz
z
z
z
z
z
z
z
z
z
z
s
s
s
s
PROBLEMS PLUS
y
1.
7. s
z
z
45.
49.
59.
Exercises
1.
vw
z
31.
33.
35.
37.
43.
47. s
51.
53.
55.
57.
s
5.
sw
z
z
29.
PAGE 991
True-False Quiz
1.
3.
11.
v
w
z
27.
s
v w
z
z
z
z
13.
sw
v
N
s
s
s
s
s
s
s
PAGE 995
3.
w
w
s
y=_x-1
CHAPTER 15
5.
y
7.
y
EXERCISES 15.1
2
1
0
2
34
1
3.
7.
11.
EXERCISES 15.2
0
PAGE 1005
1.
5.
9.
15.
5
x
N
1
2
x
1.
9.
N
PAGE 1011
3.
11.
13.
5.
13.
7.
15.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
17.
s
19.
23.
43.
21.
A35
ANSWERS TO ODD-NUMBERED EXERCISES
x x
y
z
(0, 1)
(1, 1)
y=x
0
45.
x x
x
y
1
y=
x
x=
25.
33.
27.
29.
31.
0
47.
z
x x
_1
π
2
y= x
y
x
x=e †
x=2
y=0
35.
39.
37.
49.
57.
65.
EXERCISES 15.3
1.
11.
N
3.
51.
7.
y
y
1.
5.
x
N
x
0
y
3π
¨=
4
_2
0
13.
x x
19.
π
¨=
4
27.
37.
29.
31.
23.
33.
25.
35.
z
(1, 0, 0)
x
39.
41.
(0, 1, 0)
y
x
2
13.
s
s
29.
35.
1.
9.
15.
17.
15.
19.
21.
23.
27.
31.
s
33.
37.
s
s
39.
s
EXERCISES 15.5
0
1
25.
41.
(0, 0, 1)
0
_1
9.
17.
x x
21.
x x
x
7.
11.
17.
63.
R
x
s
PAGE 1026
3.
D
x x
55.
59.
67.
EXERCISES 15.4
9.
D
15.
s
PAGE 1019
5.
s
s
53.
xx
N
PAGE 1036
3.
5.
11.
7.
13.
19.
21.
s
s
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A36
APPENDIX H
ANSWERS TO ODD-NUMBERED EXERCISES
23.
33.
s
25.
35.
27.
29.
x xs x
x x zx
x x s xs
xx
33.
s
17.
s
19.
s
EXERCISES 15.7
1.
s
9.
s
s
s
13.
N
3.
15.
23.
x x xs
z
s
15.
s
s
PAGE 1049
5.
13.
25.
27.
5.
23.
x x x
x x x
x xz xz
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
39.
43.
PAGE 1040
s
3.
z
z
7.
9.
17.
11.
19.
21.
x x x
x x x
x x x
x x x
x x x
s
z
s
z
s
z
zs
z
z
49.
51.
53.
55.
z
s
z
z
45.
z
z
1.
7.
11.
z
z
47.
N
z
x x x
x x x
x x z xs
41.
s
EXERCISES 15.6
z
z
z
x x x
x xz x
x x xz
37.
31.
z
EXERCISES 15.8
PAGE 1055
N
1.
z
z
π
”2, _ 2 , 1’
1
0
29.
31.
x
x
x
x
x
x
x
x
s
x s xss
x z xss
xss
xss
xss
x
z
s
xs zx
x
x
x
x
x
x
x x
s
xs x
x z xss
x
xss
x
x
z
s
x s zx
s
s
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
y
x
π
s
z
3.
5.
7.
9.
11.
s
z
z
z
z
1
z
0
”4, 3 , _2’
z
z
z
2
π
_2
y
_2
x
z
z
4
π
3
z=1
z
z
2
z
z
z
z
z
x
z
z
13.
y
2
z
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
z
15.
17.
ANSWERS TO ODD-NUMBERED EXERCISES
A37
s
z
3
π
6
x
x
y
17.
25.
27.
31.
19.
19.
xxx
21.
s
23.
21.
x x x
z
23.
25.
29.
33.
s
35.
EXERCISES 15.9
z
27. s
31.
29.
t
y
37.
43.
PAGE 1061
N
39.
45.
s
s
41.
1.
z
z
π π
”6, 3, 6 ’
0
π
6 6
π
2
x
0
π
3
x
s
y
3
EXERCISES 15.10
π 3π
”3, 2 , 4 ’
y
s
s
s
1.
7.
9.
11.
v
s
v
v
v
v
v
v
15.
21.
23.
z
∏=4
˙= 3
∏=2
x
y
3π
˙=
4
∏=1
19.
25.
Exercises
1.
9. x x
11.
N
15.
21.
23.
31.
v
PAGE 1073
5.
7.
3.
13.
29.
s
27.
True-False Quiz
1.
3.
y
z
17.
CHAPTER 15 REVIEW
π
x
PAGE 1071
5.
7.
13.
N
3.
13.
3.
5.
9.
11.
15.
3π
4
9.
5.
7.
17.
19.
25.
27.
33.
35.
37.
39.
s
s
s
s
s
41.
43.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A38
APPENDIX H
ANSWERS TO ODD-NUMBERED EXERCISES
19.
45.
47.
x x
z
x
s
s
z
PROBLEMS PLUS
1.
N
49.
z
51.
PAGE 1077
3.
7.
13.
s
21.
23.
CHAPTER 16
EXERCISES 16.1
N
1.
y
25.
PAGE 1085
z
s
s
z
z
s
z
z
y
2
2
_6
_4
0
_1
4
6
x
_2
1
_2
0
_2
x
1
27.
4
_1
_4
3.
4
y
2
_2
2
29.
35.
x
31.
_4
33.
y
_2
0
x
5.
EXERCISES 16.2
N
PAGE 1096
1.
7.
9.
z
9. s
17.
21.
z
3.
11.
5.
s
7.
13.
19.
23.
15.
25.
27.
11.
x
y
x
13.
15.
y
17.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
29.
EXERCISES 16.6
2.1
N
A39
PAGE 1132
1.
3.
5.
7.
{ (1)}
1
” ” œ„2 ’’
{ (0)}
0
ANSWERS TO ODD-NUMBERED EXERCISES
2
2.1
√
_0.2
z 0
s
31.
35.
x
x
z
x
z
0
33.
z
_2
z
z
x
0
x
y
u
1
z
9.
37.
39.
43.
45.
1
u
41.
1
47.
51.
√
z 0
EXERCISES 16.3
1.
5.
9.
11.
15.
17.
21.
23.
31.
33.
N
PAGE 1106
3.
_1
7.
y
13.
z
z
z
19.
27.
N
3.
13.
0
1 1
_1
_1
y
PAGE 1113
5.
7.
9.
15.
17.
u
19.
21.
13.
19.
21.
23.
15.
z
z
s
z
z
_1
x
v
z
s
z
29.
z
z
5.
7.
z
z
z
s
z
z
z
z
z
z
9.
11.
13.
17.
25.
PAGE 1121
0
1 1
17.
v
z
1.
3.
0
11.
27.
N
x
z 0
23.
EXERCISES 16.5
_1
1
√
1.
0
11.
z
z
25.
EXERCISES 16.4
_1
z
z
z
z
z
z
15.
19.
31.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A40
APPENDIX H
33.
s
35.
z
37.
43.
s
s
s
55.
5.
15.
17.
41. s
45.
51.
EXERCISES 16.9
z
s
39.
z
47. s
57.
ANSWERS TO ODD-NUMBERED EXERCISES
s
s
49.
7.
9.
s
s
s
True-False Quiz
1.
3.
7.
9.
s
7.
5.
11.
3.
61.
v
1.
s
s
5.
11. s
s
17.
25.
27.
s
7.
13.
s
19.
29.
21.
N
xx
xx
z
43.
xx
7.
s
s
10
EXERCISES 16.8
5.
N
f
_3
s
3
_10
47.
17.
21.
PAGE 1151
7.
l
g
z
z
39.
PAGE 1172
z
45.
37.
3.
13.
15.
31.
s
27.
CHAPTER 17
35.
39.
3.
11.
33.
1.
5.
9.
11.
PAGE 1144
N
5.
13.
s
25.
EXERCISES 17.1
3.
15.
41.
v
63.
EXERCISES 16.7
37.
v
s
11.
29.
s
13.
PAGE 1160
9.
17.
23.
33.
N
Exercises
1.
z
9.
11.
s
19.
CHAPTER 16 REVIEW
59.
x x
PAGE 1157
21.
53.
s
N
9.
19.
23.
25.
27.
29.
31.
33.
35.
z
z
4
z
2
0
_2
_2
17.
y
0
2
2
0
x
_2
EXERCISES 17.2
N
PAGE 1179
1.
3.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
APPENDIX H
ANSWERS TO ODD-NUMBERED EXERCISES
5.
CHAPTER 17 REVIEW
7.
True-False Quiz
1.
3.
9.
11.
3
s
3.
8
yp
PAGE 1193
Exercises
1.
l
_3
N
A41
l
5.
s
7.
_3
9.
13.
15.
17.
11.
15.
19.
13.
17.
21.
23.
25.
19.
21.
27.
EXERCISES 17.3
1.
7.
N
s
0.02
0
APPENDIXES
PAGE 1187
3.
c=10
c=15
5.
EXERCISES G
1.4
c=20
c=25
c=30
PAGE A12
N
1.
3.
9.
5.
11.
17.
13.
19.
15.
21.
s
23.
7.
25.
s
27.
_0.11
29.
13.
31.
15.
EXERCISES 17.4
1.
s
N
PAGE 1192
s
33.
35.
37.
s
s
39.
s
3.
5.
7.
9.
11.
_i
41.
47.
43.
s
45.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Index
A43
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A44
INDEX
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
INDEX
A45
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A46
INDEX
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
INDEX
A47
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A48
INDEX
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
INDEX
A49
z
z
z
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A50
INDEX
z
z
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
R E F E R E N C E PA G E 5
D I F F E R E N T I AT I O N R U L E S
General Formulas
1.
2.
t
3.
t
5.
7.
t
t
t
t
t
4.
t
6.
t
t
t
t
t
t
8.
Exponential and Logarithmic Functions
9.
10.
11.
12.
Trigonometric Functions
13.
14.
15.
16.
17.
18.
Inverse Trigonometric Functions
19.
20.
s
22.
23.
s
21.
s
24.
s
Hyperbolic Functions
25.
26.
27.
28.
29.
30.
Inverse Hyperbolic Functions
31.
34.
32.
s
s
35.
33.
s
s
36.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
R E F E R E N C E PA G E 6
TA B L E O F I N T E G R A L S
Basic Forms
1.
v
v
11.
v
12.
2.
13.
3.
14.
4.
15.
5.
16.
6.
s
17.
7.
18.
8.
s
19.
9.
10.
20.
Forms Involving s
21.
s
s
s
22.
23.
s
24.
s
25.
26.
27.
28.
29.
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
R E F E R E N C E PA G E 7
TA B L E O F I N T E G R A L S
Forms Involving s
30.
s
s
s
31.
32.
s
33.
s
34.
s
s
s
s
35.
s
s
s
s
36.
s
s
s
37.
38.
s
Forms Involving s
39.
s
41.
s
42.
s
44.
45.
46.
s
s
40.
43.
s
s
s
s
s
s
s
s
s
s
s
s
s
s
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
R E F E R E N C E PA G E 8
TA B L E O F I N T E G R A L S
Forms Involving
47.
48.
49.
50.
51.
52.
53.
54.
55.
56.
57.
s
s
s
s
s
s
s
s
s
s
s
s
58.
s
59.
s
60.
61.
62.
s
s
s
s
s
s
s
s
s
s
s
s
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
R E F E R E N C E PA G E 9
TA B L E O F I N T E G R A L S
Trigonometric Forms
63.
76.
64.
77.
65.
78.
66.
79.
67.
80.
68.
69.
81.
70.
82.
71.
83.
72.
84.
73.
85.
74.
86.
75.
Inverse Trigonometric Forms
87.
88.
s
92.
s
93.
89.
90.
91.
s
s
94.
s
s
95.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
R E F E R E N C E PA G E 1 0
TA B L E O F I N T E G R A L S
Exponential and Logarithmic Forms
96.
100.
97.
101.
98.
102.
99.
Hyperbolic Forms
103.
108.
104.
109.
105.
110.
106.
111.
107.
112.
Forms Involving s
113.
114.
s
s
s
115.
s
116.
s
117.
118.
119.
120.
s
s
s
s
s
s
s
s
s
s
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.