15.2 SOLUTIONS
269
(n) Looking at the contributions to the integral of the function f (x; y) = xex , we can see that any
contribution made by the point (x; y), where x > 0, is greater than the corresponding contribution
made by (,x; y), since ex > 1 > e,x for x > 0. Thus, the integral of f in the region D is positive.
(o) The function f (x; y) = xy2 is odd with respect to x and thus, using the same reasoning as in part (d),
has integral zero in the region D.
(p) The function f (x; y) is odd with respect to x and thus, using the same reasoning as in part (d), has
integral zero in region B , which is symmetric with respect to x.
14. Take a Riemannian sum approximation to
Z
R
X
fdA i;j
f (xi ; yi)∆A:
Then, using the fact that ja + bj jaj + jbj repeatedly, we have:
Z
R
fdA j
X
i;j
X
f (xi ; yi )∆Aj jf (xi ; yi )∆Aj:
Now jf (xi ; yj )∆Aj = jf (xi ; yj j∆A since ∆A is non-negative, so
Z
R
fdA X
i;j
jf (xi ; yi )∆Aj =
X
i;j
jf (xi ; yj )j∆A:
But the last expression on the right is a Riemann sum approximation to the integral
Z
R
fdA Thus,
X
i;j
f (xi ; yj )∆A Z
R
fdA X
i;j
Z
R
jf (xi ;j )j∆A Z
R
Z
R
jf jdA:
px + y dA = Z 2 Z 1 px + y dx dy
Z
0
0
2
=
0
=
2
3
Z
1
2
3
(x + y ) 2
3
2
0
dy
0
y , y ) dy
3
((1 + ) 2
3
2
2
=
=
=
R
jf jdA:
Solutions for Section 15.2
1.
R
2 2
5
5
[(1 + y ) 2 , y 2 ]
3 5
0
4
5
5
((3 2 , 2 2 ) , (1 , 0))
15
p
4 p
(9 3 , 4 2 , 1) = 2:38176
15
jf jdA, so we have
270
CHAPTER FIFTEEN /SOLUTIONS
2. In the other order, the integral is
Z
1
Z
0
px + y dy dx:
2
0
First we keep x fixed and calculate the inside integral with respect to y:
Z
2
0
px + y dy = 2 (x + y)3=2 y=2
3
y =0
i
2h
3 =2
3=2
(x + 2)
:
,
x
3
=
Then the outside integral becomes
Z
1
0
i
2h
3=2
3=2
(x + 2)
,
x
3
2
5=2
(x + 2)
, 25 x5=2
5
i
2 2 h 5=2
=
3 , 1 , 25=2
3 5
= 2:38176
dx = 23
1
0
Note that the answer is the same as the one we got in Problem 1
3.
Z
R
(5
x2 + 1) sin 3y dA =
Z
Z
=
=
=
=
=
Z =3
1
(5
,1
0
1
,1
(5
Z
2
3
x
2
1
,1
(5
x2 + 1) sin 3y dy dx
+ 1)
, 13 cos 3y
=3
!
dx
0
x2 + 1) dx
2 5 3
( x + x)
3 3
2 10
(
+ 2)
3 3
32
9
1
,1
4. It would be easier to integrate first in the x direction from x = y , 1 to x = ,y + 1, because integrating first
in the y direction would involve two separate integrals.
Z
R
(2
x + 3y)2 dA =
Z
Z
1
0
1
Z
1
=
Z
0
1
=
,
,
y 1
y +1
Z
=
0
,y+1
Z
,
y 1
x + 3y)2 dx dy
(4
x2 + 12xy + 9y2 ) dx dy
4 3
x + 6x2y + 9xy2
3
8
3
,y+1
,
y 1
dy
,y + 1)3 + 9y2 (,2y + 2)] dy
[ (
0
(2
15.2 SOLUTIONS
, 23 (,y + 1)4 , 92 y4 + 6y3
=
2
9
= , (,1) ,
3
2
13
=
6
+
1
0
6
y
1
y = ,x + 1
y = x+1
,1
1
x
Figure 15.6
Z
5.
4
2
1
f dy dx
Z
2
Z
4
f dx dy
The line connecting (,1; 1) and (3; ,2) is
1
6.
Z
or
1
1
3x + 4y
or
=
1
y = 1 ,4 3x
So the integral becomes
Z
3
Z
(1
,3x)=4
,1 ,2
7. The line connecting (1; 0) and (4; 1) is
So the integral is
Z
f dy dx
or
1
Z
(1
,4y)=3
,2 ,1
f dx dy
y = 13 (x , 1)
Z
4
Z
2
,
(x 1)=3
1
8. Two of the sides of the triangle have equations x =
Z
3
1
Z
f dy dx
y , 1 and x = y , 5 . So the integral is
, 12 (y,5)
1
2 (y
,1)
,2
2
f dx dy
271
272
CHAPTER FIFTEEN /SOLUTIONS
Z
3Z 4
9.
1
0
ex+y dxdy =
Z
3
1
4
ex ey dx =
Z
1
0
See Figure 15.7.
3
ex (e4 , 1) dx = (e4 , 1)(e2 , 1)e
y
y=x
y
y=4
1
x
3
2
Figure 15.7
10.
Z
2
Z x
0
0
x
Figure 15.8
ex dydx =
2
Z
2
0
x
ex y dx =
2
Z
2
0
0
sin x y
2x
xex dx = 12 ex
2
2
2
=
0
1 4
(e , 1)
2
See Figure 15.8.
11.
Z
5
1
Z 2x
x
sin x dy dx =
Z
Z
5
1
5
=
1
= (sin
x
dx
sin x x dx
x , x cos x) 51
= (sin 5
, 5 cos 5) , (sin 1 , cos 1) ,2:68:
See Figure 15.9.
y
y = 2x
y=x
1
5
Figure 15.9
x
15.2 SOLUTIONS
12.
Z
4Z y
2 3
py x y dxdy =
1
Z
1
=
1
3
=
1
3
=
1
3
y3 x3 p dy
y
3 y
4
Z
4
(
1
y6 , y ) dy
9
2
y7 , y11=2
7
11=2
47
7
,
4
1
411=2 2
11
,
1
7
, 112
656:082
See Figure 15.10.
x = py y = x
y
4
y=4
x
2
y
+
2
y
=
3
9
R
,3 ,2
3
1
1
2
x
4
,3
Figure 15.10
Figure 15.11
13. See Figure 15.11.
Z
0
Z
Z
0
0
0
p 2xy dydx = ,2 x y2 p dx
,2 , 9,x
, 9,x
Z 0
x(9 , x2) dx
=,
2
2
,2
Z
=
0
,2
x3 , 9x) dx
(
=
x4 , 9 x2
=
,4 + 18 = 14
4
2
0
,2
x
273
274
14.
CHAPTER FIFTEEN /SOLUTIONS
y
(a)
4
x = ,(y , 4)=2 or y = ,2x + 4
2
x
Figure 15.12
R 2 R ,2x+4
(b) 0 0
g(x; y) dy dx.
15. As given, the region of integration is as shown in Figure 15.13.
y
1
x=y
x=1
x
1
Figure 15.13
Reversing the limits gives
Z
1
0
Z x
0
e dydx =
x2
1
Z
ye
0
x2
0
e
x2 1
=
2
x
=
0
dx =
e , 1:
2
Z
1
0
xex dx
2
15.2 SOLUTIONS
16. As given, the region of integration is as shown in Figure 15.14.
y
x = y2
3
x=9
9
x
Figure 15.14
Reversing the limits gives
Z
9
px
Z
0
0
y sin (x ) dydx =
Z
y2 sin (x2 )
9
2
=
2
0
Z
1
2
9
px !
dx
0
x sin (x2 ) dx
0
cos (x2 )
=,
4
0
1 cos (81)
=
, 4
4
9
=
0:056:
17. As given, the region of integration is as shown in Figure 15.15.
y
1
x = py
x=1
x
1
Figure 15.15
Reversing the limits gives
Z
1 Z x2
0
0
p
2+x
3
dydx =
Z
1
y
(
Z
0
1
=
0
p
x2
2+x
=
)
0
p
2 + x3 dx
1
=
x2
3
2
3 3
(2 + x ) 2
9
0
p
2 p
(3 3 , 2 2):
9
dx
275
276
CHAPTER FIFTEEN /SOLUTIONS
y
18.
8
y = 2x + 8
y = ,2x + 8
R
,4
4
x
Figure 15.16
,
Z
8 Z (8 y )=2
Order reversed:
0
(y
,8)=2
f (x; y) dx dy:
19. The intersection of the graph of f (x; y)
solid is shown in Figure 15.17.
=
25 , x2 , y2 and xy-plane is a circle x2 + y2
z
f (x; y) = 25 , x2 , y2
y
x
Figure 15.17
Thus the volume of the solid is
V
Z
=
Z
=
f (x; y) dA
p25,y
2
2
(25 , x , y ) dx dy:
p
,5 , 25,y
R
5
Z
2
2
=
25. The given
15.2 SOLUTIONS
20. The intersection of the graph of f (x; y)
given solid is shown in Figure 15.18.
=
25 , x2 , y2 and the plane z
=
16 is a circle, x2 + y2
=
277
32 . The
f (x; y) = 25 , x2 , y2
z
z = 16
y
x
Figure 15.18
Thus, the volume of the solid is
V
Z
=
Z
=
f x; y) , 16) dA
( (
p9,y
2
2
(9 , x , y ) dx dy:
p
,3 , 9,y
R
Z
3
2
2
21. The solid is shown in Figure 15.19, and the base of the integral is the triangle as shown in Figure 15.20.
z
y=0
-
y
y,x= 4
4
backside
2x + y + z = 4
,4
y,x=4
y
2x + y = 4
4
2
x
,4
Figure 15.19
Thus, the volume of the solid is
V
Figure 15.20
Z
=
2
ZR
=
Z
R
z dA
(4
, 2x , y) dA
,
4 Z (4 y )=2
=
0
,
y 4
(4
, 2x , y) dx dy:
x
278
CHAPTER FIFTEEN /SOLUTIONS
22.
Z
2
Z
2
xy dy dx =
Volume =
0
0
Z
2
=
0
=
x2
=
4
Z
2
0
1 2
xy
2
2
dx
0
2x dx
2
0
23. Let R be the triangle with vertices (1; 0), (2; 2) and (0; 1). Note that (3x + 2y + 1) , (x + y)
for x; y > 0, so z = 3x + 2y + 1 is above z = x + y on R. We want to find
Z
Volume =
R
((3
x + 2y + 1) , (x + y)) dA =
Z
R
(2
=
2x + y + 1 > 0
x + y + 1) dA:
We need to express this in terms of double integrals.
y
(2
2
y = 1 + 0:5x
; 2)
R2
R1
1
y = 2x , 2
y = 1,x
O
1
x
2
Figure 15.21
To do this, divide R into two regions with the line x = 1 to make regions R1 for x 1 and R2 for x 1.
See Figure 15.21. We want to find
Z
R
(2
x + y + 1) dA =
Z
R1
(2
x + y + 1) dA +
Note that the line connecting (0; 1) and (1; 0) is
y = 1 + 0:5x. So
Z
R1
(2
x + y + 1) dA =
The line between (1; 0) and (2; 2) is y
Z
R2
(2
=
y
Z
0
Z
,
1 x
(2
x + y + 1) dA:
(2
x + y + 1) dy dx:
(2
x + y + 1) dy dx:
2 Z 1+0:5x
1
R2
1 , x, and the line connecting (0; 1) and (2; 2) is
1 Z 1+0:5x
2x , 2, so
x + y + 1) dA =
=
Z
,
2x 2
15.2 SOLUTIONS
We can now compute the double integral for R1 :
Z
1 Z 1+0:5x
(2
,
1 x
0
x + y + 1) dy dx =
Z
Z
1
0
1
2xy +
=
0
=
=
and the double integral for R2 :
Z
2
1
Z
1+0:5x
(2
,
2x 2
x + y + 1) dy dx =
Z
2
(2
Z
1
1
0
1+0:5x
,
1 x
dx
dx
dx
0
xy + y2 =2 + y)
1
=
2
21 2
x + 3x
8
7 3 3 2
x + 2x
8
19
;
8
=
=
y2 + y
279
1+0:5x
,
2x 2
dx
, 398 x2 + 9x + 32 dx
, 138 x3 + 92 x2 + 32 x
29
:
8
2
1
19 29
48
+
=
= 6.
8
8
8
24. We want to calculate the volume of the tetrahedron shown in Figure 15.22.
So, Volume =
z
1=c
1=b
1=a
ax + by = 1
x
y
Figure 15.22
We first find the region in the xy-plane where the graph of ax + by + cz = 1 is above the xy-plane.
When z = 0 we have ax + by = 1: So the region over which we want to integrate is bounded by x = 0; y = 0
and ax + by = 1: Integrating with respect to y first, we have
Z 1=a Z
(1
Volume =
0
,ax)=b
0
=
Z 1=a y
0
=
1
:
6abc
Z 1=a Z
0
by2 , axy
,
c 2c
c
Z 1=a
1
0
=
z dy dx =
2bc
(1
(1
,ax)=b 1 , by , ax
c
0
y=(1,ax)=b
y =0
, 2ax + a2x2 ) dx
dx
dy dx
280
CHAPTER FIFTEEN /SOLUTIONS
25. The region bounded by the x-axis and the graph of y
region is
A=
Z
=
1
(
0
1
2
x , x2 is shown in Figure 15.23. The area of this
=
2
3 1
x , x2)dx = ( x2 , x3 )
0
,
1
3
=
1
:
6
So the average distance to the x-axis for points in the region is
R
y dA
area(R)
R
Average distance =
Z
R
y dA =
Z
Z
Z x,x2
1
0
1
=
0
Therefore the average distance is
1=60
=
1 =6
0
!
y dy dx
x2 , x3 + x4 dx = 1 , 1 + 1 = 1 :
2
2
6 4 10
60
1=10:
b
y
y
1
4
x
0
a
1
1
2
Figure 15.23
x
Figure 15.24
26. Assume the length of the two legs of the right triangle are a and b, respectively. See Figure 15.24. The line
through (a; 0) and (0; b) is given by yb + xa = 1. So the area of this triangle is
A = 12 ab:
Thus the average distance from the points in the triangle to the y-axis (one of the legs) is
Average distance =
1
A
, ab x+b
Z aZ
0
0
Z a
x dy dx
b x2 + bx dx
=
,
ab 0
a
b 2 a
2
b
3
=
+ x
x
,
ab 3a
2
2
=
2
ab
a2 b
6
a
= :
3
0
15.2 SOLUTIONS
281
Similarly, the average distance from the points in the triangle to the x-axis (the other leg) is
Average distance =
=
=
, ab y+a
Z bZ
1
A
2
ab
2
0
y dx dy
0
Z b
, ab y2 + ay dy
0
ab
ab2
b:
=
6
3
27. The function sin (x2 ) has no elementary antiderivative, so we try integrating with respect to y first. The region
of integration is shown in Figure 15.25. Changing the order of integration, we get
Z
1
0
Z
y
1
sin (x2 ) dx dy
Z
1
Z x
=
Z
0
0
1
=
Z
0
1
=
0
=
sin (x2 ) dy dx
sin (x2 ) y
x
dx
0
sin (x2 ) x dx
, cos2(x )
cos 1
=,
2
2
1
+
1
2
0
=
1
(1 , cos 1) = 0:23:
2
y
y=x
1
x
Figure 15.25
28. The region of the integration is shown in Figure 15.26. To make the integration easier, we want to change the
order of the integration and get
Z
1
0
Z e
x dx dy = Z e Z ln x x dy dx
ln x
ey ln x
1
0
Z e
x y ln x dx
=
1 ln x
0
Z e
2 e
x dx = x
=
1
=
1 2
(e , 1):
2
2
1
282
CHAPTER FIFTEEN /SOLUTIONS
y
(
e; 1)
e
1
x
Figure 15.26
29. We want to find the average value of jx , yj, over the square 0 x 1; 0 y
Z
1Z 1
Average distance between gates =
0
Let’s fix x, with 0 x 1. Then jx , yj =
Z
1
0
jx , yj dy =
Z x
=
Z
x , y) dy +
(
0
=
y,x
x,y
2
xy , y
+
x : Therefore
x
y , x) dy
y2 , xy
2
0
jx , yj dy dx:
(
x
x
2
1
for y
for y
0
1:
1
x
x2 , x2
2
=
+
1
2
, x , x2
2
+
x2
x2 , x + 12 :
So,
Z
1
Z
1
Average distance between gates =
Z
0
0
1
jx , yj dy dx
Z
1
=
0
0
x ,x
3
2
3
=
!
jx , yj dy dx =
2
+
1
x
2
1
=
0
Z
1
x2 , x + 12 ) dx
(
0
1
:
3
30. There are (n + 1)2 possible pairs (i; j ) of gates, i = 0; : : :; n; j = 0; : : :; n, so the sum given represents the
average distances apart of all such gates. The Riemann sum with ∆x = ∆y = n1 , if we choose the least x and
y-values in each subdivision is
, , i
n
X1 n
X1
i=0 j =0
j 1
n , n n2 ;
which for large n is just about the same as the other sum. For n = 5 the sum is about 0:389; for n = 10 the
sum is about 0:364: