Chapter 6
Applications of Integration
6.1
Velocity and Net Change
6.1.1 The position of an object is the coordinate of the object on the line at a given time, often denoted
s(t). The displacement over an interval [a, b] is s(b) s(a), the di↵erence of the object’s ending position and
Z b
beginning position. It can be written as
v(t) dt where v(t) is the object’s velocity at time t. The distance
a
Z b
traveled by the object is
|v(t)| dt, the sum of the distance traveled along the line to the right and the
a
distance traveled along the line to the left over the given time interval.
6.1.2 If velocity is positive, then |v(t)| = v(t), so the distance and displacement are equal.
6.1.3 The displacement is given by
Z b
v(t) dt, because this quantity is equal to s(b)
s(a).
a
6.1.4 The net change of a quantity is given by
Z b
f 0 (t) dt, if f 0 (t) is the rate of change of the quantity.
a
6.1.5 The value of Q at time t will be given by Q(t) = Q(0) +
Z t
Q0 (x) dx.
0
0
6.1.6 If Q (t) is the growth rate of a population Q at time t, then
Z b
Q0 (t) dt = Q(b)
Q(a), the net change
a
of the population over the time period [a, b].
6.1.7
a. The velocity is positive for 0  t < 1 and for 3 < t < 5, so the object is moving in the positive direction
on those intervals.
Z 3
b. The displacement is
v(t) dt = 12 16 = 4.
0
c. The distance traveled is
Z 5
1
d. The displacement is
Z 5
|v(t)| dt = 16 + 10 = 26.
v(t) dt = 12
16 + 10 = 6.
0
e. After five hours the object’s position is 6 miles from the original position in the positive direction.
573
Chapter 6. Applications of Integration
574
6.1.8
a. The velocity is negative for 0 < t < 2 and for 4 < t < 6, so the object is moving in the negative
direction on those intervals.
Z 6
b. The displacement is
v(t) dt = 14 10 = 4.
2
c. The distance traveled is
Z 6
0
d. The displacement is
Z 8
|v(t)| dt = 20 + 14 + 10 = 44.
v(t) dt =
20 + 14
10 + 6 =
10.
0
e. After eight minutes the object’s position is 10 meters from the original position in the negative direction.
6.1.9
a. The displacement is the net area, which is
b. The distance traveled is
c. s(5) = s(0) +
Z 5
1
·2·2
2
1 5
1 4
· · 1 + · · 2 = 3.
2 3
2 3
1
1 4
1 5
13
·2·2+ · ·1+ · ·2=
2 3
3
2
2 3
v(t) dt = 0 + 3 = 3.
0
8Z t
>
>
( x + 2) dx
>
>
>
>
< 0 Z t
3
d. s(t) =
+
(3x 10) dx
>
2 Z3
>
>
t
>
>
>
:2 +
( 2x + 10) dx
4
6.1.10
8 2
t
>
>
+ 2t
if 0  t  3,
>
>
< 22
if 3 < t  4, = 3t
10t + 18 if 3 < t  4, .
>
>
2
>
>
:
t2 + 10t 22 if 4 < t  5.
if 4 < t  5
if 0  t  3,
a. The displacement is the net area, which is 3 · 3 +
1
· 3 · 2 = 9 + 3 = 12.
2
b. Because v(t) > 0 on that interval, the distance traveled is the same as the displacement, so it is 12
also.
Z 5
c. s(5) = s(0) +
v(t) dt = 0 + 12 = 12.
0
d. s(t) =
6.1.11
8
>
<3t
>
:9 +
Z t✓
3
3
15
x+
2
2
◆
if 0  t  3,
dx
if 3  t  5
=
8
>
<3t
3 2 15
>
:
t + t
4
2
27
4
if 0  t  3,
.
if 3  t  5
a. Note that v(t) > 0 on [0, ⇡/2], so the distance traveled is
Z ⇡/2
0
⇡/2
3 sin 2t dt =
3
cos 2t
2
0
=
3
( 1
2
1) = 3.
b. Note that the function is alternately positive then negative (with the same enclosed area) on each
interval of length ⇡/2. The displacement on [0, ⇡] is then 3 3 = 0, the displacement on [0, 3⇡/2] is
3 3 + 3 = 3, and the displacement on [0, 2⇡] is 3 3 + 3 3 = 0.
Copyright c 2019 Pearson Education, Inc.
6.1. Velocity and Net Change
575
c. The total distance traveled is 3 + 3 + 3 + 3 = 12.
6.1.12
�
�
The velocity is positive (and so the object is moving to the right) on [0, 3) and the velocity is negative (and so the object is moving to the left) on
(3, 6].
�
a.
�
�
�
�
�
�
-�
-�
-�
b. The displacement is
Z 6
Z 6
(6
(6
2t) dt +
v(t) dt =
0
Z 3
0
36)
(9
t2 )
0
c. The total distance traveled is
0) + (36
6
2t) dt = (6t
= 0 m.
0
Z 6
3
(2t
6) dt = (6t
3
18) = 18 m.
6
t2 ) + (t2
6t)
0
= (18
9)
(0
3
6.1.13
a. v(t) = 3t(t
2) which is zero for t = 0 and t = 2. v(t) > 0 on (2, 3) and is negative on (0, 2).
Z 3
3
b. The displacement is
(3t2 6t) dt = (t3 3t2 ) = 33 3(3)2 = 0 m.
0
0
c. The total distance traveled is
Z 2
(6t
0
8) + (27
27)
(8
12) = 8 m.
3t2 ) dt +
Z 3
2
(3t2
6t) dt = (3t2
2
3
+ (t3
t3 )
3t2 )
0
= (12
2
6.1.14
a. v(t) = 4t(t2 6t + 5) = 4t(t
is negative on (1, 5).
Z 5
b. The displacement is
(4t3
5)(t
1), which is zero for t = 0, t = 1, and t = 5. v > 0 on (0, 1) and
5
24t2 + 20t) dt = (t4
8t3 + 10t2 )
0
8(53 ) + 10(52 ) =
Z 1
8t3 + 10t2 )
8t3 + 10t2 )
=
0
24t2 + 20t) dt =
1
5
(t4
125 m.
1
(4t3
24t2 + 20t) dt = (t4
0
Z 5
8 + 10 = 3 m, and the distance traveled on the time interval (1, 5) is
(4t3
c. The distance traveled on the time interval (0, 1) is
1
= 54
0
= 125 + 3 = 128 m. Therefore the total distance traveled over the time interval
1
(0, 5) is 3 + 128 = 131 m.
6.1.15
a. v(t) = 3(t2 6t + 8) = 3(t 4)(t 2), which is zero for t = 4 and t = 2. v > 0 on (0, 2) and (4, 5),
while v is negative on (2, 4).
Z 5
5
b. The displacement is given by
(3t2 18t + 24) dt = (t3 9t2 + 24t) = 53 9(52 ) + 24(5) = 20 m.
0
Copyright c 2019 Pearson Education, Inc.
0
576
Chapter 6. Applications of Integration
Z 2
2
(3t2 18t+24) dt = (t3 9t2 +24t) = 8 36+48 =
0
0
Z 4
4
2
3
20 m. On the interval (2, 4), the distance traveled is
(3t
18t + 24) dt = (t
9t2 + 24t) =
c. On the time interval (0, 2) the distance traveled is
2
is
(64
Z 5
144 + 96) + (8
36 + 48) = 20
2
16 = 4 m. On the time interval (4, 5), the distance traveled
5
(3t2
18t + 24) dt = (t3
9t2 + 24t)
4
= 20
16 = 4 m. Therefore the total distance traveled is
4
20 + 4 + 4 = 28 m.
6.1.16
a. The motion is positive for 0  t  4, because the exponential function is everywhere positive.
Z 4
4
b. The displacement is
50e 2t dt =
25e 2t
= 25(1 e 8 ) ⇡ 24.992 m.
0
0
c. Because the velocity is positive on the given interval, the distance traveled is the same as the displacement, given above.
6.1.17
a. s(t) =
Z
sin t dt =
b. s(t) = s(0) +
Z t
cos t + C, and because s(0) = 1, we must have C = 2. Thus, s(t) = 2
cos t.
t
sin x dx = 1 + ( cos x)
0
=2
cos t.
0
6.1.18
a. s(t) =
Z
( t3 + 3t2
2t) dt =
t4
+ t3 t2 + 4.
4
Z t
b. s(t) = s(0) +
( x3 + 3x2
t4
+ t3
4
t2 + C. Because s(0) = 4, we have C = 4, and thus
s(t) =
2x) dx = 4 +
0
6.1.19
a. s(t) =
Z
(6
2t) dt = 6t
b. Also, s(t) = s(0) +
Z t
a. s(t) =
Z
x4
+ x3
4
◆ t
=
0
t4
+ t3
4
t2 + 4.
t2 + C, and because s(0) = 0, we must have C = 0. Thus, s(t) = 6t
3 sin ⇡t dt =
2x) dx = 6x
x2
t2 .
= 6t
t2 .
0
3
3
cos ⇡t + C, and because s(0) = 1, we must have C = 1 + . Thus,
⇡
⇡
3
3
cos ⇡t + 1 + .
⇡
⇡
✓
◆t
Z t
3
b. s(t) = s(0) +
3 sin ⇡x dx = 1 +
cos ⇡x
=
⇡
0
0
s(t) =
x2
t
(6
0
6.1.20
✓
3
3
cos ⇡t + 1 + .
⇡
⇡
6.1.21
a. s(t) =
Z
(9 t2 ) dt = 9t
t3
+C, and because s(0) =
3
2, we must have C =
Copyright c 2019 Pearson Education, Inc.
2. Thus, s(t) = 9t
t3
2.
3
6.1. Velocity and Net Change
b. s(t) = s(0) +
Z t
(9
577
2
x ) dx =
2+
0
6.1.22
✓
x3
3
9x
◆t
t3
3
= 9t
0
2.
Z
1
dt = ln(t + 1) + C, and because s(0) = 4, we must have C =
t+1
ln(t + 1) 4.
Z t
t
1
b. s(t) = s(0) +
dx = 4 + (ln(x + 1)) = ln(t + 1) 4.
0 1+x
0
a. s(t) =
4. Thus, s(t) =
6.1.23
s
2
a. s(t) = s(0) +
Z t
0
2 sin ⇡t.
t
2⇡ cos ⇡x dx = 2 sin ⇡x|0 =
3
4
t
20t = 250, or t = 7.5, and thus s(7.5) =
Z 7.5
(400
0
1
2
2
b.
c. The mass reaches its lowest point at t = 1.5, t = 3.5 and t = 5.5.
d. The mass reaches its highest point at t = .5, t = 2.5, and t = 4.5.
6.1.24
a. s(5) =
Z 5
0
b. s(10) =
|400
Z 10
0
|400
20t| dt =
Z 5
5
(400
20t) dt = 400t
10t2
0
20t| dt =
= 1750 m.
0
Z 10
10
(400
20t) dt = 400t
10t2
0
c. Her velocity is 250 when 400
= 3000 m.
0
7.5
400t
10t2
20t) dt =
0
= 2437.5.
0
6.1.25
a. s(t) = s(0) +
Z t
t
30(16
10x3
x2 ) dx = 480x
0
= 480t
10t3 = 10t(48
t2 ).
0
b. Because the velocity is positive, this is given by s(2)
s(0) = 960 80 = 880 miles.
p
30t2 = 400, or t =
8/3. At this point the plane has traveled
c. The velocity is 400 when 480
p
p
p 3
s( 8/3) = 480 8/3 10 8/3 ⇡ 740.290 miles.
6.1.26
a. s(t) = s(0) +
Z t
0
3 sin2 (⇡x/2) dx =
3
2
Z t
0
(1
cos(⇡x)) dx =
3
2
✓
x
1
sin ⇡x
⇡
Copyright c 2019 Pearson Education, Inc.
◆t
0
=
3
2
✓
t
◆
1
sin ⇡t .
⇡
578
Chapter 6. Applications of Integration
b. Because the velocity is positive, this is given by s(0.25)
s(0) =
3
2
0.0374 miles.
✓
sin(⇡/4)
⇡
0.25
◆
=
p
3 2
⇡
4⇡
3
8
c. s(3) = 4.5 miles.
6.1.27
v
60
The velocity has a maximum of 60 for 20 
t  45. The velocity is 0 at t = 0 and at
t = 60.
a.
0
b.
Z 20
3t dt +
0
c. 1200 +
Z 30
20
t
40
60 dt = 1200 m.
20
Z 45
60 dt +
30
Z 60
60
(240
2t2
4t) dt = 1200 + 900 + 240t
45
= 2550 m.
45
Z 75
d. At time t = 60 the automobile is at position 2550. In the following 15 seconds, it moves
4t) dt = 450 feet in the opposite direction, so it is at position 2550
60
450 = 2100.
6.1.28
�
a.
The velocity is given by
8
<9.8t if 0  t  10,
v(t) =
:
10
if t > 10.
b. s(30) =
Z 10
9.8t dt +
0
Z 30
��
��
��
��
�
��
��
�
10 dt = 490 + 200 = 690 m.
10
c. We seek t so that 490 +
Z t
10 dx = 3000, which can be written 10t
100 = 2510, so t = 261 s.
10
Z t
Z t
t
6.1.29 v(t) = v(0) +
a(x) dx = 70 +
32 dx = 70 32x = 32t + 70.
0
0
Z t 0
t
2
s(t) = s(0) +
( 32x + 70) dx = 10 + ( 16x + 70x) = 16t2 + 70t + 10.
0
0
Copyright c 2019 Pearson Education, Inc.
(240
6.1. Velocity and Net Change
579
Z t
t
6.1.30 v(t) = v(0) +
32 dx = 50 + 32x = 32t + 50.
0
Z t 0
t
s(t) = s(0) +
( 32x + 50) dx = 0 + ( 16x2 + 50x) =
0
16t2 + 50t.
0
Z t
t
6.1.31 v(t) = v(0) +
9.8 dx = 20 + 9.8x = 9.8t + 20.
0
Z t 0
t
s(t) = s(0) +
( 9.8x + 20) dx = 0 + ( 4.9x2 + 20x) = 4.9t2 + 20t.
0
0
Z t
t
6.1.32 v(t) = v(0) +
e x dx = 60 e x = 60 (e t 1) = 61 e t .
0
Z t 0
t
s(t) = s(0) +
(61 e x ) dx = 40 + (61x + e x ) = 39 + 61t + e t .
0
0
Z t
t
6.1.33 v(t) = v(0) +
0.01x dx = 10 + ( 0.005x2 ) = 10 0.005t2 .
✓
◆0 t
Z t 0
1 3
1 3
2
s(t) = s(0) +
(10 .005x ) dx = 10x
x
= 10t
t .
600
600
0
0
Z t
t
20
20
20
dx
=
20
= 30
.
6.1.34 v(t) = v(0) +
2
+
2)
x
+
2
t
+2
(x
0
0
◆
Z t✓
t
20
s(t) = s(0) +
30
dx = 10 + (30x 20 ln |x + 2|) = 10 + 20 ln 2 + 30t
x+2
0
0
✓
◆ t
1
1
cos 2x dx = 5 +
sin 2x
= sin 2t + 5.
6.1.35 v(t) = v(0) +
2
2
0
◆
✓
◆ t
Z t✓0
1
1
sin 2x + 5 dx = 7 +
cos 2x + 5x
=
s(t) = s(0) +
2
4
0
0
Z t
20 ln |t + 2|.
1
29
cos 2t + 5t + .
4
4
Z t
2x
6.1.36 v(t) = v(0) +
dx. Let u = x2 + 1 so that du = 2x dx. Then we have 0 +
2
2
0 (x + 1)
✓
◆ t2 +1
1
1
=
+ 1.
2
t +1
u 1
✓
◆
Z t
t
1
1
s(t) = s(0) +
+
1
dx
=
tan
(x)
+
x
= tan 1 t + t.
x2 + 1
0
0
Z t2 +1
1
1
du =
u2
6.1.37
a. The velocity is given by v(0) +
Z t
t
88 dx = 88x
0
The position is given by s(0) +
Z t
= 88t ft/s.
0
t
88x dx = 44x2
0
= 44t2 ft.
0
b. The car travels s(4) = 44 · 16 = 704 feet.
c. Because a quarter mile is 1320 feet, we need 44t2 = 1320, so t =
p
30 ⇡ 5.477 seconds.
d. We need 44t2 = 300, so t ⇡ 2.611 seconds.
e. It reaches that speed when 88t = 178, or t = 89/44 seconds. At that time the racer has traveled
s(89/44) = 44(89/44)2 = 892 /44 ⇡ 180.023 feet.
Copyright c 2019 Pearson Education, Inc.
580
Chapter 6. Applications of Integration
6.1.38
a. The velocity is given by v(0) +
Z t
t
15 dx = 60 + ( 15x)
=
0
The position is given by s(0) +
Z t
15t + 60.
0
t
( 15x + 60) dx = ( 7.5x2 + 60x)
0
=
7.5t2 + 60t.
0
b. The car comes to rest when v(t) = 0, which occurs for t = 4. At that time s(4) = 120 feet.
✓
◆ t
Z t
1280
80
80
.
dx
=
80
+
=
6.1.39 v(t) = v(0) +
(1 + 8x)2 0
(1 + 8t)2
(1 + 8x)3
0
✓
◆
Z t
t
80
10
10
dx
=
= 10
.
s(t) = s(0) +
2
1 + 8x 0
1 + 8t
0 (1 + 8x)
10
50
80
= 10
=
⇡ 6.154 miles.
Then in the first 0.2 seconds the train travels s(0.2) s(0) = 10
13
◆ 13
✓ 2.6
10
10
50 50
400
Between time 0.2 and 0.4 the train travels s(0.4) s(0.2) = 10
10
=
=
⇡ 1.465
2.6
13 21
273
4.2
miles.
6.1.40
a. 55 +
Z 6
(t/5)) dt = 55 +
(20
0
b. P (200) = 55 +
Z 200
0
(20
✓
20t
t2
10
t
) dt = 55 +
5
✓
◆6
857
⇡ 171.4.
5
=
0
20t
t2
10
◆ 200
= 55.
0
6.1.41
a. P (20) = 250+
0
people.
b. P (t) = 250 +
Z 20
Z t
0
⇣
⌘ 20
p
p
p
(30+30 t) dt = 250+ 30t + 20t3/2
= 250+600+800 5 = 850+800 5 ⇡ 2639
0
⇣
⌘t
p
(30 + 30 x) dx = 250 + 30x + 20x3/2
= 250 + 30t + 20t3/2 people.
0
6.1.42
a. P (15) = 35 +
Z 15
(5 + 10 sin(⇡t/5)) dt = 35 + (5t
0
(50/⇡)) = 110 + 100/⇡ ⇡ 142 foxes.
Z 35
P (35) = 35 +
(5 + 10 sin(⇡t/5)) dt = 35 + (5t
0
(50/⇡)) = 210 + 100/⇡ ⇡ 242 foxes.
Z t
b. P (t) = 35 +
(5 + 10 sin(⇡x/5)) dx = 35 + (5x
(50/⇡) foxes.
0
15
(0
35
(0
(50/⇡) cos(⇡t/5))|0 = 35 + (75 + (50/⇡))
(50/⇡) cos(⇡t/5))|0 = 35 + (175 + (50/⇡))
t
(50/⇡) cos(⇡x/5))|0 = 35 + 5t
(50/⇡) cos(⇡t/5) +
6.1.43
a. N (20) = 1500+
1897 cells.
N (40) = 1500+
1900 cells.
Z 20
100e 0.25t dt = 1500+
400e 0.25t
20
= 1500+(
0
400e 5 +400) = 1900
400
⇡
e5
Z 40
100e 0.25t dt = 1500+
400e 0.25t
40
= 1500+(
0
400e 10 +400) = 1900
400
⇡
e10
0
0
Copyright c 2019 Pearson Education, Inc.
6.1. Velocity and Net Change
b. N (t) = 1500+
1900 cells.
Z t
581
t
= 1500+(
0
400e 0.25x
100e 0.25x dx = 1500+
0
400e 0.25t +400) =
400e 0.25t +
6.1.44
0.1485
59.4
a. r(t) = 0 when 0.0025e0.25t = 0.1485e 0.15t , which occurs when e0.40t = 0.0025
= 59.4. So t0 = ln0.40
⇡
10.21 days.
�
� = �(�)
����
�
� � � � � � � � � �� �� �� �� ��
-����
-���
b.
-����
The tumor decreases in size for the first 10.21 days and then starts to increase in size.
Z 10.21
10.21
c.
(0.0025e0.25t 0.1485e 0.15t ) dt = 0.01e0.25t + 0.99e 0.15t
⇡ 0.658 cm3 . The tumor
0
0
decreases in size by 0.658 cm3 in the first 10.21 days.
6.1.45
a. In the first 35 days the number of barrels produced is
Z 30
800 dt+
0
27250.
Z 35
(2600 60t) dt = 24000+3250 =
30
b. In the first 50 days the number of barrels produced is 27250 +
Z 40
(2600
60t) dt +
35
27250 + 1750 + 2000 = 31000.
Z 50
200 dt =
40
c. A constant 200 barrels per day times 20 days yields 4000 barrels.
6.1.46
✓
◆ 30
Z 30
0.25 3 37.46 2
a.
(0.25t2 + 37.46t + 722.47) dt =
t +
t + 722.47t
= 40781.1.
3
2
0
0
✓
◆ 30
Z 30
0.90 3 69.06 2
2
b.
(0.90t
69.06t + 2053.12) dt =
t +
t + 2053.12t
= 38616.6.
3
2
0
0
c. 40781.1 + 38616.6 = 79397.7 millions of cubic feet, or 7.93977 ⇥ 1010 cubic feet. There are 52803 =
147197952000 cubic feet in a cubic mile, so the Spokane River contains 0.67 · 147197952000 = 9.86226 ·
7.93977 ⇥ 1010
1010 cubic feet of water. So the percentage can be calculated by
⇡ 80.507%.
9.86226 · 1010
6.1.47
a. Q(t) = Q(0) +
Z t
107 e kx dx =
0
e kt )
107 kx
e
k
◆ t
0
=
e kt )
107 (1
k
.
107
. This represents the total number of barrels extracted if the
t!1
t!1
k
k
7
nation extracts the oil indefinitely where it is assumed that the nation has at least 10k barrels of oil in
reserve.
b. lim Q(t) = lim
107 (1
✓
c. We seek k so that
=
107
1
= 2 ⇥ 109 , which gives k =
= 0.005.
k
200
Copyright c 2019 Pearson Education, Inc.
582
Chapter 6. Applications of Integration
d. We want T so that (2 ⇥ 107 )
Z T
0
so T = 200 ln 2 ⇡ 138.629 years.
e 0.005t dt = 2 ⇥ 109 , so
200e 0.005t
T
= 100, so 1
0
e T /200 = 1/2,
6.1.48
Z 60
⇣
⌘ 60
p
a.
3 t dt = 2t3/2
⇡ 929.52 L.
0
0
b. Q(t) =
Z t
0
⇣
⌘t
p
3 x dx = 2x3/2
= 2t3/2 L, where t is measured in minutes.
0
c. The tank will be full when 2t3/2 = 2000, or t =
p
3
10002 = 100 minutes.
6.1.49
✓
◆2
Z 2
240
240 1
120
sin(⇡t/12)
= 40 +
· = 40 +
⇡ 78.197 m3 .
a.
20(1 + cos(⇡t/12)) dt = 20t +
⇡
⇡
⇡
2
0
0
Z t
b. Q(t) =
20(1 + cos(⇡x/12)) dx, which is equal to
0
✓
◆t
240
240
20x +
sin(⇡x/12)
= 20t +
· sin(⇡t/12) m3 .
⇡
⇡
0
c. The reservoir is full when 20T +
6.1.50
✓
Z 1
a.
70(1 + sin(2⇡t)) dt = 70t
0
240
sin(⇡T /12) = 2500, which occurs for T ⇡ 122.6 hours.
⇡
35
cos(2⇡t)
⇡
◆1
= 70
0
35 35
+
= 70 mL.
⇡
⇡
b.
Z t
◆t
35
cos(2⇡x)
= 70t
⇡
0
35
35
cos(2⇡t) mL.
=
+ 70t
⇡
⇡
70(1 + sin(2⇡x)) dx =
0
✓
70x
35
35
cos(2⇡t) +
⇡
⇡
c.
Z t+1
t
◆ t+1
35
cos(2⇡x)
= 70t + 70
70(1 + sin(2⇡x)) dx = 70x
⇡
t
35
35
=
+ 70t
cos(2⇡t) = 70 mL.
⇡
⇡
✓
35
cos(2⇡(t + 1))
⇡
(70t
35
cos(2⇡t))
⇡
6.1.51
a. V (t) = V (0) +
Z t
0
⇣
⇡
⇡x
⇡x ⌘
⇡t
sin
dx = 6 + cos
= 5 + cos .
2
2
2
2
0
t
⇡t
is periodic with period 4, the breathing cycle repeats every 4 seconds, so there are
b. Because sin
2
60
= 15 breaths per minute.
4
c. The lungs are full at t = 0, at which time V (0) = 6 L, so this is the capacity of the lungs. The tidal
volume is the di↵erence between this amount and the amount in the lungs after each exhalation, which
is the minimum value of V (t). This minimum occurs at the first positive zero of V 0 (t), which is at t=2.
Because V (2) = 4, the tidal volume is 6 4 = 2L.
Copyright c 2019 Pearson Education, Inc.
6.1. Velocity and Net Change
583
Note that the general solution for N (t) is
6.1.52
N (t) = N (0) +
Z t
0
Z t
(A sin(2⇡x/P ) + r) dx
0
◆ t
PA
cos(2⇡x/P ) + rx
2⇡
0
PA
(1 cos(2⇡t/P )).
= N (0) + rt +
2⇡
= N (0) +
✓
N 0 (x) dx = N (0) +
100
a. Using the general solution above, we have N (t) = 10 +
(1
⇡
is always 10 or more.
cos(⇡t/5)). This is never 0, because it
100
(1
b. Using the general solution above, we have that N (t) = 100 +
⇡
never extinct, because it is always 100 or more.
250
(1
c. Using the general solution above, we have N (t) = 10 + 5t +
⇡
is always at least 10, so the population never becomes extinct.
cos(⇡t/5)). The population is
cos(⇡t/5)). Again, for t
0 this
250
(1 cos(⇡t/5)). Suppose t = 5k
⇡
for a positive even integer k. Then cos(⇡t/5) = cos(k⇡) = 1, so N (t) = N (0) 25k + 0, which grows
negatively without bound as k ! 1. Thus, there is no choice of N (0) which will ensure that the
population won’t become extinct.
d. Using the general solution above, we have N (t) = N (0)
5t +
6.1.53
✓
◆ 24
2400
a. E =
(300 200 sin(⇡t/12)) dt = 300t +
cos(⇡t/12)
= 7200 MWh. This is equivalent to
⇡
0
0
6
6
13
7.2 ⇥ 10 · 3.6 ⇥ 10 = 2.592 ⇥ 10 Joules.
Z 24
7.2 ⇥ 106 KWh
= 16, 000 kg coal needed.
450 Kwh/kg
For one year, 16000 kg ⇥ 365 = 5, 840, 000 kg coal needed.
b. For one day,
7.2 ⇥ 106 KWh
= 450 g U-235 need.
1.6 ⇥ 104 Kwh/g
For one year, 450 ⇥ 365 = 164, 250 g needed.
c. For one day,
d.
7.2 ⇥ 106 KWh/day
= 1500 turbines.
(200 KW/turbine) · (24 hours/day)
6.1.54
a. Photosynthesis occurs during daylight hours. So from approximately 5 AM to 5 PM, plants and trees
take in CO2 to produce oxygen, reducing the amount of CO2 in the atmosphere.
Z 24
b.
N (t) dt ⇡ 2(0.13) + 2(0.14) + 2(0) + 2( 0.4) + 2( 0.42) + 2( 0.39) + 2( 0.3) + 2( 0.21) + 2(0) +
0
2(0.24) + 2(0.17) + 2(0.16) =
1.76.
About 1.76 grams of carbon per square meter are taken out of the coniferous forest during an average
July day. (Answers will vary.)
6.1.55
a. The additional cost is
Z 150
100
(2000
.5x) dx =
✓
2000x
x2
4
◆ 150
= 96875 dollars.
100
Copyright c 2019 Pearson Education, Inc.
584
b. The additional cost is
Chapter 6. Applications of Integration
Z 550
(2000
Z 150
(200
.5x) dx =
500
✓
2000x
✓
200x
x2
4
◆ 550
= 86875 dollars.
500
6.1.56
a. The additional cost is
.05x) dx =
100
b. The additional cost is
Z 550
(200
✓
.05x) dx =
500
x2
40
x2
40
200x
◆ 150
= 9687.5 dollars.
100
◆ 550
= 8687.5 dollars.
500
6.1.57
a. The additional cost is
Z 150
(300 + 10x
✓
2
.01x ) dx =
100
b. The additional cost is
Z 550
(300 + 10x
Z 150
(3000
✓
2
.01x ) dx =
500
x3
300
2
300x + 5x
x3
300
2
300x + 5x
◆ 150
= 69583.33 dollars.
100
◆ 550
= 139583.33 dollars.
500
6.1.58
a. The additional cost is
x
.001x2 ) dx =
100
b. The additional cost is
Z 550
(3000
x
2
.001x ) dx =
500
✓
3000x
.5x2
3000x
2
✓
.5x
x3
3000
x3
3000
◆ 150
= 142958.33 dollars.
100
◆ 550
= 109958.33 dollars.
500
6.1.59
a. False. This would only be the case if the motion was all in the same direction. If the object changes
direction at all, then the distance traveled is greater than the displacement.
b. True. This is because v(t) = |v(t)| in this case.
c. True. This is because R(t) > 0 for 0 < t < 10, but R(t) < 0 for t > 10.
Z B
d. True. The cost of increasing production from A to B is given by
C 0 (t) dt, which is geometrically
A
the area under the curve y = C 0 (x) from A to B. If C 0 is positive and decreasing, there is more area
under the curve from A to 2A than from 2A to 3A.
Z 8
8
6.1.60 The distance traveled is
(2t + 6) dt = t2 + 6t
= 112. So the same distance could have been
0
0
112
= 14.
8
◆
✓
◆ 4
Z 4✓
t2
t3
8
1
dt = t
= . So the same distance could have been
6.1.61 The distance traveled is
3
16
48
0
0
8/3
2
= .
traveled over the given time period at a constant velocity of
4
3
Z ⇡
⇡
6.1.62 The distance traveled is
2 sin t dt = ( 2 cos t)
= 4. So the same distance could have been
traveled over the given time period at a constant velocity of
0
0
traveled over the given time period at a constant velocity of
Z 5 p
6.1.63 The distance traveled is
t 25
Z 25
p
✓
u du =
1 3/2
u
3
◆ 25
125
. So the same distance
3
0
0
0
125/3
25
could have been traveled over the given time period at a constant velocity of
=
.
5
3
t2 dt =
1
2
4
.
⇡
Copyright c 2019 Pearson Education, Inc.
=
6.1. Velocity and Net Change
585
6.1.64
�
�
��
��
��
�
��
�
�
�
�
�
�
�
a.
R t 15
sK (t) = 0 (x+1)
2 dx = 15
�
�
�
�
�
b. R
t
20
sS (t) = 0 (x+1)2 dx = 20
15
15t
t+1 = t+1 .
�
�
20
20t
t+1 = t+1 .
15t
20t
4
+
= 20, which occurs for t = . (Which represents 1:20 PM.) At this
t+1
3
t+1
20
60
80
time, Kelly has gone sK (4/3) =
=
km, and Sandy has gone sS (4/3) =
km.
7/3
7
7
c. They meet when
D
At
Bt
+
= D to have a solution. If we solve for t, we obtain t =
A+B
t+1 t+1
we need A + B > D.
d. We would need
e. The maximum distances are A and B respectively, because lim
At
t!1 t + 1
= A and lim
Bt
t!1 t + 1
D
, so
= B.
6.1.65
v
15
vS
10
vT
5
0
1
2
b. After 1 hour, Theo has ridden 1·10 = 10 miles,
and Sasha has ridden 12 ·15 = 7.5 miles, so Theo
has ridden farther.
t
a.
c. After 2 hours, Theo has ridden 2 · 10 = 20 miles, and Sasha has ridden 7.5 + 15 · 1 = 22.5 miles, so
Sasha has ridden farther.
d. The times when they arrive at the various mile markers are in the following table:
10
15
20
Theo
1
3/2
2
Sasha
7/6
3/2
11/6
Note that Theo hits the 10 mile marker first, then they are
tied as they hit the 15 mile marker, and Sasha hits the 20
mile marker first. The area under vS is the same as the
area under vT for t = 1.5, for t < 1.5 the area under vT is
greater, and for t > 1.5, the area under vS is greater.
e. Theo will then hit the 20 mile mark in 18.8/10 = 1.88 hours. Sasha hits the 20 mile mark at t =
11/6 ⇡ 1.833 hours, so Sasha will win.
f. A head start of 0.2 hours is equivalent for Theo of 10 · 0.2 = 2 miles. It will take him 18/10 = 1.8
hours to ride the other 18 miles, while it still takes Sasha about 1.83 hours to cover 20 miles, so Theo
will win.
Copyright c 2019 Pearson Education, Inc.
586
Chapter 6. Applications of Integration
6.1.66
Z t
t
4
dx = (4 ln(x + 1)) = 4 ln(t + 1).
x
+
1
0
0
Z t
t
2
sB (t) = 2 +
dx = 2 + (2 ln(x + 1)) = 2 + 2 ln(t + 1).
0 x+1
0
a. sA (t) =
b. This would occur if 4 ln(t + 1) = 2 + 2 ln(t + 1), or ln(t + 1) = 1, which occurs for t = e
about 1 hour and 43 minutes.
1, which is
6.1.67 Let D(t) equal the depth of the snow t hours after it started snowing and let t = T correspond to
noon. This means that D(0) = 0 and because the snow piles up on the road at a constant rate, D0 (t) = k
for some positive constant k. It follows that
D(t) = D(0) +
Z k
t
= kx
0
= kt.
0
Let v(t) equal the velocity of the plow t hours after it started snowing. The plowing rate is inversely
C
for some constant C and for t
T.
proportional to the depth of the snow which means that v(t) =
kt
Because the plow goes twice as far from noon to 1 PM as from 1 PM to 2 PM,
Z T +2
Z T +1
C
C
dt.
dt = 2
kt
kt
T
T +1
✓
◆
✓
◆
✓
◆
✓
◆2
C
T +1
2C
T +2
T +1
T +2
ln
=
ln
, or ln
= ln
. This
Evaluating the integrals, we have
T +1
T
T +1
k
T
k
3
2
implies that (T + 1) = T (T + 2) , Expanding both sides
p and collecting terms gives the quadratic equation
1
+
5
⇡ 0.618 hours ⇡ 37 minutes. So it started snowing
T 2 + T 1 = 0, which has a positive root of T =
2
at approximately 11:23 AM.
6.1.68
a. y(t) is the position of the projectile, and the derivative of position is velocity, and the derivative of
velocity is acceleration. Note that the acceleration force is due to gravity, and only depends on the
position y of the projectile.
b.
1
dv
dy dv
dv
1 d
v 2 = · 2v
=
·
=
.
2 dy
2
dy
dt dy
dt
1 d 2
1 d 2
dv
dv
=
(v ), we must have
(v ) = a(y).
= a(y) and
2 dy
2 dy
dt
dt
Z
Z
Z
1 2
g
1
1
d 2
dy = gR(
) + D. Now when t = 0, we
d.
(v ) dy = a(y) dy, so (v ) =
1 + y/R
2
dy
2
(1 + y/R)2
1
1
have v = v0 and y = 0, so v02 = gR + D, so D = gR + v02 , and we can write
2
2
✓
◆
1 2
1
2
v0 ) = gR
1 .
(v
2
1 + y/R
c. Because
e. When v = 0 we have
1
2gR v02
2gR
1 2 1
1
v0 ·
+1 =
, so
=
, so 1 + y/R =
, and
1 + y/R
2gR
2gR v02
2
gR
1 + y/R
y=
2gR2
2gR v02
R=
2gR2
2gR v02
2gR2
2gR
Rv02
Rv02
=
.
2
v0
2gR v02
Copyright c 2019 Pearson Education, Inc.
6.2. Regions Between Curves
587
�
f.
ymax (500) ⇡ 12, 780 m.
ymax (1500) ⇡ 116, 893 m.
ymax (5000) ⇡ 1, 592, 990 m.
��� ���
�
��
����
g. The denominator of the expression for ymax is 0 when 2gR = v02 , so when v0 =
ymax ! 1.
6.1.69 Using the Fundamental Theorem, we have
Z b
f 0 (x) dx = f (b) f (a) = g(b)
a
g(a) =
Z b
p
2gR. As v0 !
p
2gR,
g 0 (x) dx.
a
6.1.70 Let f (t) represent the position of the first runner at time t and let g(t) represent the position of the
second runner over the time interval [a, b]. We are given that f (a) = g(a) and f (b) = g(b). Then by problem
69, we have
Z b
Z b
f 0 (t) dt =
g 0 (t) dt,
a
a
so the displacements of the two runners is the same, even though it might not be the case that f 0 (t) = g 0 (t)
for every t.
6.1.71 If f (x) is the elevation of trail one at position x and g(x) is the elevation of trail two at position x,
then because we are given that f (a) = g(a) and f (b) = g(b), we must have that
Z b
Z b
0
f (x) dx =
g 0 (x) dx,
a
a
so the two trails have the same net change in elevation.
6.1.72 Let f (x) = 12 sin(⇡x2 ) and g(x) = x10 (2 x)2 . Note that f (0) = 0 = g(0) and f (2) = 0 = g(2).
Z 2
Z 2
Thus by problem 69,
f 0 (x) dx =
g 0 (x) dx, which is equivalent to the statement we are trying to prove.
0
6.2
6.2.1
Regions Between Curves
Z b
(f (x)
g(x)) dx +
a
6.2.2
a.
Z 4✓
p
Z 2
Z c
(g(x)
f (x)) dx.
b
0
b.
0
(2y
x
◆
1
x dx.
2
y 2 ) dy.
0
Copyright c 2019 Pearson Education, Inc.
588
Chapter 6. Applications of Integration
6.2.3
6.2.4
�
�
�
�
�
�
�
�
�
�
�
-�
-���
���
-���
6.2.5
a. The area is given by
���
Z 1
���
((2
x)
0
�
-�
x) dx =
Z 1
1
(2
2x) dx = 2x
x2
0
1
· 2 · 1 = 1.
2
b. The area is that of a triangle with base 2 and height 1, so its area is
6.2.6
a. The area is given by
Z 1
((y + 1)
0
2y) dy =
Z 1
0
(1
✓
y) dy = y
y2
2
◆ 1
b. The area is that of a triangle with base 1 and height 1, so its area is
6.2.7
Z 1
0
6.2.8
Z 1
0
y dy +
Z 2
(2
= 1.
0
=
0
1
.
2
1
1
·1·1= .
2
2
y) dy.
1
x
dx +
2
Z 2⇣
1
x
2
⌘
x + 1 dx.
6.2.9 The curves intersect when x = x2 2, or (x + 1)(x 2) = 0, so at x = 1 and x = 2. The area is
✓
◆ 2
Z 2
x2
x3
= 4.5.
(x (x2 2)) = 2x +
2
3
1
1
6.2.10 The curves intersect when x2 2x = x2 + 4x, or 2x2 6x = 0. This can be written as 2x(x 3) = 0,
so x = 0 and x = 3 are the x-values of the points of intersection. The area is given by
✓
◆ 3
Z 3
Z 3
2x3
2
2
2
2
( x + 4x (x
2x)) dx =
( 2x + 6x) dx =
+ 3x
= 18 + 27 = 9.
3
0
0
0
6
= 3x2 , or 3x2 (x2 +1) = 6. This can be written as 3x4 +3x2 6 = 0,
x2 + 1
or (x4 + x2 2) = 0. Factoring gives (x2 + 2)(x2 1) = (x2 + 2)(x 1)(x + 1) = 0, so the curves intersect
at x = 1 and x = 1. Using symmetry, the area is given by
◆
✓
◆
Z 1✓
1
6
6⇡
2
1
3
2
3x
dx = 2 6 tan x x
=2
1 = 3⇡ 2.
4
x2 + 1
0
0
6.2.11 The curves intersect when
6.2.12 The curves intersect where sin x = cos x, which is where tan x =
given by
p
p
Z 3⇡/4
3⇡/4
2
2
=
+
(sin x ( cos x)) dx = ( cos x + sin x)
2
2
⇡/4
⇡/4
Copyright c 2019 Pearson Education, Inc.
1, or
p
2
2
⇡
3⇡
and
. The area is
4
4
p !
p
2
= 2 2.
2
6.2. Regions Between Curves
589
6.2.13 By inspection, the curves intersect at x = 1. The area is given by
✓
◆ 1
✓
◆
Z 1
2x
1
2
1
5
x
2
=3
=
(3 x 2 ) dx = 3x x /2
2
ln
2
2
ln
2
2
ln
0
0
1
.
ln 2
6.2.14 The given curve is 0 when sin x = 0 (because cos x 2 is never zero), which occurs for x = 0 and
x = ⇡. The area is given by
Z ⇡
Z ⇡
(0 (cos x 2) sin x) dx =
(cos x 2) sin x dx.
0
Let u = cos x
0
2. Then du =
sin x dx, and substituting gives
Z 3
3
u2
9 1
=
= 4.
u du =
2 1
2 2
1
6.2.15 The curves intersect at ⇡/4, so the area is given by
Z ⇡/4
sin x dx +
0
⇡/2
(sin x)
= (1
p
p
2/2) + (1
2/2) = 2
p
Z ⇡/2
⇡/4
cos x dx = ( cos x)
⇡/4
+
0
2.
⇡/4
6.2.16 The curves intersect when x = x3 , or x(x 1)(x + 1) = 0, so at x = 0, x = 1 and x = 1. The area
is given by
Z 0
Z 1
Z 1
1
1
3
3
(x
x) dx +
(x x ) dx = 2
(x x3 ) dx = 2 x2 /2 x4 /4
= .
2
1
0
0
0
6.2.17 The curves intersect when x2 4x = 2x, or x2 2x = x(x
is given by
✓
Z 2
Z 2
( 2x (x2 4x)) dx =
(2x x2 ) dx = x2
0
2) = 0, so x = 0 and x = 2. The area
x3
3
0
◆ 2
=4
0
8
4
= .
3
3
2
sec x
1
1
6.2.18 The curves intersect when
= 4 cos2 x, which can be written as cos4 x =
= 4 . This occurs
4
16
2
1
when cos x = ± , which occurs at ⇡/3 and ⇡/3. Using symmetry, the area is given by
2
Z ⇡/3
Z ⇡/3
2
2
2
(4 cos x (1/4) sec x) dx = 2
(2(1 + cos 2x) (1/4) sec2 x) dx
0
0
⇡/3
= 2 (2x + sin 2x
(1/4) tan x)
0
2
4⇡ p
=
+ 3
3
p
p
3
4⇡
3
=
+
.
3
2
2
2
6.2.19 The curves intersect when 2y = y
3, or y
2y 3 = (y 3)(y + 1) = 0. This occurs for y = and
y = 1. The area is given by
✓
◆ 3
Z 3
y3
1
32
(2y (y 2 3)) dy = y 2
+ 3y
= (9 9 + 9) (1 +
3) =
.
3
3
3
1
1
6.2.20 The area is given by
Z 4⇣
0
p
y
y⌘
dy =
4
✓
2y 3/2
3
y2
8
◆ 4
0
=
16
3
2=
2
10
.
3
6.2.21 Note that the two curves intersect where y = 2 y , so for y 2 + y 2 = (y + 2)(y 1) = 0. So they
intersect for y = 2 and y = 1. The area is given by
✓
◆ 1
✓
◆ ✓
◆
Z 1
2 3
2
16
2
2
(4 2y 2y ) dy = 4y y
y
= 4 1
8 4+
= 9.
3
3
3
2
2
Copyright c 2019 Pearson Education, Inc.
590
Chapter 6. Applications of Integration
sin 2y and cos y intersect where
sin 2y = cos y, but sin 2y = 2 sin y cos y, so we
1
must solve 2 sin y cos y = cos y. So either cos y = 0 or sin y =
. The first positive value of y satisfying
2
⇡
⇡
1
where sin y =
.
either of these conditions is y = , where cos y = 0; the first negative value of y is y =
2
6
2
So the area is
✓
◆ ⇡/2
Z ⇡/2
1
cos 2y
(cos y ( sin 2y)) dy = sin y
2
⇡/6
⇡/6
⇣ ⇡⌘ 1
⇣ ⇡⌘
⇡ 1
+ cos
= sin
cos ⇡ sin
2
3
2
2
6
1 1 1
9
=1+ + + = .
2 2 4
4
6.2.23 The area is given by
◆
✓
◆ 5
Z 5✓
(y 2)2
y2
(y 2)3
25
81
8 y
dy = 8y
= 40
3 ( 32 8 + 24) =
.
9
2
2
3
2
4
4
6.2.22 The curves
6.2.24 The curves intersect where 2 sin 2y = 2 cos y, or 4 sin y cos y = 2 cos y. If cos y 6= 0, this reduces to
1
⇡
⇡
sin y = , which occurs for y = . If cos y = 0, then we have y = ± . The area is therefore given by
2
2
6
Z ⇡/2
Z ⇡/6
(2 cos y 2 sin 2y) dy +
(2 sin 2y 2 cos y) dy
⇡/2
⇡/6
⇡/6
= (2 sin y + cos 2y)
=
✓
1+
◆
⇡/2
+ ( cos 2y
⇡/2
✓
1
+2+1 + 1
2
2+
1
+1
2
2 sin y)
◆
⇡/6
= 5.
6.2.25 The two curves meet at x = 0; at that point y = 1. So the integration is from y = 0 to y = 1. Solving
for x gives the curves x = 2y 2 2 and x = 1 y 2 . Thus the area of the region is
Z 1
Z 1
1
2
2
((1 y ) (2y
2)) dy =
( 3y 2 + 3) dy =
y 3 + 3y
= 2.
0
0
0
6.2.26 Note that (by inspection) the curves intersect for y = 1 and y = 1. Using symmetry, the area is
✓
◆ 1
✓
◆
Z 1
y3
y2
1 1
2
7
=2 2
=3
= .
2
((2 y 2 ) y) dy = 2 2y
2
3
3
3
2
3
0
0
6.2.27 The curves intersect when x2 2 = x, or x2 x 2 = (x 2)(x + 1) = 0, or x = 1 and x = 2. The
relevant point for the shaded region is ( 1, 1).
The area is
p
p
✓ 2
◆ 0
✓
◆
Z 0
Z 0
p
p
y
2(y + 2)3/2
2 8
1 2
8 2 7
y + 2)) dy =
(y + y + 2) dy =
+
=
+
=
.
(y (
2 3
6
2
3
3
1
1
1
6.2.28 Note that sin 2x = 2 sin x cos x, so the curves intersect where 2 sin x cos x = sin x, so the curves
1
intersect at x = 0 and x = ⇡, and x = ⇡/3, because cos ⇡/3 = . The area is given by
2
Z ⇡
Z ⇡/3
(sin 2x sin x) dx +
(sin x sin 2x) dx
0
⇡/3
⇡/3
= ( cos(2x)/2 + cos x)
=
✓
1 1
+
4 2
✓✓
1
2
◆
⇡
+ ( cos(x) + cos(2x)/2)
0
◆◆ ✓
✓ ◆
1
+1
+ 1+
2
✓✓
1
2
◆
⇡/3
+
✓
1
4
Copyright c 2019 Pearson Education, Inc.
◆◆◆
=
1 9
5
+ = .
4 4
2
6.2. Regions Between Curves
591
6.2.29 The curves x = y 2 3y + 12 and x = 2y 2 6y + 30 intersect where y 2 3y + 12 = 2y 2 6y + 30,
or 3y 2 + 3y 18 = 3(y + 3)(y 2) = 0. So the points of intersection are y = 3 and y = 2. Thus the area is
Z 2
Z 2
(( 2y 2 6y + 30) (y 2 3y + 12)) dy =
( 3y 2 3y + 18) dy
3
=
✓
=
3
3 2
y + 18y
2
y3
8
◆ 2
3
27
125
27 +
+ 54 =
.
2
2
6 + 36
6.2.30 These two curves intersect where y 3 4y 2 + 3y = y 2 y, or y 3 5y 2 + 4y = y(y 4)(y 1) = 0.
The three intersection points occur where y = 0, y = 4, and y = 1. Note that from y = 0 to y = 1 we have
y 2 y  y 3 4y 2 + 3y; from y = 1 to y = 4 this is reversed. Thus the area between the curves is
Z 4
Z 1
(y 3 4y 2 + 3y (y 2 y)) dy +
(y 2 y (y 3 4y 2 + 3y)) dy
0
1
=
Z 1
3
(y
2
5y + 4y) dy +
0
✓
( y 3 + 5y 2
4y) dy
1
◆ 4
1 4 5 3
1 4 5 3
2
2
=
y
y + 2y
+
y + y
2y
3
4
3
4
0
1
1 5
320
1 5
71
=
+ 2 64 +
32 +
+2=
.
4 3
6
4 3
3
6.2.31
a. The area is given by
Z 1
p
( x
◆ 1
Z 4
✓
x3 ) dx.
0
b. The area can also be written
Z 1
p
(3y
y 2 ) dy.
0
6.2.32
a. The area is given by
Z 2
( (x
2
4x)) dx +
0
Z 4
( (2x
8)) dx.
2
Z 0
p
y + 4)) dy. Note that in order to solve y = x2
p
for x, we needed to complete the squarepto obtain y + 4 = x2 4x + 4 = (x 2)2 , so y + 4 + |x
and the part of this we need is x = 2
y + 4.
✓
◆ 1
Z 1⇣
⌘
p
20 3/2 7t2
20 7
6.2.33 The area is
10 t 7t dt =
t
=
⇡ 3.17 km.
2
3
2
3
0
0
b. The area can also be written as
(y/2 + 4
4
(2
4x
2|,
The area under the curve v1 equals the distance traveled by the first runner and the area under the curve
v2 is the distance traveled by the second runner. So the area between the curves is the di↵erence in the
distance covered by the faster runner and the slower runner, which means that the faster runner ran about
3.17 km further than the slower runner.
6.2.34
a. The area of R1 is
Z 1
0
(3
x
2
2x ) dx =
✓
3x
x2
2
2x3
3
◆ 1
0
=3
1
2
2
11
=
.
3
6
9
b. The area of the R1 and R2 together is the area of a triangle with base 3 and height 3 so it is .
2
9 11
16
8
Therefore, the area of R2 is
=
= .
3
2
6
6
Copyright c 2019 Pearson Education, Inc.
592
Chapter 6. Applications of Integration
6.2.35
a. The area of R1 is given by
Z 1
x2 )2 dx. Let u = 2
6x(2
x2 . Then du =
2x dx, and substituting
0
yields
Z 1
1
3u2 du =
u3
2
=
1
( 8) = 7.
2
b. The area of R1 minus the area of a triangle with base 1 and height 6 is equal to the area of R2 . Thus,
the area of R2 is 7 3 = 4.
6.2.36
a. The area of R1 is
Z ⇡/2
⇡/2
cos y dy = sin y
0
= 1.
0
b. Together, R1 and R2 comprise a 1 ⇥
⇡
⇡
rectangle. So the area of R2 is
2
2
1.
�
The nonvertical lines intersect when 4x + 4 = 6x +
6, or x = 1. The vertical line x = 4 intersects
both nonvertical
lines when x = 4. The area is
Z
��
��
4
6.2.37
(6x + 6 (4x + 4)) dx =
given by
1
Z 4
4
(2x + 2) dx = x2 + 2x
=
1
16 + 8
��
��
�
1
(1
2) = 25.
�
-�
�
�
�
�
1
6.2.38
cos x = sin x when tan x = 1, or x = tan (1) =
⇡/4. They also intersect at ⇡/4 + ⇡ = 5⇡/4.
Z 5⇡/4
The area is given
(sin x
cos x) dx =
⇡/4
5⇡/4
( (cos x + sin x))
=
⇡/4
p
p
p
( 2/2 + 2/2)) = 2 2.
((
p
2/2
p
2/2)
���
�
�
�
�
�
-���
-���
�
6.2.39
The curves ex and e 2x intersect when ex = e 2x ,
or e3x = 1, which occurs only for x = 0. The
vertical line x = ln 4 clearly intersects the curves
Z ln 4
for x = ln 4. The area is given by
(ex
ln 4
e 2x ) dx = ex + e 2x /2
(1/2)) = 81/32.
�
�
0
= 4 + (1/32)
�
(1 +
0
-���
Copyright c 2019 Pearson Education, Inc.
���
���
�
6.2. Regions Between Curves
593
�
6.2.40
��
The curves intersect for 3x2 6x = 6x, or 3x2
12x = 3x(x 4) = 0, so for x = 0 and x = 4.
Z 4
Z 4
(6x (3x2 6x) dx =
(12x
The area is
0
2
3x ) dx = (6x
0
4
2
3
x )
��
= 96
��
64 = 32.
0
�
�
�
�
�
�
6.2.41
2
The curves intersect when
= 1, or x2 +
1 + x2
1 = 2, or x2
1 = (x 1)(x + 1) = 0. So
the intersections occur when
Z 1 ✓ x = ±1. ◆ Using
2
1 dx =
symmetry, the area is 2
1 + x2
0
���
���
���
1
2 2 tan
1
x
x
=⇡
2.
0
-���
-���
���
-���
�
���
�
6.2.42
p
The curves intersect when 24 x = 3x2 , so when
either x = 0 or when 8 = Z
x3/2 . Thus happens for
4
p
x = 82/3 = 4. We have
24 x 3x2 dx =
⇣
16x3/2
x3
⌘ 4
��
��
��
0
= 128
��
64 = 64.
��
0
�
�
�
�
�
�
6.2.43
The curves intersect when x = 1/x, or x2 = 1.
This occurs in the first quadrant when x = 1. The
Z 1
Z 2
1
area is given by
x dx+
1/x dx = x2 /2
+
2
(ln x)
0
= 1/2 + ln 2.
1
�
�
0
�
1
���
Copyright c 2019 Pearson Education, Inc.
���
���
�
594
Chapter 6. Applications of Integration
Note that the curves intersect when 4x x2 =
4x 4, or x2 4 = 0. We see that the curves
intersect for x > 0 when x = 2. The line
y = 4x 4 intersects
the x-axis atZ x = 1. The
Z
1
6.2.44
0
1
x3 /3
2
+ 4x
x3 /3
1/3) =
10
.
3
0
(8
(4
8/3)
�
2
area is given by
(4x x2 ) dx +
(4x x2
0
1
Z 1
Z 2
2
(4x x ) dx +
(4 x2 ) dx =
(4x 4)) dx =
2x2
�
�
�
1
= (2
�
1/3) +
1
���
���
���
���
���
�
�
6.2.45
The curves intersect at x = ±1. Using symmetry,
the area is
✓
◆ 1
Z 1
1 2 1 2
7
2
(2 x x2 ) dx = 2 2x
x
x
= .
2
3
3
0
0
���
���
���
-���
���
-���
���
�
�
��
Using symmetry, the area is 2
6.2.46
4
2 9x /2
x /4
��
x3 ) dx =
(9x
0
3
2
Z 3
= 2(81/2
-�
-�
�
-�
81/4) = 81/2.
�
�
-��
0
-��
-��
Z 6
6.2.47
Z2 6
(x/2
|x
(x/2
(x
3
x/2) dx =
�
Z 3
3|) dx =
(x/2 (3 x)) dx +
Z 6
Z 32
3)) dx =
(3x/2 3) dx +
(3
2
2
3x /4
3
3
3x
+ 3x
2
6
2
x /4
�
�
=
3
27/4 9 (3 6)+(18 9 (9 9/4)) = 3/4+9/4 =
3.
�
�
Copyright c 2019 Pearson Education, Inc.
�
�
�
�
�
�
6.2. Regions Between Curves
595
�
The curves intersect for 3x x3 = x, or x3 4x =
x(x2 4) = x(x 2)(x + 2) = 0, so for x = 0 and
x = ±2. Using symmetry, the area is given by
6.2.48
2
Z 2
✓
(3x x3 +x) dx = 2 2x2
0
x4
4
◆ 2
�
-�
�
�
-�
= 16 8 = 8.
-�
0
-�
6.2.49
�
Z 8
The area is given by
⇣
4x
3x
5/3
/5
⌘ 8
(4
x2/3 ) dx
=
0
= 32
�
�
96/5 = 64/5.
0
�
�
�
�
�
6.2.50
�
Z ⇡/2
The area is given by
2 sin x) dx =
0
⇡/2
(2x + 2 cos x)
(2
=⇡
���
���
2.
0
���
���
�
���
6.2.51
�
The area is given by
ln 2
2e x + x
ln 2.
ex
Z ln 2
ex ) dx =
0
=
0
���
(2e x + 1
1+ln 2 2 ( 2 1) =
���
���
���
���
���
Copyright c 2019 Pearson Education, Inc.
���
���
���
�
596
Chapter 6. Applications of Integration
6.2.52
�
���
Z ⇡/4
The area is given by
tan x)
���
0
⇡/4
(2x
���
sec2 x) dx =
(2
=
0
⇡
2
���
1.
���
���
���
���
���
���
���
���
���
�
6.2.53
�
Z p3/2
���
p
The area is given by
(2x 1 x2 x) dx =
0 p
✓
◆ 3/2
2
x2
2 3 2
5
2 3/2
(1 x )
=
+ =
.
24
8
3
24
3
2
0
���
���
���
���
���
���
���
�
6.2.54
�
The area is given by
4
3y 2 /2
Z 4
(3y
y 3 /3 + 4y
4) = 44
65/3
4)) dy =
�
= 24 64/3 +16 (3/2 +
1
1/3
(y 2
1
3/2 =
�
125
.
6
��
�
-�
-�
6.2.55
�
The curves intersect at the point (9, 3). The
◆
Z 3✓
y 15
+
y 2 dy =
area is given by
2
2
✓ 2
◆ 3 0
3
9 45
63
y
15y y
+
= +
9=
.
4
2
3
4
2
4
0
�
�
�
�
�
Copyright c 2019 Pearson Education, Inc.
�
�
�
�
6.2. Regions Between Curves
597
6.2.56
�
1
Note that the curves intersect for 2 = p
,
1 x2
p
1
3
. Using symmetry,
or 1 x2 = , or x = ±
4
2✓
p
◆
Z 3/2
1
2 p
dx =
the area is given by 2
1 x2
p 0
3/2
p
2 2x sin 1 (x)
= 2 3 2⇡/3.
���
���
���
���
���
���
0
-���
���
-���
���
�
6.2.57
�
The curves intersect for x2 2x + 1 = 5x 9, or
x2 7x + 10 = (x 5)(x 2) = 0, so for x = 2
Z 5
and x = 5. The area is given by
(5x 9 (x
2
2
2
1) ) dx = 5x /2
45
64/3
(10
9x
18
5
3
(x
1) /3
(1/3)) = 4.5.
= 125/2
��
��
��
�
2
�
�
�
�
�
�
-�
6.2.58
�
Using symmetry, this is 2
y 3 /3 + y 2 /2
���
( y(y
���
1)) dy =
0
1
2
Z 1
���
���
= 1/3.
���
0
-���
���
-���
���
�
-���
6.2.59
�
This is given by
Z 4
0
2
y ) dy = 2y
2
(y 2
(3y
y /3
= 32
0
Z 4
�
(4y
0
4
3
y)) dy =
�
64/3 = 32/3.
�
�
�
Copyright c 2019 Pearson Education, Inc.
�
�
�
��
�
598
Chapter 6. Applications of Integration
6.2.60
�
The area is given by
Z ⇡
0
cos x
3
2
x2 + ⇡x) dx =
(sin x
���
���
⇡
x /3 + ⇡x /2
3
=1
⇡ /3 + ⇡ /2
���
���
���
���
�
-���
0
( 1) = 2 + ⇡ 3 /6.
���
-���
3
-���
-���
-���
6.2.61
�
This area is given by
Z 2
(5/2
1/x
x) dx =
1/2
���
2
5x/2
ln x
2
x /2
���
=5
ln 2
2
(5/4
���
1/2
ln(1/2)
1/8) = 15/8
2 ln 2.
���
���
���
���
���
�
6.2.62
�
This area is given by
Z 1/2
Z 2
(4x
x/4) dx +
(1/x
0
x/4) dx
1/2
1/2
15x2 /8
+ ln x
(ln 1/2
=
���
2
x2 /8
= 15/32+ln 2
0
1/2
���
1/2
���
���
1/32) = 2 ln 2.
���
���
���
�
6.2.63 y = 8x and y = 9 x2 intersect when x2 + 8x 9 = (x + 9)(x 1) = 0, so at x = 1 in the first
quadrant. y = 52 x and y = 9 x2 intersect when x2 + 52 x 9 = 0, or 2x2 + 5x 18 = (2x + 9)(x 2) = 0,
so at x = 2 in the first quadrant. Thus the area of the region is
◆
◆
Z 2✓
5
5
x dx +
9 x2
x dx
2
2
0
1
◆
Z 2✓
Z 1
11
5
=
x dx +
9 x2
x dx
2
2
0
1
✓
◆ 1 ✓
◆ 2
11 2
x3
5 2
=
x
+ 9x
x
4
4
3
0
1
11
8
1 5
17
5 9+ + =
.
=
+ 18
3 4
3
4
3
A=
Z 1✓
8x
Copyright c 2019 Pearson Education, Inc.
6.2. Regions Between Curves
599
6.2.64
A=
Z 1
p
(4 2x
( 4x + 6)) dx +
1/2
=
p
8 2 3/2
x + 2x2
3
6x
Z 2 p
(4 2x
1
!
1
+
1/2
2x2 ) dx
p
8 2 3/2
x
3
3
2x /3
!
2
= 19/6.
1
6.2.65
a. False. ThisZ can be done either with respect to x or with respect to y. For the latter, the relevant
1
integral is
y 2 ) dy.
(y
0
b. False. On the interval (0, ⇡/4) the cosine function is greater, but on (⇡/4, ⇡/2) the sine function is
Z ⇡/2
Z ⇡/4
greater. The area is
(cos x sin x) dx +
(sin x cos x) dx.
0
⇡/4
c. True. They both represent the area of the region in the first quadrant under y = x and above y = x2 .
Z a
Z a
6.2.66 Given
(f (x) g(x)) dx = 10, we have (by symmetry) that
(f (x) g(x)) dx = 5. Then,
a
0
Z pa
Z a
1
5
x(f (x2 ) g(x2 )) dx =
(f (u) g(u)) du = . (Where we used the substitution u = x2 .)
2 0
2
0
6.2.67 A = 2
Z 1
y
2
0
6.2.68 An =
Z 1
p
1
y 3 dy =
2
3
Z 1
p
u du (where u = 1
0
1
(x
xn ) dx = x2 /2
xn+1 /(n + 1)
0
6.2.69 An =
(x
Z 1
(x1/n
1/n
x) dx =
0
6.2.70 An =
=
1
2
=
n
n+1
0
Z 1
✓
xn ) dx =
0
2
y ). So A =
3
3
◆ 1
nx(n+1)/n
n+1
x /2
✓
xn+1
n+1
2
nx(n+1)/n
n+1
0
◆ 1
✓
2 3/2
u
3
◆ 1
0
=
4
.
9
1
n 1
=
.
2n + 2
n+1
=
0
1
n 1
=
.
2
2n + 2
n
n+1
6.2.71 Using the result of the previous problem, lim An = lim
1
n 1
=
.
n+1
n+1
n
1
n!1 n + 1
n!1
= 1. As n ! 1, the region in
question approaches the 1 ⇥ 1 square over the interval [0, 1], which has area 1.
1
6.2.72 R is a right triangle with both legs equal to 1, so its area is . Integrating with respect to y, we are
2
Z k
(1 y) dy is half the area of the triangle, or 14 . Now,
looking for k so that
0
Z k
0
(1
y) dy =
✓
y
y2
2
◆ k
0
=k
k2
.
2
p
2± 2
4k + 1 = 0, which has roots k =
. Only the negative sign choice
Setting k
2
p
p
2
2
2
=1
.
gives a value between 0 and 1, so we want k =
2
2
k2
1
= , we have 2k 2
2
4
Copyright c 2019 Pearson Education, Inc.
600
Chapter 6. Applications of Integration
6.2.73 This is a triangle with base 2 and height 1, so it has area 1. Note that for 0  x  1 the line which
forms the top of the triangle is y = x, and for 1  x  2, the line is y = 2 x. Integrating with respect to
Z k
1
(2 y y) dy = . Now,
y, we are looking for k so that
2
0
Z k
k
(2
2y) dy = 2y
0
y2
k2 .
= 2k
0
p
2± 2
4k + 1 = 0, which has roots k =
. Only the negative sign choice
Setting 2k
2
p
p
2
2
2
=1
.
gives a value between 0 and 1, so we want k =
2
2
1
k = , we have 2k 2
2
2
6.2.74
a. The proportion of the whole board which has the desired property is the same as the proportion of
each “quarter board” with the desired property, so we can consider only the quarter board rather than
the whole board.
b. Let P (x, y) be a point on the curve C. Let Q be the point on line segment AB so that QP ? AB.
Then, because the distance from O to P is the same as the distance from P to Q, we must have
p
1
x2 + y 2 = 1 y, so y = (1 x2 ).
2
c. The area of the region R is 1, and the area of R1
2
Z p
0
=
R is
p
◆
2 1
1
2
3
2
(1 x ) x dx = x x /3 x
2
0
p
p
p
p
p
1 ( 2 1)3 /3 ( 2 1)2 = 2 1 + 7/3 (5/3) 2 + 2 2
2 1✓
p
2
So the probability of landing in R2 is 1
p
(4 2/3
5/3) =
p
34 2/3
5/3.
p
4 2
⇡ .781.
3
8
6.2.75
a. The point (n, n) on the curve y = x would represent the notion that the lowest p% of the society owns
p% of the wealth, which would represent a form of equality.
b. The function must be increasing and concave up because the poorest p% cannot own more than p% of
the wealth.
�
���
c.
y = x1.1 is closest to y = x, and y = x4 is
furthest from y = x.
���
���
���
���
Z 1
1
L(x) dx, and A + B = 1/2, so A =
d. Note that B =
2
Z 1 0
L(x) dx.
2A = 1 2
Z 1
���
���
���
L(x) dx. Then G =
0
0
Copyright c 2019 Pearson Education, Inc.
�
A
A
=
=
A+B
1/2
6.2. Regions Between Curves
601
p
e. For L(x) = x , we have G = 1
2
Z 1
p
x dx = 1
2
0
p
1.1
1.5
2
3
4
G
1/21
1/5
1/3
1/2
3/5
f. For p = 1 we have G = 0. Because lim
p
1
p!1 p + 1
✓
xp+1
p+1
◆ 1
2
p 1
=
. So we have
p+1
p+1
=1
0
= 1, the largest possible value of G approaches 1.
g. For L(x) = 5x2 /6 + x/6, note that L(0) = 0, L(1) = 1, L0 (x) = 5x/3 + 1/6 > 0 on [0, 1], and
L00 (x) = 5/3 > 0 as well. The Gini index is
G=1 2
Z 1
1
(5x2 /6+x/6) dx = 1 2 5x3 /18 + x2 /12
0
= 1 2 (5/18 + 1/12) = 1 5/9 1/6 = 5/18.
0
6.2.76
a. In this case lQ is y = 2ax a2 , lR is y = 0, and lP is the line y = 2ax a2 . We have P 0 = (a/2, 0),
R0 = (0, a2 ), and Q0 = ( a/2, 0). The area of triangle P QR is 12 · 2a · a2 = a3 and the area of triangle
P 0 Q0 R0 is 12 · (2(a/2)) · (a2 ) = a3 /2.
b. In this case lQ is y = 2bx b2 , lR is y = 0, and lP is y = 2ax a2 . We have P 0 = (b/2, 0),
R0 = ((b a)/2, ab), and Q0 = ( a/2, 0). The area of triangle P QR is ab(a + b), and the area of
P 0 Q0 R0 is ab · a+b
2 .
c. Let the coordinates of R be (c, c2 ). Assume 0 < a < c < b (the other cases can be handled similarly.)
c2 . We have P 0 = ((b + c)/2, bc),
1
Q0 = ((c a)/2, ca), and R0 = ((b a)/2, ab). The area of P QR is (a2 b+b2 c+ab2 ac2 a2 c bc2 ).
2
1
The area of triangle P 0 Q0 R0 is (a2 b + b2 c + ab2 ac2 a2 c bc2 ).
4
b2 , lP is y =
In this case lQ is y = 2bx
2ax
a2 , and lR is y = 2cx
6.2.77
y
y
1
1
y
x1/3
x1/4
y
y
y
y
x2
0
x3
y
x1/2
1
x4
1
0
1
x
1
x
a.
b. An (x) is the net area of the region between the graphs of f and g from 0 to x.
c. Note that
c so that
R1
(f (s)
0
Rc
1
(f (s)
⇣ n+1
s(1/n)+1
(1/n)+1
⌘ 1
n
1 n
1
= n+1
n+1 = n+1 < 0. So we seek the smallest
0
⇣ n+1
⌘ c
n+1
(n+1)/n
n 1
s(1/n)+1
= nn+11 , or cn+1 = c(n+1)/n ,
g(s)) ds = n+1 . This occurs when sn+1
(1/n)+1
g(s)) ds =
s
n+1
or cn+1 = nc(n+1)/n , or cn (1/n) = n, so c = nn/(n
2
1
1)
. As n increases, this root increases as well.
Copyright c 2019 Pearson Education, Inc.
602
Chapter 6. Applications of Integration
6.2.78
�
�
�
���
���
���
���
���
a.
���
���
���
���
���
���
�
���
���
a = 1/2.
�
���
���
a = 1.
���
�
a = 2.
x
1
b. We seek a root of a sin 2x = (sin x)/a, or 2 sin x cos x = sin
a2 , so cos x = 2a2 . This has a solution as
p
long as 2a12  1, which occurs when a
1/2.
c.
A=
=
Z x?
✓
(sin 2x
0
1
(2 cos2 x
2
cos x
= cos x?
✓
sin x) dx = cos x
cos2 x? +
1
2
◆ x?
1
cos 2x
2
0
◆ x?
✓
◆ x?
1
1)
= cos x cos2 x +
2 0
◆
✓ 0
1
1
1
2a2 1
1 1+
= 2
=
.
2
2a
4a4
4a4
When a = 1 this is equal to 14 .
p
2
d. Using the result of the previous calculation, we simply note that 2a4a4 1 ! 0 as a ! 1/ 2.
6.3
Volume by Slicing
6.3.1 A(x) is the area of the cross section through the solid at the point x.
6.3.2 The general slicing method would be used.
6.3.3
p
a. A(x) = ( 3 x)2 = 3
Z 2
b.
(3 x) dx.
x.
0
6.3.4 The cross sections are disks and A(x) is the area of a disk.
6.3.5
a. The radius of a cross section is
p
b. A(x) = ⇡( cos x)2 = ⇡ cos x.
Z 2
c. V =
⇡ cos x dx.
0
p
6.3.6
cos x.
a. The radius of a cross section is cos y.
b. A(y) = ⇡ cos2 y.
Z ⇡/2
⇡ cos2 y dy.
c. V =
0
Copyright c 2019 Pearson Education, Inc.
6.3. Volume by Slicing
603
6.3.7
a. The outer radius is given by
p
6.3.8
x + 1.
a. The outer radius is 4.
b. The inner radius is 1.
⇣p
⌘
2
c. A(x) = ⇡ ( x + 1)
1 .
Z 4 ⇣
p
2
d. V =
⇡
x+1
0
6.3.9
1)2 .
b. The inner radius is (y
c. A(y) = ⇡(16 (y 1)4 ).
Z 3
⇡(16 (y 1)4 ) dy.
d. V =
⌘
1 dx.
1
6.3.10
p
a. The radius is x.
p
b. A(x) = ⇡( x)2 = ⇡x.
Z 4
c. V =
⇡x dx.
a. The radius is 4
1)2 .
(y
b. A(y) = ⇡(4 (y 1)2 )2 .
Z 3
c. V =
⇡(4 (y 1)2 )2 dy.
0
1
p
6.3.11 The p
curves intersect when 1 x2 = 0, or x = ±1.
A(x) = ( 1 x2 )2 = 1 x2 . The volume is given by
Z 1
1
(1
x2 ) dx = x
x3 /3
=
1
1
✓
1
1
3
◆
✓
1+
1
3
◆
=
4
.
3
6.3.12 The curves intersect when 2 x2 = x2 , or x = ±1.
A(x) = ((2 x2 ) x2 )2 = (2(1 x2 ))2 = 4(1 2x2 + x4 ). The volume is given by
Z 1
1
4(1
2
4
3
2x + x ) dx = 4 x
5
2x /3 + x /5
1
1
✓✓
=4
1
2 1
+
3 5
◆
✓
2
1+
3
1
5
◆◆
=
64
.
15
p
cos x
1 p
2
cos x =
. The volume is
6.3.13 The curves intersect when cos x = 0, or x = ±⇡/2. A(x) =
2
2
given by
✓
◆
Z ⇡/2
⇡/2
sin x
1
1
cos x
dx =
=
= 1.
2
2
2
2
⇡/2
⇡/2
p
6.3.14 A(x) = (2 25
V =
Z 5
x2 )2 = 100
4x2 .
Z 5
(100
A(x) dx =
0
5
4x2 ) dx = 100x
4x3 /3
0
500
1000
=
.
3
3
= 500
0
6.3.15 For each value of x, the height of the triangle is 2 x, which is the diameter of the semicircle, so the
area of that semicircle is
✓
◆2
⇡ 2 x
⇡
A(x) =
= (4 4x + x2 ).
2
2
8
Thus the volume is
V =
Z 2
A(x) dx =
0
6.3.16 The volume is V =
✓
◆
2 1
16
2 1
+
=
.
3 5
15
Z 2
0
Z 1
1
(1
2 2
⇡
(4
8
4x + x2 ) dx =
x ) dx = 2
Z 1
0
(1
2
⇡
4x
8
4
2
2x2 + x3 /3
=
0
✓
2x + x ) dx = 2 x
Copyright c 2019 Pearson Education, Inc.
⇡
.
3
2 3 1 4
x + x
3
5
◆ 1
0
=
604
Chapter 6. Applications of Integration
6.3.17 V = ⇡
Z 3
3
4x2 dx = 4⇡ x3 /3
= 36⇡.
0
6.3.18 V = ⇡
Z 1
0
1
2x)2 dx = 4⇡ x
(2
x2 + x3 /3
0
0
6.3.19 V = ⇡
Z ln 4
e 2x dx =
6.3.20 V = ⇡
Z 1/2
p
0
Z 1
6.3.22 V = ⇡
Z 2
Z 4
=
0
p
1 + y2
!2
1) =
⇡2
.
6
= ⇡(⇡/4 + ⇡/4) = ⇡ 2 /2.
1
2
e2y
⇡(e4 1)
=
.
2 0
2
p
((2 x)2
x2 ) dx = ⇡ 2x2
4
x3 /3
= ⇡(32
0
6.3.26 V = ⇡
Z ln 3
((ex/2 )2
(e x/2 )2 ) dx = ⇡
Z 4
((x + 2)2
x2 ) dx = ⇡
Z 1
Z 6
Z ln 3
y 2/3 ) dy = ⇡
(1
(y 2
Z ⇡/2
0
Z 4
y 2 /4) dy =
cos2 x dx =
⇡
2
✓
e x ) dx = ⇡ ex + e x
x2 ) dx =
= 48⇡.
0
◆ 1
0
=
2⇡
.
5
6
= 54⇡.
0
(1 + cos 2x) dx =
0
⇡
2
✓
x+
sin 2x
2
◆ ⇡/2
x2 )2 dx = 2⇡ 25x
=
0
5
x3 /3
=
0
4
500⇡
⇡ · 53 =
.
3
3
Z
Z ⇡
⇡ ⇡
2
6.3.31 V = ⇡
sin x dx =
(1
2 0
0
0
0
⇡
cos 2x) dx =
2
✓
x
sin 2x
2
◆ ⇡
=
0
⇡/4
sec2 x dx = ⇡ tan x
= ⇡.
0
Copyright c 2019 Pearson Education, Inc.
⇡2
.
4
500⇡
. The volume of a sphere
3
of radius 5 is
Z ⇡/4
(2 +
4
3⇡ 3
y /3
4
Z ⇡/2
= ⇡((3 + 1/3)
ln 2
(4x + 4) dx = ⇡ 2x2 + 4x
3 5/3
y
5
y
Z 5 p
( 25
6.3.30 Using symmetry, V = 2⇡
6.3.32 V = ⇡
p
( x
ln 3
(ex
0
0
6.3.29 V = ⇡
Z 1
0
ln 2
0
6.3.28 V = ⇡
x2 ) dx = ⇡
1/3) = ⇡/3.
0
6.3.27 V = ⇡
p 2
(( 4 x )
0
ln 2
1/2)) = 5⇡/6.
Z 1
32⇡
.
3
0
⌘ 1
x3 /3
= ⇡(2/3
6.3.25 V = ⇡
64/3) =
0
6.3.24 The curves intersect at x = 0 and x = 1. We have V = ⇡
⇣
⇡ 2x3/2 /3
15⇡
.
32
1
dy = ⇡ tan 1 y
e2y dy = ⇡
0
6.3.23 V = ⇡
0
dx = ⇡ sin 1 x
x2
1
1
⇡
((1/16)
2
=
1/2
1
1
ln 4
⇡
e 2x
2
0
6.3.21 V = ⇡
4⇡
.
3
=
⇡2
.
2
6.3. Volume by Slicing
6.3.33 V = ⇡
Z ⇡/4
0
6.3.34 V = ⇡
Z ⇡/2
605
⇡
sin y dy =
2
2
Z ⇡/4
(1
0
⇡
cos 2y) dy =
2
✓
y
sin 2y
2
◆ ⇡/4
0
⇡
=
2
✓
⇡
4
1
2
◆
=
⇡(⇡
2)
8
.
⇡/2
(1
sin x) dx = ⇡ ( x + cos x)
0
Z ⇡/2
(sin x
6.3.35 V = ⇡
0
⇣
⇡ ⌘ 4⇡ ⇡ 2
=⇡ 1
=
.
4
4
= ⇡(⇡/2
1).
0
sin2 x) dx = ⇡
Z ⇡/2
0
(sin x
1
(1
2
cos 2x)) dx = ⇡
Z 0
✓
sin 2x
4
cos x
2 2
2
x
2
◆ ⇡/2
0
Z 1
6.3.36 Note that the curves intersect at x = ±1. We have V = ⇡
((2 x )
( x) ) dx + ⇡
((2
1
0
✓
◆
Z 1
Z 1
1
5x3
x5
x2 ) 2
x2 ) dx = ⇡
(4 4x2 + x4
x2 ) dx = ⇡
(4 5x2 + x4 ) dx = ⇡ 4x
+
=
3
4
1
1
✓✓
◆ ✓ 1
◆◆
76⇡
5 1
5 1
⇡
4
+
4+
=
.
15
3 5
3 5
✓
◆ 2
Z 2
y5
128⇡
(16 y 4 ) dy = ⇡ 16y
=
.
6.3.37 V = ⇡
5
5
0
0
Z 4⇣
Z 4
4
⌘
p
y2
6.3.38 V = ⇡
22 ( 4 y)2 dy = ⇡
y dy = ⇡
= ⇡(8 0) = 8⇡.
2 0
0
0
◆2
Z 6✓
6
1
p
6.3.39 V = ⇡
dx = ⇡ ( ln x) = ⇡ ln 3.
x
2
2
✓ 3
◆ 1
Z 1
Z 1
y
5y 2
p
6.3.40 V = ⇡
(2 y)2 ( y)2 dy = ⇡
(y 2 5y + 4) dy = ⇡
+ 4y
=
3
2
0
0
✓
◆0
1 5
11⇡
.
⇡
+4 =
3 2
6
Z 2
2
⇡ 2x
⇡
6.3.41 V = ⇡
e2x dx =
e
= (e4 1).
2
2
0
0
✓
◆
Z ln 4
ln 4
⇡ 2x
⇡
1
225⇡
6.3.42 V = ⇡
(e2x e 2x ) dx =
e + e 2x
=
16 +
2 =
.
16
32
2
2
0
0
✓
✓
◆◆
Z ln 8
ln 8
1
49⇡
2y
y
2y
y
6.3.43 V = ⇡
(e
e ) dy = ⇡ e /2 e
= ⇡ 32 8
1
=
.
2
2
0
0
Z p
p
⇡
⇡
6.3.44 V (p) = ⇡
e 2x dx =
e 2x
=
1 e 2p . The volume is bounded by ⇡/2 as p ! 1.
2
2
0
0
Z 5
5
500⇡
4x2 dx = ⇡ 4x3 /3
=
.
6.3.45 About the x-axis: Vx = ⇡
3
0
Z 10 0
10
500⇡
(25 y 2 /4) dy = ⇡ 25y y 3 /12
=
.
About the y-axis: Vy = ⇡
3
0
0
The volumes are the same.
Z 2
2
32⇡
6.3.46 About the x-axis: Vx = ⇡
(4 2x)2 dx = 4⇡ 4x 2x2 + x3 /3
=
.
3
0
0
Z 4
4
16⇡
About the y-axis: Vy = ⇡
(2 (y/2))2 dy = ⇡ 4y y 2 + y 3 /12
=
.
3
0
0
The volume Vx is bigger.
Copyright c 2019 Pearson Education, Inc.
606
Chapter 6. Applications of Integration
Z 1
Z 1
1
9⇡
6.3.47 About the x-axis: Vx = ⇡
(1 x ) dx = ⇡
(1 2x3 + x6 ) dx = ⇡ x x4 /2 + x7 /7
=
.
14
0
✓ 0
◆ 1
Z 1 p0
3
3⇡
( 3 1 y)2 dy = ⇡
(1 y)5/3
= ⇡(0 3/5) =
. The volume Vx
About the y-axis: Vy = ⇡
5
5
0
0
is bigger.
Z 2
2
48⇡
6.3.48 About the x-axis: Vx = ⇡
(8x x4 ) dx = ⇡ 4x2 x5 /5
=
.
5
0
0
Z 4
4
24⇡
About the y-axis: Vy = ⇡
(y y 4 /64) dy = ⇡ y 2 /2 y 5 /320
=
.
5
0
0
The volume Vx is bigger.
Z 1
Z 1
1
p 2
x2
⇡
6.3.49 V = ⇡
1
1
x
dx = ⇡
x dx = ⇡
= .
2
2
0
0
0
6.3.50 V = ⇡
Z 1
(e
x/2
2
+2
2) dx = ⇡
0
6.3.51 V = ⇡
Z ⇡/3
3 2
Z 1
0
(2
(2
2
sec y)) dy = ⇡
0
Z 1
1
e x dx =
6.3.52 V = ⇡
(1 (1
✓
◆0
1
4
8⇡
1+
=
.
⇡
15
5
3
y) ) dy = ⇡
Z 1
0
Z ⇡/3
(y
= ⇡(1
e 1 ).
0
⇡/3
sec2 y dy = ⇡ tan y
0
2 2
⇡e x
4
p
= ⇡ 3.
0
✓ 5
y
3
2
4y + 4y ) dy = ⇡
5
4y 3
y +
3
4
◆ 1
=
0
p
6.3.53 Using the disk method, a disk located at x has a radius of 1
x when revolved about the line
y = 1, so the volume is
✓
◆ 1
✓
◆
Z 1
Z 1
p 2
p
4 3/2 1 2
4 1
⇡
V =⇡
(1
x) dx = ⇡
(1 2 x + x) dx = ⇡ x
x + x
=⇡ 1
+
= .
3
2
6
3
2
0
0
0
6.3.54 We use the washer method. Note that x = y 2 , and for each value of y, the washer has outer radius
4 and inner radius 4 y 2 , so that the volume is
✓
◆ 2
✓
◆
Z 2
Z 2
64 32
224
8 3 1 5
V =⇡
(42 (4 y 2 )2 ) dy = ⇡
(8y 2 y 4 ) dy = ⇡
y
y
=⇡
=
⇡.
5
15
3
5
3
0
0
0
6.3.55 We use the washer method. For each x, the washer has outer radius 2 + 2 sin x and inner radius 2,
so the volume is
Z ⇡
Z ⇡
V =⇡
(2 + 2 sin x)2 22 dx = ⇡
(8 sin x + 4 sin2 x) dx
0
0
Z ⇡
⇡
=⇡
(8 sin x + 2(1 cos 2x)) dx = ⇡( 8 cos x + 2x sin 2x)
0
0
= ⇡(8 + 2⇡ + 8) = 2⇡(⇡ + 8).
6.3.56 We use the washer method. First solve for x to obtain x = ey ; then for each value of y, the outer
radius of a washer is ey + 1 and the inner radius is 1, so the volume is
✓
◆ 1
Z 1
Z 1
1
V =⇡
(ey + 1)2 12 dy = ⇡
(2ey + e2y ) dy = ⇡ 2ey + e2y
2
0
0
0
✓
◆
1 2 5
⇡ 2
=
e + 4e 5 .
= ⇡ 2e + e
2
2
2
Copyright c 2019 Pearson Education, Inc.
6.3. Volume by Slicing
607
6.3.57 We use the washer method. For each x, the outer radius is 1 + sin x and the inner radius is 1 + (1
sin x) = 2 sin x. Thus the volume is
Z 5⇡/6
Z 5⇡/6
2
2
V =⇡
(1 + sin x)
(2 sin x) dx = ⇡
(6 sin x 3) dx
⇡/6
5⇡/6
= ⇡( 6 cos x
3x)
⇡/6
✓
p
=⇡ 3 3
⇡/6
p
5⇡
⇡
+3 3+
2
2
◆
⇣ p
=⇡ 6 3
⌘
2⇡ .
6.3.58 The lines y = 1 + x2 and y = x meet at the point (2, 2). For each x, we obtain a washer with outer
radius 3 x and inner radius 3
1 + x2 = 2 x2 . Hence the volume is
◆
◆
✓
◆ 2
Z 2✓
Z 2✓
⇣
x ⌘2
3 2
1 3
2
2
2
dx = ⇡
5 4x + x
dx = ⇡ 5x 2x + x
= 4⇡.
V =⇡
(3 x)
2
4
4
0
0
0
6.3.59 The lines x = 2 y and x = 1 y2 meet at the point (0, 2). For each y, we obtain a washer whose
outer radius is 3 (1 y2 ) = 2 + y2 and whose inner radius is 3 (2 y) = y + 1. Thus the volume is
◆
◆
✓
◆ 2
Z 2 ✓⇣
Z 2✓
y ⌘2
3 2
1 3
2
2+
(y + 1)
dy = ⇡
y + 3 dy = ⇡
y + 3y
= 4⇡.
V =⇡
2
4
4
0
0
0
6.3.60 We would expect the volume to be greater when revolved about the line y = 2, because the wider
portion of the figure then moves through a larger radius, so it traces out a larger volume. When revolving
about y = 0, we obtain a disk with radius 2x(2 x), so the volume is
✓
◆ 2
Z 2
Z 2
16 3
4 5
64⇡
2
2
3
4
4
V =⇡
(2x(2 x)) dx = ⇡
(16x
16x + 4x ) dx =
x
4x + x
=
.
15
3
5
0
0
0
When revolving about the line y = 2, we have washers with outer radius 2 and inner radius 2
so the volume is
Z 2
Z 2
V =⇡
4 (2 2x(2 x))2 dx = ⇡
(8x(2 x) 4x2 (2 x)2 ) dx
0
=⇡
Z 2
0
4
( 4x + 16x
3
2
24x + 16x) dx = ⇡
0
✓
4 5
x + 4x4
5
3
8x + 8x
2
◆ 2
0
=
2x(2
x),
32
⇡.
5
The second calculation yields a larger result.
6.3.61
a. False. The cross sections are not disks or washers.
Z R
b. True. It is given by V = ⇡
(R2 x2 ) dx.
0
c. True. This is because if we shift the sine function horizontally by ⇡/2 units, we obtain the cosine
function.
p
6.3.62 The relationship between the height h of tetrahedron and the edge length l is h = l 2/3, which can
be deduced using triangle geometry and the Pythagorean theorem.
Let z be the distance from the top vertex of the tetrahedron down toward the base perpendicularly. The
cross
sections perpendicular to this axis are all equilateral triangles with height z, so their side length is
p
3/2z, and their area is
r !2
p
p
3
3
3 3z 2
z
=
.
A(z) =
4
2
8
The volume is thus
p
Z 4p2/3 p
3 3 2
3 3 3
z /3
z dz =
8
8
0
4
0
p
2/3
p
16 2
=
.
3
Copyright c 2019 Pearson Education, Inc.
Chapter 6. Applications of Integration
608
6.3.63 The volume VS is ⇡
Z pa
p
(y
2
2
5
a) dy = ⇡ y /5
3
2
2ay /3 + a y
0
The volume VT is
6.3.64 V = ⇡
Z 2
0
a
=
0
8⇡ 5/2
a .
15
1 2 1/2
1
8/15
8
⇡a · a
= ⇡a5/2 . The ratio of VS /VT =
= .
1/3
5
3
3
x2 dx + ⇡
Z 5
2
(2x
2)2 dx + ⇡
Z 6
(18
2x)2 dx =
5
8
148⇡
⇡ + 84⇡ +
= 136⇡.
3
3
6.3.65
y
1
a.
This comes from revolving the region in the
first quadrant bounded by sin x and the line
y = 0 between x = 0 and x = ⇡ around the
x axis.
x
y
3
b.
This comes from revolving the region in the
first quadrant bounded by y = x + 1 and the
line y = 0 between x = 0 and x = 2 around
the x axis.
Copyright c 2019 Pearson Education, Inc.
2
x
6.3. Volume by Slicing
609
6.3.66
�
�
�
�
�
�
-�
-�
-�
�
�
�
�
By symmetry, the cup can be thought of as the result of revolving the line pictured about the y-axis.
y
The line is given by y = 6x 6 or x = + 1. Using disks, we have
6
◆
✓ 3
◆ 6
Z 6✓ 2
Z 6⇣
⌘
2
y
y
y
y
y2
⇡
+ 1 dy = ⇡
+ + 1 dy = ⇡
+
+y
= ⇡(2 + 6 + 6) = 14⇡.
108
6
6
36 3
0
0
0
6.3.67 At a given point x in [0, 4], let A(x) equal the area of a cross section of the solid of revolution that
n
X
is perpendicular to the x-axis. Use the formula V ⇡
A(x⇤k ) x to estimate the volume.
k=1
Left Riemann sum: V ⇡ ⇡(12 + 42 + 72 + 102 ) = 166⇡.
Right Riemann sum: V ⇡ ⇡(42 + 72 + 102 + 122 ) = 309⇡.
Midpoint Riemann sum: V ⇡ ⇡(32 + 52 + 82 + 112 ) = 219⇡.
6.3.68 The volume is approximately
2
2
2
2
2
⇡ (12.6/2) + (14.0/2) + (16.8/2) + (25.2/2) + (36.4/2) + (42.0/2)
2
✓
50
6
◆
⇡ 28542.7 cm3 .
6.3.69
a. Think of the cone as being obtained by revolving the region under y = x in the first quadrant from
Z R
1
x2 dx = ⇡R3 /3 = VC .
x = 0 to x = R around the x-axis. The volume is ⇡
3
0
p
b. Think of the hemisphere as being obtained by revolving the region under y = R2 x2 between x = 0
Z R
R
2⇡R3
2
and x = R around the x-axis. The volume is ⇡
(R2 x2 ) dx = ⇡ R2 x x3 /3
=
= VC .
3
3
0
0
Z 8+h p
8+h
⇡h3
6.3.70 V (h) = ⇡
( 64 y 2 )2 dy = ⇡ 64y y 3 /3
= 8⇡h2
. Note that V (0) = 0 and
3
8
8
2⇡ 3
8 .
V (8) =
3
Z 2p
Z 2
p
p
2
2
2
2
6.3.71 V = ⇡
((3 + 4 y )
(3
4 y ) ) dy = 24⇡
4 y 2 dy = 24⇡(⇡) = 24⇡ 2 . Note that
2
0
the last integral evaluated represents 1/4 the area of a circle of radius 2
1
R1
6.3.72 Around the x-axis: Vx = ⇡ 0 (x x4 ) dx = ⇡ x2 /2 x5 /5
= ⇡(1/2 1/5) = 3⇡
10 .
0
R1
R
p
p
1
Around the line y = 1: V1 = ⇡ 0 ((1 x2 )2
(1
x)2 ) dx = ⇡ 0 (x4
2x2
x + 2 x) dx =
1
⇡ x5 /5
2x3 /3
x2 /2 + 4x3/2 /3
= ⇡(1/5
0
2/3
1/2 + 4/3) = 11⇡
30 . Note that V1 > Vx .
Copyright c 2019 Pearson Education, Inc.
610
Chapter 6. Applications of Integration
6.3.73
a. By the general slicing method, V (x) =
Z b
A(x) dx. Because the two figures have the same cross
a
sections A(x), they must therefore have the same volumes.
b. According to Cavelieri’s principle, the volume is V = ⇡r2 h (the angle ⇡/4 is irrelevant).
6.3.74
a.
V (n) = ⇡
Z 1
(x2/n
0
=⇡
✓
n
n+2
✓
◆ 1
n
1
x2n ) dx = ⇡
x(n+2)/n
x2n+1
2n + 1
n+2
0
◆
2⇡(n + 1)(n 1)
2⇡(n2 1)
1
=
=
.
2n + 1
(n + 2)(2n + 1)
(n + 2)(2n + 1)
2n2 2
· ⇡ = ⇡.
n!1 2n2 + 5n + 2
b. lim V (n) = lim
n!1
6.4
Volume by Shells
6.4.1 V = 2⇡
Z b
x(f (x)
g(x)) dx.
a
6.4.2 . . . revolved about the y axis. . . . using the disk/washer method and integrating with respect to y or
using the shell method and integrating with respect to x.
6.4.3 . . . revolved about the x axis. . . . using the disk/washer method and integrating with respect to x or
using the shell method and integrating with respect to y.
6.4.4 In order to use washers, one would have to find the inner and outer radii of the washers, which would
involve finding inverse functions for y = x2 x3 over di↵erent portions of the interval [0, 1].
6.4.5
a. The radius is x.
x2 x.
Z 1
c. The volume is 2⇡
x(2
b. The height is 2
x2
x) dx.
0
6.4.6
a. The radius is y.
(2 y)2 = 4 (4
Z 2
c. The volume is 2⇡
y(4y y 2 ) dy.
4y + y 2 ) = 4y
y2 .
(2 y)2 = 4 (4 4y + y 2 ) = 4y
Z 2
c. The volume is 2⇡
(2 y)(4y y 2 ) dy.
y2 .
b. The height is 4
0
6.4.7
a. The radius is 2
y.
b. The height is 4
0
Copyright c 2019 Pearson Education, Inc.
6.4. Volume by Shells
611
6.4.8
a. The radius is 4
x.
b. The height is 2
(2
Z 4
c. The volume is 2⇡
p
x) =
p
(4
p
x) x dx.
x.
0
6.4.9 V = 2⇡
Z 1
x(x
2
x ) dx = 2⇡
0
6.4.10 V = 2⇡
✓ 3
x
x ) dx = 2⇡
3
2
3
(x
0
Z 2
x
dx = ⇡ ln(1 + x2 ) = ⇡ ln 5.
1 + x2
0
Z 1
x(3
0
6.4.11 V = 2⇡
Z 1
x4
4
◆ 1
= 2⇡
0
✓
1
3
1
4
◆
=
⇡
.
6
2
1
x3
3x) dx = 2⇡ 3x2 /2
0
= ⇡.
0
The volume of a cone is
1 2
1
⇡r h = ⇡ · 12 · 3 = ⇡.
3
3
Z 4
Z 4
6.4.12 V = 2⇡
x(( x2 + 4x + 2) (x2 6x + 10)) dx = 2⇡
( 2x3 + 10x2 8x) dx =
1
1
✓ 4
◆ 4
✓✓
◆ ✓
◆◆
x
10x3
640
1 10
2
2⇡
+
4x
= 2⇡
128 +
64
+
4
= 45⇡.
3
2
3
3
2
1
6.4.13 V = 2⇡
Z 2
2
y(4
y 2 ) dy = 2⇡ 2y 2
y 4 /4
= 8⇡.
0
0
Z 1
6.4.14 V = 2⇡
y(2 y
✓
◆0
1 1
5⇡
2⇡ 1
=
.
3 4
6
6.4.15 V = 2⇡
Z 4
y(4
Z 1
(2y
2
y3
3
y 2 ) dy = 2⇡
✓
y 3 ) dy = 2⇡ y 2
y2
0
y) dy = 2⇡
2
✓
2y
◆ 4
=
2
y3
3
y4
4
◆ 1
=
0
32⇡
.
3
◆
◆
✓
Z p3 ✓
Z p3 ✓
4
1
4
y
p
p
6.4.16 V = 2⇡
y
dy = 2⇡
dy = 2⇡ 4 tan 1 y
3
2
y
+
y
1
+
y
3
3
1
1
✓
◆
✓
◆
p ⌘
4⇡
3
1
⇡
1
2⇡ ⇣
p ) (⇡
p ) = 2⇡
p
=
⇡
3 .
2⇡ (
3
3
3
2 3
2 3
3
6.4.17 V = 2⇡
Z p⇡/2
2
p
2
6.4.18 V = 2⇡
x(6
= ⇡.
Z 1
x(1
0
x) dx = 2⇡
2
6.4.19 V = 2⇡
Z 1
0
✓
x2 ) dx = 2⇡
0
6.4.20 V = 2⇡
⇡/2
x cos x dx = ⇡ sin x
0
Z 4
y2
p
2 3
x3
3
2
3x
Z 1
◆ 4
2
p
x x dx = 2⇡
◆ 1
5/2
2x
5
104⇡
.
3
x3 ) dx = 2⇡
(x
0
✓
=
0
=
✓ 2
x
2
x4
4
◆ 1
= 2⇡
0
4⇡
.
5
Copyright c 2019 Pearson Education, Inc.
✓
1
2
1
4
◆
=
⇡
.
2
◆ p3
1
=
612
Chapter 6. Applications of Integration
✓ 4◆ 3
y
6.4.21 V = 2⇡
y(y ) dy = 2⇡
y dy = 2⇡
= 81⇡/2.
4
0
0
0
Z 3
6.4.22 V = 2⇡
Z 1
Z 3
2
p
3
y( y) dy = 2⇡
0
✓
3
3 7/3
y
7
◆ 1
6⇡
.
7
=
0
6.4.23 Note that the lines intersect at (0, 0), (2, 0) and (1, 1).
✓
Z 1
Z 1
2
V = 2⇡
y((2 y) y) dy = 2⇡
(2y 2y ) dy = 2⇡ y 2
0
2y 3
3
0
6.4.24 V = 2⇡
Z p
2
x
0
p
⇡
2
6.4.25 V = 2⇡
Z 2
1
2x2 , so that du =
2x2 dx. Let u = 4
4
x
(x2 + 1)2
Z 0
◆ 1
=
0
4x dx. Substituting gives
0
⇡ 2 3/2
8⇡
· u
=
.
3
2 3
4
u1/2 du =
4
dx. Let u = x2 + 1, so that du = 2x dx. Substituting gives
V =⇡
Z 5
5
u 2 du =
⇡·
2
1
=
u 2
⇡
✓
1
5
1
2
◆
=
3⇡
.
10
6.4.26 Note that the line x = 4 intersects the curve x = y 2 at (4, 2). V = 2⇡
y 4 /4
Z 2
y(4
y 2 ) dy = 2⇡
0
2
y 3 ) dy = 2⇡ 2y 2
2⇡
.
3
Z 2
(4y
0
= 8⇡.
0
6.4.27 V = 2⇡
3⇡(25
Z 16
y(2/y)2/3 dy = 25/3 ⇡
2
2) = 90⇡.
Z p
Z 16
2
⇣
⌘ 16
y 1/3 dy = 25/3 ⇡ 3y 4/3 /4
= 25/3 ⇡(3 · 210/3
⇡/2
y sin y 2 dy =
6.4.28 V = 2⇡
⇡ cos y 2
0
6.4.29 V = 2⇡
Z 2
Z 3
1
2
p
⇡/2
=
1) = ⇡.
2
x(ex /x) dx = 2⇡ ex
= 2⇡(e2
e).
1
✓
ln x
dx = 2⇡
x
6.4.31 Note that y =
p
p
cos
ln2 x
2
◆ 3
= ⇡(ln 3)2 .
1
1 x intersects the axes at (0,
⇡/2
⇡ sin y 2
⇡(0
0
1
6.4.30 V = 2⇡
3 · 2 2/3 ) =
= ⇡(1
p
⇡/2) and (1, 0). V = 2⇡
Z p⇡/2
y cos y 2 dy =
0
0) = ⇡.
0
6.4.32
Z 5p 2 p
V = 2⇡
y (50
0
where u = 50
2
y . Thus,
V =
p
y 2 )/2 dy =
p
Z 5p 2 p
2⇡
y 50
0
2⇡ ⇣ 3/2 ⌘
2u /3
=
2
0
50
y 2 dy =
Z
p
1 50 1/2
2⇡ ·
u du,
2 0
p
⌘ 500⇡
p
2⇡ ⇣
2 · 50 · 5 2/3 =
.
3
2
Copyright c 2019 Pearson Education, Inc.
6.4. Volume by Shells
613
6.4.33 Note that the line y = 1 intersects the curve y = x3 x8 + 1 at x = 0 and x = 1. V = 2⇡
✓ 5
◆ 1
✓
◆
Z 1
x
x10
1
1
⇡
x8 + 1 1) dx = 2⇡
(x4 x9 ) dx = 2⇡
= 2⇡
= .
5 10
5
5
10
0
0
6.4.34
V = 2⇡
Z 2
y(ey
⇣ 2
2
ey /3 ) dy = ⇡ ey
2
0
6.4.35 With washers we have
V =⇡
Z 1
(x
2/3
2
3ey /3
⌘ 2
V = 2⇡
Z 1
2
x ) dx = ⇡
y 3 ) dy = 2⇡
y(y
0
6.4.36 Using washers we have
Z 2 ✓⇣
x ⌘2
1
V =⇡
3
0
Using shells we have
Z 1 ✓
V = 2⇡
y 3
1/3
✓
= 2⇡ 0
✓
✓
3x5/3
5
x3
3
1
(x + 1)2
◆
✓
◆ 1
= 2⇡
y3
3
y5
5
dx = ⇡
✓
3)) = ⇡(2 + e4
=
0
4⇡
.
15
✓
1
3
1
5
◆
=
=
1) dy = 2⇡ 2y 2
y3
0
10
= ⇡ 3u5/3 /5 + 3u4/3 + 4u
⌘ 27
8
8
= ⇡(36 /5 + 35 + 108
( 96/5 + 48
2)3 + 2) dx = 2⇡
0
= 2⇡
✓
x5
3x4
+
5
2
4x3 +
6.4.38 Using shells, we have
Z 2
V = 2⇡
x(8
35x2
2
= 2⇡
0
(6x
(27x
2
2)3 ) dx = 2⇡
x(x
0
y
32) = ⇡
Z 5
✓
2484 16
+
5
5
x4 + 6x3
= 500⇡
0
✓
Z 2
x(6
0
2
2x ) dx = 2⇡ 3x
2x3
3
2x) dx
◆ 2
✓
16
3
= 2⇡ 12
0
Using disks, we have
◆2
Z 8
Z 6✓ 2
Z 6✓
1
y
y 1
dy + ⇡
4 dy = ⇡
V =⇡
2
4
2
6
2
✓ 3
◆ 6
✓
2
y
y
+ 8⇡ = ⇡ (18 18 + 6)
=⇡
+y
12
2
2
16⇡
40⇡
.
=
+ 8⇡ =
3
3
y+1
✓
8
12
Copyright c 2019 Pearson Education, Inc.
◆
◆
=
◆
= 500⇡.
12x2 + 35x dx
0
(2x + 2)) dx = 2⇡
0
Z 2
◆ 5
Z 5
8⇡
.
27
1/3
8
Using shells we have
Z 5
V = 2⇡
x(25 (x
3e4/3 ).
1
3y 2
6.4.37 Using washers we have
Z 27
Z 27
Z 25 p
V =⇡
( 3 y + 2 + 2)2 dy = ⇡
(u1/3 + 2)2 du = ⇡
(u2/3 + 4u1/3 + 4) du
⇣
0
4⇡
.
15
◆ 2
x2
x3
1
+
+
3
27 (x + 1)
x
◆◆
Z 1
1
1
dy = 2⇡
(4y
y
1/3
◆◆
1
8⇡
.
=
3
27
1
27
(1
◆ 1
0
✓
3y
2
9
3e4/3
x(x3
0
0
With shells we have
= ⇡(e4
Z 1
40⇡
.
3
dy + 8⇡
◆◆
2+2
+ 8⇡
614
Chapter 6. Applications of Integration
6.4.39 V = 2⇡
Z 1
(x + 2x ) dx = ⇡
Z 1
(x2
Z 1
(2x2
y)(1
p
Z 1
(x + 2)x dx = 2⇡
Z 1
(1
Z 1
(2
2
0
6.4.40 V = 2⇡
2
0
x)x2 dx = 2⇡
0
6.4.41 V = 2⇡
3
x3 ) dx = 2⇡
0
x)x2 dx = 2⇡
0
✓
x4
2x3
+
4
3
✓
x3
3
x4
4
✓
2x3
3
(1
p
x3 ) dx = 2⇡
0
◆ 1
=
◆ 1
⇡
.
6
0
=
0
x4
4
◆ 1
11⇡
.
6
= 2⇡
0
✓
2
3
1
4
◆
=
5⇡
.
6
6.4.42
V = 2⇡
Z 1
(1
y) dy = 2⇡
0
✓
= 2⇡ y
2y 3/2
3
y2
2y 5/2
+
5
2
◆
Z 1
0
1
0
y
✓
= 2⇡ 1
y + y 3/2 ) dy
2
3
1 2
+
2 5
◆
=
7⇡
15
6.4.43
V = 2⇡
Z 1
p
(y + 2)(1
y) dy = 2⇡
◆
4
23⇡
=
.
3
15
0
✓
1
+2
2
Z 1
p
= 2⇡
2
5
Z 1
(y + 2
y
3/2
2y
1/2
) dy = 2⇡
0
✓
y2
+ 2y
2
2y 5/2 /
5
4y 3/2
3
◆ 1
◆ 1
=
0
6.4.44
V = 2⇡
(1
y)(2
y) dy = 2⇡
0
Z 1
(2
y
2y
1/2
+y
3/2
) dy = 2⇡
0
✓
y2
2
2y
4y 3/2
2y 5/2
+
3
5
0
6.4.45 Using washers, we have
V =⇡
Z 1
(32
(x2 + 2)2 ) dx = ⇡
0
Z 1
(9
x4
4x2
0
=⇡
✓
5x
x5
5
4x3
3
4) dx = ⇡
◆ 1
Z 1
(5
x4
0
✓
1
=⇡ 5=
5
0
4
3
◆
4x2 ) dx
=
52⇡
.
15
Using shells, we have
Z 1
p
(y + 2) y dy = 2⇡
0
✓
◆
2 4
52⇡
=
= 2⇡
+
5 3
15
V =2⇡
Z 1
(y 3/2 + 2y 1/2 ) dy = 2⇡
0
✓
2y 5/2
4y 3/2
+
5
3
◆ 1
0
6.4.46 Using washers, we have
V =⇡
Z 1
p
(( y + 1)2
12 ) dy = ⇡
0
Z 1
p
(y + 2 y) dy = ⇡
0
✓
y2
4y 3/2
+
2
3
◆ 1
=
0
11⇡
.
6
✓ 2
x
2
x4
+x
4
x3
3
Using shells, we have
V = 2⇡
Z 1
(x + 1)(1
0
= 2⇡
✓
1
2
1
+1
4
x2 ) dx = 2⇡
◆
11⇡
1
=
.
6
3
Z 1
0
(x
x3 + 1
x2 ) dx = 2⇡
Copyright c 2019 Pearson Education, Inc.
◆ 1
0
17⇡
.
15
6.4. Volume by Shells
615
6.4.47 Using washers, we have
V =⇡
Z 1
x2 ) 2
((6
52 ) dx = ⇡
0
Z 1
(x4
12x2 + 11) dx = ⇡
0
✓
x5
5
4x3 + 11x
2y 5/2
5
◆ 1
8y 3/2
3
y2
2
◆ 1
=
0
36⇡
.
5
Using shells, we have
V = 2⇡
Z 2
p
y) y dy = 2⇡
(6
0
Z 1
✓
y 3/2 ) dy = 2⇡ 4y 3/2
(6y 1/2
0
0
✓
= 2⇡ 4
2
5
◆
◆ 1
13⇡
.
6
=
36⇡
.
5
6.4.48 Using washers, we have
V =⇡
Z 1
(4
p
(2
y)2 ) dy = ⇡
0
Z 1
p
(4 y
y) dy = ⇡
0
✓
=
0
Using shells, we have
V = 2⇡
Z 1
(2
x)(1
x ) dx = 2⇡
0
✓
= 2⇡ 2x
6.4.49
a. The volume is V = 2⇡
Z r
y(2
2⇡
Z 0
0
Z 1
2
(2
2x2
x + x3 ) dx
✓
2
3
1 1
+
2 4
0
◆ 1
2x3
3
x2
x4
+
4
2
p
y 2 ) dy. Let u = r2
r2
= 2⇡ 2
0
Z r2
u
Z r p
( r2
V =⇡
x2 )2 dx = ⇡
Z r
= ⇡ r3
r3
p
u du = 2⇡
r2
◆
=
y 2 , so that du =
13⇡
.
6
2y dy. Substituting gives
r2
1/2
0
2u3/2
4
du = 2⇡
= ⇡r3 .
3 0
3
b. The volume is
r
✓
3
r
3
(r
( r)
3
3
◆◆
x3
3
2
x ) dx = ⇡ r x
r
✓
✓
2
2
✓
= 2⇡ r3
r
3
◆
✓
1
a
3
✓
3
=
◆ r
r
4 3
⇡r .
3
6.4.50
a. Using washers, the volume is
◆
✓
Z a ✓
x2
2
2
b 1
dx = ⇡b x
V =⇡
a2
a
Using shells, the volume is V = 2⇡
Z b
x3
3a2
y 2a
0
Substituting gives
V =
2⇡ab2
Z 0
1
p
u du = 2⇡ab2
◆ a
r
1
Z 1
0
2
= ⇡b
a
y2
b2
!
a
dy. Let u = 1
u1/2 du = 2⇡ab2
✓
2u3/2
3
( a)3
3
a
y2
b2 .
◆ 1
0
=
◆◆
=
4
⇡ab2 .
3
Then du =
2y
dy.
b2
4
⇡ab2 .
3
4
b. For the rugby union football, we have ⇡ · 6 · (3.82)2 ⇡ 366.75 in3 . For the American football, we have
3
4
⇡ · 5.62 · (3.38)2 ⇡ 268.94 in3 .
3
Copyright c 2019 Pearson Education, Inc.
616
c.
Chapter 6. Applications of Integration
366.74
⇡ 1.36.
269.94
6.4.51
a. V = 2⇡
Z 5
1
b. V = 2⇡
Z 2
p
2x
4
p
((3 +
y 2 )2
4
0
3)2 dx.
(x
p
(3
y 2 )2 ) dy = 24⇡
4
Z 2p
y 2 dy.
4
0
1
c. Note that the last integral in part b represents of the area of the circle of radius 2 centered at the
4
1
origin, so its value is ⇡ · 4 = ⇡. So the volume is 24⇡ · ⇡ = 24⇡ 2 .
4
6.4.52
a. V = ⇡
Z h
2
(rx/h) dx =
0
b. V = 2⇡
Z r
h
⇡r2 3
x /3
h2
=
0
⇡r2 h
.
3
r
(h/r)y) dy = 2⇡h y 2 /2
y(h
y 3 /(3r)
=
0
0
⇡r2 h
.
3
6.4.53 First, note that the curve and the line intersect at (0, 0) and (1, 0).
We choose the washer method:
V =⇡
Z 1
4 2
(x
x ) dx = ⇡
0
Z 1
(x2
2x5 + x8 ) dx
0
1
= ⇡ x3 /3
x6 /3 + x9 /9
= ⇡/9
0
6.4.54 First, note that the curve and the line intersect at (0, 0) and (1, 0).
We choose the shell method:
Z 1
Z 1
V = 2⇡
x(x x4 ) dx = 2⇡
(x2 x5 ) dx
0
0
1
= 2⇡ x3 /3
x6 /6
= 2⇡(1/3
1/6) = ⇡/3.
0
6.4.55 Using washers:
Z 1
V = 2⇡
((2 x2 )2
(x2 )2 ) dx = 2⇡
0
Z 1
1
(4
4x2 ) dx = 2⇡ 4x
4x3 /3
=
0
0
16⇡
.
3
6.4.56 Using washers:
V =⇡
Z 5⇡/6
2
(sin x
(1
2
sin x) ) dx = ⇡
⇡/6
=
5⇡/6
⇡ ( 2 cos x + x)
2
x) dx = 2⇡
(2 sin x
1) dx
⇡/6
⇡/6
6.4.57 Using shells:
Z 6
V = 2⇡
x(2x + 2
Z 5⇡/6
Z 6
2
p
= 2 3⇡
2⇡ 2
.
3
6
(x2 + 2x) dx = 2⇡ x3 /3 + x2
=
2
Copyright c 2019 Pearson Education, Inc.
608⇡
.
3
6.4. Volume by Shells
617
6.4.58 Using disks:
Z 2
V =⇡
(x3 )2 dx = ⇡ x7 /7
2
=
0
0
128⇡
.
7
6.4.59 Using shells:
V = 2⇡
Z 1
y(ey
⇣ 2
2
ey /2 ) dy = 2⇡ ey /2
2
0
6.4.60 Using shells:
Z 6
V = 2⇡
x(2x + 2
2
ey /2
⌘ 1
= ⇡(e
p
2 e
(1
p
2)) = ⇡( e
1)2 .
0
6
2) dx = 2⇡ 2x3 /3
= 288⇡.
0
0
6.4.61 The curves intersect when x2 = 2 x, or x2 + x 2 = (x 1)(x + 2) = 0, so for x = 1 and x =
Only x = 1 is in the first quadrant; the corresponding y coordinate is y = 1. Also, 2 x x2 on [0, 1].
Using shells, the height of each shell is 2 x x2 , so the volume is
✓
◆ 1
Z 1
x3
x4
5⇡
2
2
V = 2⇡
x(2 x x ) dx = 2⇡ x
=
.
3
4
6
0
0
6.4.62 Using disks:
Z 4
p
V =⇡
( x)2 dx = ⇡ x2 /2
2.
4
= 8⇡.
0
0
6.4.63
a. True. Otherwise, we wouldn’t have shells!
b. False. Either method can be used when revolving around either axis.
c. True.
6.4.64 Consider the rectangle in the first quadrant bounded by x = 2, x = 4, y = 6 and y = 0. The solid in
question is formed by revolving this region around the y-axis. The volume is
V = 2⇡
Z 4
4
6x dx = 2⇡ 3x2
2
= 72⇡.
2
6.4.65 Consider the region in the first quadrant bounded by the coordinate axes and the line y = 8 (8/3)x.
We can generate the desired cone by revolving this region around the y-axis. We then have
V = 2⇡
Z 3
x(8
(8/3)x) dx =
0
16⇡
3
Z 3
(3x
x2 ) dx =
0
16⇡
3x2 /2
3
3
x3 /3
= 24⇡.
0
6.4.66 Consider the region in the first and second quadrants bounded by x = 3, the circle x2 + y 2 =
36.Z The solid in question is formed by revolving this region around the y-axis. The volume is V =
6 p
4⇡
x 36 x2 dx. This integral can be computing using the substitution u = 36 x2 . We then have
3
V = 2⇡
Z 27
0
⇣
⌘ 27
p
= 108 3⇡.
u1/2 du = 2⇡ 2u3/2 /3
0
6.4.67 Consider the triangle in the first quadrant bounded by y = 0, x = 3, and y = 9
in question is formed by revolving this region around the y-axis. The volume is
V = 2⇡
Z 6
6
x(9
(3/2)x) dx = 2⇡ 9x2 /2
x3 /2
3
Copyright c 2019 Pearson Education, Inc.
= 54⇡.
3
(3/2)x. The solid
618
Chapter 6. Applications of Integration
6.4.68
V = 2⇡
Z R
r
R
x2 /R2 ) dx = 12⇡ x2 /2
x · 6(1
x4 /4R2
=
r
3⇡(R2 r2 )2
.
R2
6.4.69
a. V1 = ⇡
Z 1
(ax2 + 1)2 dx = ⇡
0
V2 = 2⇡
Z 1
Z 1
(a2 x4 + 2ax2 + 1) dx = ⇡(a2 /5 + 2a/3 + 1).
0
x(ax2 + 1) dx = 2⇡(a/4 + 1/2) = ⇡a/2 + ⇡.
0
b. These are equal when ⇡a/2 + ⇡ = ⇡a2 /5 + 2a⇡/3 + ⇡, or when ⇡a2 /5 + a⇡/6 = 0, which occurs when
a = 0 and when a = 5/6.
6.4.70
a. V = ⇡
Z r
r
(r2
y 2 ) dy = ⇡ r2 y
r h
y 3 /3
= ⇡(2r3 /3
(r2 (r
h)
h)3 /3)) =
(r
r h
⇡h2
(3r
3
h).
b.
Z p2rh h2
✓
1 p 2
( r
x2 ) 3
V = 2⇡
x( r2 x2 (r h)) dx = 2⇡
3
0
✓
◆
1 p 2
1 1
= 2⇡
( r
2rh + h2 )3 (r h)(2rh h2 ) + r3
3
2 3
= 2⇡
(r
p
h)3 /3
r2 h + rh2 /2 + rh2
h3 /2 + r3 /3 = ⇡h2 r
(r
2
h)x /2
0
⇡h3
.
3
c. If Zwe take slices perpendicular to the y-axis, we have circles whose area A(y) = ⇡(r2
r
(r2
⇡
◆ p2rh h2
y 2 ). So V =
y 2 ) dy, which is exactly the integral computed in part (a) above.
r h
Note that all three approaches led to the same result, and the result is consistent with other formulas.
2⇡r3
, the
For example, when h = 0 we have no figure, so our volume is 0. When h = r, we have V =
3
volume of a hemisphere.
6.4.71 V = 2⇡
Z 2
0
xf (x2 ) dx = ⇡
Z 4
f (u) du, where u = x2 . Thus, V = 10⇡.
0
6.4.72
a. Consider the region in the first quadrant bounded by the coordinate axes and y = 8 2x. The integral
on the left represents the volume of the solid obtained when this region is revolved around the x-axis,
using the disk/washer method, and the integral on the right represents the same volume calculated
using the shell method. Hence, they are equal.
b. Consider the region in the first quadrant bounded by the line x = 0, the line y = 5, and the curve
y = x2 +1. The integral on the left represents ⇡1 times the volume of the solid obtained when this region
is revolved around the x-axis, using the disk/washer method, and the integral on the right represents
1
⇡ times the same volume calculated using the shell method. Hence, they are equal.
6.4.73
a. The longest diagonal of the cube is equal to the diameter of
pthe sphere, which is
3
Thus, if R is the radius of the sphere, we must have R =
r.
2
Copyright c 2019 Pearson Education, Inc.
p
p
r2 + r2 + r2 = 3r.
6.4. Volume by Shells
619
the cone which contains the sphere, we see that the height h of the cone is 3R, so
Now analyzing
p
3 3
r.
h=
2
p
p !
✓ ◆2
3r
27 3⇡r3
3 3r
=
.
Now the volume of the cylinder is V = ⇡
2
2
8
b. Imagine the cone with its vertex up. Consider the plane which contains the vertex of the cone and
two non-adjacent vertices
p of the cube’s bottom face. The cross section of this plane with the cone and
cube consists of an r ⇥ 2r rectangle (where r is the side length of the cube) and an isosceles triangle
of base 2 and height 3, with the rectangle inscribed in the triangle, and the longerpside of the rectangle
p
r
1 (r/ 2)
3 2
p ,
, so r =
lying on the base of the triangle. Using similar triangles, we have =
3
1
3+ 2
p
54 2
and the volume of the cube is
p 3.
3+ 2
c. Imagine the sphere with the hole as being obtained
by revolving the region pictured around the x-axis,
p
where the relevant curves are y = r and y = R2 x2 where r is the radius of the hole and R is the
radius of the sphere. Note that if you draw the triangle with vertices (0, 0), (5, 0), and (5, r), you have
a right triangle with legs of length 5 and r, and hypotenuse of length R, so 25 + r2 = R2 .
The volume we are interested in is
Z 5
Z 5
2
2
2
x ) r ) dx = 2⇡
(25
V = 2⇡
((R
0
5
x2 ) dx = 2⇡ 25x
0
x3 /3
=
0
500⇡
.
3
Note that (surprisingly!), the result doesn’t depend on the radius of the original sphere.
�
�
�
�
�
�
-�
-�
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-�
�
�
6.4.74 Vertical slices of the wedge are triangles with area 12 xh, where x is the base and h is the height. Note
that h = x tan ✓, so the triangles have area 12 x2 tan ✓. Now if we think of the curved part of the base of the
wedge as having equation x2 + y 2 = a2 , then we have x2 = a2 y 2 . So the volume of the wedge is given by
Z a
Z a
a
1
2a3
tan ✓(a2 y 2 ) dy =
tan ✓(a2 y 2 ) dy = tan ✓ a2 y y 3 /3
=
tan ✓.
3
a 2
0
0
6.4.75
a. The cross sections of such an object are washers, with inner radius given b g(x) y0 and outer radius
Z b
given by f (x) y0 , so the volume is given by ⇡
((f (x) y0 )2 (g(x) y0 )2 ) dx.
a
b. The cross section of this object are also washers, and this time the inner radius is y0 f (x) and the
Z b
outer radius is y0 g(x), so the volume is given by ⇡
((y0 g(x))2 (y0 f (x))2 ) dx.
a
Copyright c 2019 Pearson Education, Inc.
620
Chapter 6. Applications of Integration
6.4.76
a. The result of revolving a slice around the line x = x0 is a shell with the same height as before, but with
the radius being xk x0 rather than xk . Thus, when the volumes of the shells are added and the limit
Z b
Z b
is taken, the resulting integral is
2⇡(x x0 )(f (x) g(x)) dx instead of
2⇡(x)(f (x) g(x)) dx.
a
a
b. When x0 > b, the radius of a typical shell is given by x0
Z b
2⇡(x0
x)(f (x)
xk , so the volume is given by
g(x)) dx.
a
6.5
Length of Curves
Z bp
0
6.5.1 Given f (x) and a and b, compute f (x) and then compute
1 + f 0 (x)2 dx.
a
0
6.5.2 Given g(y) and c and d, compute g (y) and then compute
Z dp
1 + g 0 (y)2 dy.
c
6.5.3 Because f 0 (x) = 3x2 , the arc length is
Z 5p
1 + f 0 (x)2 dx =
2
6.5.4 Because f 0 (x) =
2
Z ⇡ p
6 sin 3x, the arc length is
Z 5p
1 + 9x4 dx.
1 + f 0 (x)2 dx =
⇡
6.5.5 Because f 0 (x) =
2e 2x , the arc length is
⇡
Z 2p
Z 2p
1 + f 0 (x)2 dx =
1 + 4e 4x dx.
0
1
6.5.6 Because f (x) = , the arc length is
x
0
6.5.7 L =
Z ⇡ p
1 + 36 sin2 3x dx.
Z 10 p
0
1 + f 0 (x)2 dx =
1
Z 10 r
1
1
1 + 2 dx =
x
Z 10 p 2
x +1
dx.
x
1
Z 5p
p
1 + 22 dx = 4 5. Using the fact that this is a straight line segment between the points
Z 2p
1 + (y 0 )2 =
Z 2p
Z 2p
p
1 + ( 3)2 dx =
10 dx = 5 10. Using the fact that this is a straight
Z 6p
1 + (y 0 )2 dx =
1
(1, 3) and (5, 11), we can calculate the arc length using the distance formula in the plane:
p
p
p
p
L = (5 1)2 + (11 3)2 = 16 + 64 = 80 = 4 5.
6.5.8 L =
3
3
3
line segment between the points ( 3, 13) and (2, 2), we can calculate the arc length using the distance
formula in the plane:
p
p
p
p
L = (2 ( 3))2 + ( 2 13)2 = 25 + 225 = 250 = 5 10.
6.5.9 L =
2
6.5.10 y 0 =
Z 6p
1+(
8)2 dx =
2
1 x
(e
2
e x ), so 1 + (y 0 )2 = 1 +
L=
Z ln 2
Z 6p
p
65 dx = 8 65.
2
1 2x
e
4
1 x
1 x
(e + e x ) dx =
e
2
ln 2 2
2 + e 2x =
ln 2
e x
=
ln 2
✓
◆2
. Thus,
(1/2
2)) =
1 x
(e + e x )
2
1
(2
2
1/2
Copyright c 2019 Pearson Education, Inc.
3
.
2
6.5. Length of Curves
6.5.11 y 0 =
621
p
x/2, so 1 + (y 0 )2 = 1 + x/4. Thus,
Z 60 p
L=
1 + (x/4) dx =
0
Z 60
0
x/(12), so 1 + (y 0 )2 = 1 +
6.5.12 y 0 = 3/x
L=
Z 6✓
1
3
x
+
x 12
◆
1p
x + 4 dx =
2
✓
✓
1
x2
+
=
2 144
9
x2
1
(x + 4)3/2
3
3
x
+
x 12
◆2
◆ 60
0
. Thus,
6
dx = 3 ln x + x2 /24
504
= 168.
3
=
= 3 ln 6 +
1
35
.
24
6.5.13 y 0 = x(x2 + 2)1/2 , so 1 + (y 0 )2 = 1 + x2 (x2 + 2) = x4 + 2x2 + 1 = (x2 + 1)2 . Thus,
L=
Z 1
1
(x2 + 1) dx = x3 /3 + x
0
p
6.5.14 y = x/2
L=
Z 16
4
6.5.16 y = x
1
1
+
=
2 16x6
0 2
6
1
4x3 , so 1 + (y ) = 1 + x
Z 2
(x3 +
1
1/2
(1/(4x
1
) dx =
4x3
1/2
Z 9✓
L=
x1/2 +
1
dx
= y3
dy
L=
Z 2
(y 3 +
1+
x4 /4 +
1 2
x
8
◆2
◆2
. Thus,
(8/3 + 2) = 2 + 56/3 =
1
4x1/2
◆
✓
✓
◆2
. Thus,
(1/32)
(1/4
x3 +
=4
1
4x3
(1/8)) =
1
1
1/2 +
=
16x
)), so 1 + (y ) = 1 + x
✓
x
1/2
1
+ 1/2
4x
⇣
⌘ 9
p
dx = 2x3/2 /3 + x/2
= 18 + 3/2
◆2
dx
dy
◆2
1
· y 3 ) dy =
4
= 1 + y6
✓
y 4 /4
1/2 + 1/(16y 6 ) =
1
8y 2
◆2
=4
62
.
3
123
.
32
. Thus,
(2/3 + 1/2) =
1
p
p p
dx
= 2 2e 2y
dy
✓
✓
0 2
1/(4y 3 ), so 1 +
1
6.5.18
x
1
+ p
2
2 x
4
L=
6.5.17
1
1
+
=
2 4x
✓p
⇣
⌘ 16
= 64/3 + 4
(x1/2 /2 + (1/2)x 1/2 ) dx = x3/2 /3 + x1/2
6.5.15 y 0 = x3
0
0
p
x
1/(2 x), so 1 + (y 0 )2 = 1 +
4
0
4
.
3
=
55
.
3
✓
◆2
1
y 3 + 3 . So
4y
(1/32)
(1/4
1/8) =
1
123
.
32
2 p2y
e
, so
16
dx
dy
◆2
= 1 + 8e
p
2 2y
p p
2 2e 2y +
p
2
(1/2) + 2 e 2 2y =
16
p
2 p2y
e
16
!2
(1/32)
(2
.
Then
p
✓ p
Z (ln 2)/p2 p p
2 p2y
2y
L=
(2 2e
+
e
) dy = 2e 2y
16
0
6.5.19
dx
= 2, so 1 +
dy
✓
dx
dy
◆2
= 1 + 4 = 5, so L =
1 p2y
e
16
◆ (ln 2)/p2
= (4
0
Z 4p
p
5dy = 7 5.
3
Copyright c 2019 Pearson Education, Inc.
(1/16)) =
65
.
32
622
Chapter 6. Applications of Integration
6.5.20 y = ln(x
p
x2
p
1), so ey = x
1, and e y =
x2
ey + e y
= x.
2
✓ ◆2
dx
ey e y
dx
, so 1 +
= 1 + e2y /4
We have
=
dy
2
dy
1
p
x2
x
1
=
p
x + x2
x2 (x2
Thus
Z 0
ey + e y
L=
dy =
p
2
ln( 2 1)
✓
ey
e y
2
◆0
p
ln( 2 1)
1/2 + e
p
2
=0
2y
2
p
2
·p
2
1
=
1
6.5.21
a. y 0 = 2x, so 1 + (y 0 )2 = 1 + 4x2 , so L =
Z 1p
1 + 4x2 dx.
1
Z 1p
b. L =
1 + 4x2 dx ⇡ 2.958.
1
6.5.22
a. y 0 = cos x, so 1 + (y 0 )2 = 1 + cos2 x, so L =
Z ⇡p
1 + cos2 x dx.
0
Z ⇡p
b. L =
1 + cos2 x dx ⇡ 3.820.
0
6.5.23
x2 + 1
, so L =
a. y = 1/x, so 1 + (y ) = 1 + (1/x) =
x2
Z 4p 2
x +1
dx ⇡ 3.343.
b. L =
x
1
0 2
0
2
Z 4p 2
x +1
dx.
x
1
6.5.24
a. y 0 = x2 , so 1 + (y 0 )2 = 1 + x4 , so L =
Z 1p
1 + x4 dx.
1
Z 1p
b. L =
1 + x4 dx ⇡ 2.179.
1
6.5.25
1
1
4y
a. x0 = p
, so 1 + (x0 )2 = 1 +
=
4y
4(y 2)
2 y 2
r
Z 4
4y 7
dy ⇡ 1.083.
b. L =
4y 8
3
7
, so L =
8
Z 4r
3
4y
4y
6.5.26
a. y 0 =
Z 4
162
x6 + 162
0 2
16
,
so
1
+
(y
)
=
1
+
=
,
so
L
=
x3
x6
x6
1
Z 4p 6
x + 162
b. L =
dx ⇡ 8.708.
x3
1
r
1.
✓ y
◆2
e +e y
/4 =
. Thus,
2
p1
2 1
1
p
1
= x + x2
1)
x6 + 162
dx.
x6
Copyright c 2019 Pearson Education, Inc.
7
dy.
8
p
2+2 2 1+1
p
= 1.
2( 2 1)
6.5. Length of Curves
623
6.5.27
0 2
0
a. y =
2
2 sin 2x, so 1 + (y ) = 1 + 4 sin 2x, so L =
Z ⇡p
1 + 4 sin2 2x dx.
0
b. L =
Z ⇡p
1 + 4 sin2 2x dx ⇡ 5.270.
0
6.5.28
2x, so 1 + (y 0 )2 = 1 + 16
a. y 0 = 4
16x + 4x2 = 4x2
16x + 17, so L =
Z 4p
4x2
16x + 17 dx.
0
b. L =
Z 4p
4x2
0
6.5.29
a. y 0 =
16x + 17 dx ⇡ 9.294.
1/x2 , so 1 + (y 0 )2 = 1 +
1
x4 + 1
=
. Thus, L =
4
x
x4
Z 10 p 4
x +1
dx ⇡ 9.153.
b. L =
x2
1
Z 10 p 4
x +1
dx.
x2
1
6.5.30
2y
4y 2
(y 2 + 1)4 + 4y 2
a. x = 2
, so 1 + (x0 )2 = 1 + 2
=
, so L =
2
4
(y + 1)
(y 2 + 1)4
(y + 1)
Z 5 p 2
(y + 1)4 + 4y 2
dy ⇡ 10.369.
b. L =
(y 2 + 1)2
5
0
6.5.31 f 0 (x) = 0.00074x, so L =
Z 640 p
640
Z 5 p 2
(y + 1)4 + 4y 2
dy.
(y 2 + 1)2
5
1 + (0.00074x)2 dx ⇡ 1326.4 meters.
630
ex e x
sinh(x/239.2), where sinh(x) =
.
6.5.32 f 0 (x) =
239.2
2
So
s
✓
◆2
Z 315
630
L=
1+
sinh(x/239.2) dx ⇡ 1472.17 ft.
239.2
315
6.5.33
p
a. False. For example, if f (x) = x2 , the first integrand is 1 + 4x2 and the second is 1 + 2x, which clearly
yield di↵erent values for (for example) a = 0 and b = 1.
Z bp
1 + f 0 (x)2 dx.
b. True. They are both equal to
a
c. False. Because
0
p
1 + f 0 (x)2 > 0, arc length can’t be negative.
0 2
2
6.5.34 y = m, so 1 + (y ) = 1 + m , and L =
Z bp
a
1 + m2 dx = (b
a)
p
1 + m2 .
To see that this is the same result as the one given by the distance formula, consider the points (a, ma+c)
and (b, mb + c). The distance between them is
p
p
p
p
p
(b a)2 + (mb + c (ma + c))2 = (b a)2 + m2 (b a)2 = (b a)2 1 + m2 = (b a) 1 + m2 .
Copyright c 2019 Pearson Education, Inc.
624
Chapter 6. Applications of Integration
6.5.35
a. We are seeking functions f (x) so that f 0 (x) = ±4x2 , so any function of the form f (x) = ±4x3 /3 + C
will work.
b. We are seeking functions f (x) so that f 0 (x) = ±6 cos(2x), so any function of the form f (x) =
±3 sin(2x) + C will work.
±2
6.5.36 Because f 0 (x) = ±4x 3 , we have f (x) = 2 + C. Because f (1) = 5, we must have either f (x) =
x
2
2
f
(x)
=
3
+
.
7
or
x2
x2
Z 1p
6.5.37 The length of the parabola is
1 + 4x2 dx ⇡ 2.9597.
1
Z 1r
⇡2
The length of the given cosine function is
1+
sin2 (⇡x/2) dx ⇡ 2.924, so the parabola is longer.
4
1
Z ⇡p
6.5.38 f 0 (x) = sin x, so the arc length is given by
1 + sin2 x dx.
0
6.5.39
Z bp
L
1 + f 0 (u)2 du = .
2
a
Z bp
1
L
1 + f 0 (u)2 du = .
b. Let u = cx. Then the given integral is equal to
c a
c
a. Let u = 2x. Then the given integral is equal to
1
2
6.5.40 If f is odd, then the symmetry of f assures that the portion of the curve from 0 to b matches exactly
the portion of the curve from b to 0, as can be seen by rotating the original curve about the y-axis and
then about the x-axis. The same is true for f even, although the two portions match up after just rotating
about the y-axis.
Suppose f is even, and recall that this means that f 0 is odd. Consider
Z 0p
1 + f 0 (x)2 dx.
L =
b
Let u =
x. The integral becomes
Z bp
Z bp
Z bp
1 + f 0 ( x)2 dx =
1 + ( f 0 (x))2 dx =
1 + f 0 (x)2 dx = L+ .
L =
0
0
If f is odd, so that f 0 is even, we have L =
0
Z 0p
1 + f 0 (x)2 dx. Let u =
x. Then the integral becomes
b
Z bp
Z bp
0
2
L =
1 + f ( x) dx =
1 + f 0 (x)2 dx = L+ .
0
6.5.41
0
✓
◆2
1
1
1
1
a. f 0 (x) = Aaeax
e ax , so 1 + f 0 (x)2 = 1 + (Aaeax )2
+
e ax
= (Aaeax )2 + +
4Aa
2
4Aa
2
◆2
✓
◆2 ✓
1
1
e ax
= Aaeax +
e ax . So the arc length for c  x  d is
4Aa
4Aa
L=
Z d
c
(Aae
ax
1
+
e ax ) dx =
4Aa
✓
ax
Ae
1
e ax
4Aa2
Copyright c 2019 Pearson Education, Inc.
◆d
c
.
6.5. Length of Curves
625
b. Applying the previous result with c = 0 and d = ln 2, we have
✓
1
2 a
4Aa2
2n + 1 1/(2n)
x
, so 1 + (y 0 )2 = 1 +
2n
✓
2n + 1
2n
◆2
x1/n .
L=
Z a
s
✓
2n + 1
2n
Aeax
1
e ax
4Aa2
◆ ln 2
= A2a
L=
0
A+
1
= A(2a
4Aa2
◆2
x1/n dx.
1)
1
(2 a
4a2 A
1).
6.5.42
a. y 0 =
So
0
1+
p
2n + 1
, and let d = 1 + r2 a1/n . If we let u2 = 1 + r2 x1/n , then
2n
b. For the sake of simplicity, let r =
✓ 2
◆n
u
1
n
x=
, and dx = 2n (u2
2
r
r
1)n 1 · 2u du, and our integral becomes
Z d
n
u · 2n (u2
r
1
1)
n 1
Z d
2n
· 2u du = 2n
r
u2 (u2
1)n 1 du.
1
c. The binomial theorem assures us that
(u2
1)n 1 = u2n 2
2
n 1
✓
n
1
◆
✓
n
1 2n 4
u
+
2
◆
1 2n 6
u
· · · + ( 1)n 1 .
Thus
2
u (u
1)
=u
2
✓
u
2n 2
✓
n
= u2n
✓
n
◆
2n 4
/(2n + 1)
✓
n
1
✓
n
1
◆
2n 6
n 1
u
+
u
· · · + ( 1)
1
2
◆
✓
◆
n 1 2n 4
1 2n 2
u
+
u
· · · + ( 1)n 1 u2 .
1
2
◆
Then
2n
L = 2n
r
✓
n
+
✓
2
u
2n+1
◆
1 2n 3
u
/(2n
3)
1
1
◆
u2n 1 /(2n
1)
◆d
· · · + ( 1)n+1 u3 /3
.
1
p
d. For n = 2 and a = 1p we have r = (5/4) and d = 41/4. Using the previous result, we have
p
41/4
45 5
2048 1763 41
3
L = 4 u /5 u /3
=
+
⇡ 1.423.
9375
9375
5
1
p
For n = 3 and a = 1 we have r = 7/6 and d = 85/6. Using the previous results, we have
67
L = 6 u7 /7
7
5
3
2u /5 + u /3
p
1
85/6
p
142885 85
=
823543
Copyright c 2019 Pearson Education, Inc.
746496
⇡ 1.418.
4117715
626
Chapter 6. Applications of Integration
e.
�
����
����
����
����
����
�
6.6
�
�
�
�
�
�
�
Surface Area
6.6.1 S = ⇡r
p
p
r2 + h2 = 3⇡ 9 + 16 = 15⇡.
p
p
6.6.2 By the formula developed in the section: S = ⇡(4)(62 22 ) 1 + 42 = 128 17⇡. Or it can be computed
via integration:
Z 6 p
6
p
p
p
S = 2⇡
4x 1 + 42 dx = 8 17⇡ x2 /2
= 8 17⇡(18 2) = 128 17⇡.
2
2
6.6.3 Evaluate
Z b
2⇡f (x)
a
Z d
2⇡g(y)
p
1 + g 0 (y)2 dy.
2⇡(2
p
1)2 dx = 2
6.6.4 Evaluate
c
6.6.5
a. S =
Z 2
p
1 + f 0 (x)2 dx.
0
x)
p
1+(
2⇡
Z 2
(2
0
p
✓
p
x) dx = 2 2⇡ 2x
x2
2
◆ 2
p
= 4 2⇡.
0
b. The surface is a conepwith ` = 2 2 and r = 2; using the formula for the lateral surface area of a cone,
we have s = ⇡r` = 4 2⇡.
6.6.6
a. S =
Z d
c
Z 8
Z 8
p
p
2⇡g(y) 1 + g 0 (y)2 dy =
2⇡ · 3 1 + 0 dy = 6⇡
dy = 48⇡.
0
0
b. This surface is right circular cylinder of radius r = 3 and height h = 8; using the formula for the lateral
surface area of a cylinder we have S = 2⇡rh = 2⇡ · 3 · 8 = 48⇡.
Z 6
6
p
p
p
p
6.6.7 S = 2⇡
(3x + 4) 1 + 9 dx = 2 10⇡ 3x2 /2 + 4x
= 2 10⇡(54 + 24) = 156 10⇡.
0
6.6.8 S = 2⇡
Z 3
0
(12
3
p
p
3x) 1 + 9 dx = 2 10⇡ 12x
3x2 /2
1
p
= 2 10⇡(36
27/2
1
6.6.9
S = 2⇡
Z 20
9
= 16⇡
✓
p p
8 x 1 + (16/x) dx = 16⇡
2(x + 16)
3
◆ 20
3/2
9
=
Z 20
32⇡
(216
3
p
x + 16 dx
9
125) =
Copyright c 2019 Pearson Education, Inc.
2912⇡
.
3
(12
p
3/2)) = 24 10⇡.
6.6. Surface Area
627
Z 1 p
6.6.10 S = 2⇡
x3 1 + 9x4 dx. Let u = 1 + 9x4 so that du = 36x3 dx. Substituting gives
0
✓
◆ 10
Z 10
p
p
⇡ 2u3/2
⇡
⇡
S=
u du =
=
(10 10 1).
3
27
18 1
18
1
Z 2
p
6.6.11 Note that x = y 3 /3 for 0  y  2. The surface area is S = 2⇡
(y 3 /3) 1 + y 4 dy. Let u = 1 + y 4
0
Z 17
17
p
1 1/2
⇡ ⇣ 3/2 ⌘
⇡
3
u
u du =
= (17 17 1).
so that du = 4y dy. Substituting yields S = 2⇡
12
9
9
1
1
p
6.6.12 Note that x = 2 y for 1  y  4. The surface area is S = 2⇡
Z 4p
⌘ 4
p
8⇡ ⇣
8⇡ p
(1 + y)3/2
=
(5 5 2 2).
4⇡
1 + y dy =
3
3
1
1
Z 4
p p
2 y 1 + 1/y dy =
1
6.6.13
S = 2⇡
Z 1/2 p
x2
1
1/2
= 4⇡
Z 1/2 p
x2
1
0
s
1+
r
✓
p
x
1
x2
◆2
dx
Z 1/2 p
x2
1+
dx = 4⇡
1 x2
x2
1
0
r
1
1
x
dx = 4⇡
2
Z 1/2
dx = 2⇡.
0
6.6.14
S = 2⇡
Z 3.5 p
y 2 + 6y
3
2⇡
Z 3.5 p
y 2 + 6y
3
= 2⇡
Z 3.5 p
v
u
u
t
8 1+
v
u
u
8t 1 +
y 2 + 6y
8 + y2
2
p
p
2y + 6
y 2 + 6y
y+3
y 2 + 6y
8
!2
6y + 9 dy = 2⇡
3
6.6.15 Note that x =
8
!2
dy
dy
Z 3.5
1 dy = ⇡.
3
y+1
for 3  y  15. The surface area is
4
r
p Z
1
⇡ 17 15
dy =
(y + 1) dy
S = 2⇡
1+
8
16
3
3
p
p
p
15
p
⇡ 17 2
⇡ 17
⇡ 17
=
y /2 + y
=
(225/2 + 15 (9/2 + 3) =
(120) = 15 17⇡.
8
8
8
3
Z 15
y+1
4
6.6.16
S = 2⇡
Z 5
0
p
4x + 6(
q
1 + (2/
p
4x + 6)2 ) dx = 2⇡
Z 5
0
⌘
p
4⇡ ⇣
⇡
= (30 30
=
(4x + 10)3/2 /4
3
3
0
5
p
s✓
4x + 6
4x + 6 + 4
4x + 6
p
10 10⇡ p
10 10) =
(3 3
3
p
1).
Copyright c 2019 Pearson Education, Inc.
◆
dx = 2⇡
Z 5
0
p
4x + 10 dx
628
Chapter 6. Applications of Integration
6.6.17
s
r
Z
2
2e 2x )
⇡ 2 2x
16 + 4e4x 8 + 4e 4x
2x
S = 2⇡
(1/4)(e + e
) 1+
dx =
(e + e
)
dx
16
16
2
2
2
Z
Z
p
⇡ 2 2x
⇡ 2 2x
=
(e + e 2x ) (2e2x + 2e 2x )2 dx =
(e + e 2x )(2e2x + 2e 2x ) dx
8
8
2
2
✓ 4x
◆ 2
Z
4x
⇡ 2 4x
⇡
e
e
4x
=
(e + 2 + e
) dx =
+ 2x
4
4
4
4
2
2
✓ 8
✓ 8
◆◆
✓ 8
◆
8
8
⇡ e
e
e
e
⇡ e
e 8
4
=
+8
.
=
+4
4
4
4 2
2
4 4
4
Z 2
2x
2x
(2e2x
6.6.18
S = 2⇡
Z 4p
s
s
Z 4p
2
(5
2x)
20x 4x2 + 25 20x + 4x2
2
x2 1 +
dx
=
2⇡
5x
x
dx
4(5x x2 )
4(5x x2 )
1
s
Z 4
4
25
2
x
dx
=
5⇡
(x)
= 15⇡
dx
=
2⇡
5/2
4(5x x2 )
1
1
5x
1
= 2⇡
Z 4p
5x
1
6.6.19
Z 10 p
S = 2⇡
y2
12y
2
Z 10
= 2⇡
s
(12 2y)2
1+
dy = 2⇡
4(12y y 2 )
6 dy = 12⇡(10
Z 10 p
12y
y2
2
s
48y
4y 2 + 144 48y + 4y 2
dy
4(12y y 2 )
2) = 96⇡
2
6.6.20 Note that x =
p
1
Z 3/2 p
S = 2⇡
2y
= 2⇡
1)2 =
s
(2 2y)2
1+
dy = 2⇡
4(2y y 2 )
y2
1
Z 3/2
p
(y
1 dy = 2⇡(3/2
1
(y 2
2y + 1) =
p
2y
Z 3/2 p
2y
1
y2 .
y2
s
8y
4y 2 + 4 8y + 4y 2
dy
4(2y y 2 )
1) = ⇡.
1
6.6.21 S = 2⇡
Z 7p
8x
x2
Z 7p
1
s
s
s
Z 7p
(8 2x)2
32x 4x2 + 64 32x + 4x2
2
1+
dx = 2⇡
8x x
dx =
2
4(8x x )
4(8x x2 )
1
Z 7
dx = 2⇡
4 dx = 8⇡(7 1) = 48⇡. Because the surface area is 48⇡ square
64
4(8x x2 )
1
1
meters, the volume of paint required to cover the surface to a thickness of 0.0015 meters is 48⇡(.0015) ⇡ .226
cubic meters. This is about 59.75 gallons.
2⇡
8x
x2
6.6.22
S = 2⇡
Z 8p
100
x2
8
Z 8p
= 2⇡
100
8
x2
s
s
1+
( 2x)2
dx
4(100 x2 )
400 4x2 + 4x2
dx = 2⇡
4(100 x2 )
Z 8
10 dx = 20⇡(8
( 8)) = 320⇡.
8
Because the surface area is 320⇡ square meters, the volume of paint required to cover the surface to a
thickness of 0.0015 meters is 320⇡(0.0015) ⇡ 1.5 cubic meters. This is about 398.36 gallons.
Copyright c 2019 Pearson Education, Inc.
6.6. Surface Area
629
6.6.23
a. False. One would need to find x = f
1
(y) and compute the corresponding integral using f
1
.
b. False. For example, the curve given in number 14 above isn’t one-to-one on the given interval, but the
surface is still defined.
c. True. Because the curve is symmetric about the y-axis, the surface generated by revolving half the
curve is half the surface generated by revolving the whole curve.
d. False. This curve is symmetric about the y-axis, so the surface generated by revolving the whole curve
is the same as the surface generated by revolving the portion over [0, 4].
6.6.24
a. S = 2⇡
Z 1
x5
0
b. S =⇡ 3.362.
p
1 + 25x8 dx.
6.6.25
a. S = 2⇡
Z ⇡/2
0
b. S ⇡ 7.21.
6.6.26
a. S = 2⇡
Z e
p
cos x 1 + sin2 x dx.
ln y
1
r
1+
1
dy.
y2
b. S ⇡ 7.05.
6.6.27
a. S = 2⇡
Z ⇡/4
tan x
0
b. S ⇡ 3.84.
p
1 + sec4 x dx.
6.6.28
a. S = 2⇡
Z pe
ln x2
1
b. S ⇡ 3.845.
p
1 + 4/x2 dx.
p
x2/3 )3/2 . Note that f 0 (x) = x 1/3 a2/3 x2/3 , so
r
r
2/3
2/3
p
a
x
x2/3 + a2/3 x2/3
a1/3
=
= 1/3 .
1 + f 0 (x)2 = 1 +
2/3
2/3
x
x
x
Z a
Thus (using symmetry) S = 4⇡
(a2/3 x2/3 )3/2 (a1/3 x 1/3 ) dx. Let u = (a2/3 x2/3 ), so that du =
6.6.29 Let y = f (x) = (a2/3
( 2x 1/3 /3) dx. Substituting gives
S=
6⇡a
1/3
Z 0
a2/3
u
3/2
0
1/3
du = 6⇡a
Z a2/3
0
12⇡ 1/3 ⇣ 5/2 ⌘
du =
a
u
5
0
a2/3
3/2
u
=
12⇡a2
.
5
6.6.30 The surface area of the described cylinder is 2⇡rh. The surface area of the described cone is
p
2h
⇡r h2 + r2 . The cone’s area is p
of the cylinder’s.
h2 + r 2
Copyright c 2019 Pearson Education, Inc.
630
Chapter 6. Applications of Integration
6.6.31
s
✓
◆2
p ◆q
Z 2
p
p
x
9x 1
3/2
3/2
1/2
2
p
S = 2⇡
x
1 + (3 x/2 x
/6) dx = 2⇡
(x
x/3) 1 +
dx
3
6 x
1
1
s✓
r
◆
Z 2
Z 2
p
p
36x + 81x2 18x + 1
(1 + 9x)2
3/2
3/2
= 2⇡
(x
x/3)
dx = 2⇡
(x
x/3)
dx
36x
36x
1
1
◆
Z 2✓
Z
(1 + 9x)x 1 + 9x
2⇡ 2
(3x + 27x2 1 9x) dx
= 2⇡
dx =
18 1
6
18
1
Z
2
2⇡ 2
⇡
⇡
53⇡
=
(27x2 6x 1) dx =
9x3 3x2 x
= (72 12 2 (9 3 1)) =
9
18 1
9
9
1
Z 2✓
6.6.32
S = 2⇡
Z 2
(x4 /8 + x 2 /4)
1
= 2⇡
Z 2
(x4 /8 + x 2 /4)
1
= 2⇡
Z 2
p
p
1 + (x3 /2
x 3 /2)2 dx = 2⇡
x6 /4 + 1/2 + x 6 /4 dx = 2⇡
(x4 /8 + x 2 /4)(x3 /2 + x 3 /2) dx = 2⇡
1
Z 2
Z 2
1
Z 2
= 2⇡
= 2⇡
1/2
Z 2
1/2
Z 2
p
1 + x6 /4
p
1/2 + x 6 /4 dx
(x3 /2 + x 3 /2)2 dx
(x7 /16 + x/16 + x/8 + x 5 /8) dx
1
1/2
= 2⇡
(x4 /8 + x 2 /4)
1
Z
⇡ 2 7
⇡ 8
=
(x + 3x + 2x 5 ) dx =
x /8 + 3x2 /2 x 4 /2
8 1
8
⇡
1179⇡
= (32 + 6 (1/32) (1/8 + 3/2 1/2)) =
8
256
6.6.33
Z 2
p
S = 2⇡
(x3 /3 + x 1 /4) 1 + (x2 x 2 /4)2 dx
Z 2
(x4 /8 + x 2 /4)
(x3 /3 + x 1 /4)
3
(x /3 + x
1
/4)
p
p
1 + x4
1/2 + x 4 /16 dx = 2⇡
(x2 + x 2 /4)2 dx = 2⇡
(x5 /3 + x/12 + x/4 + x 3 /16) dx = 2⇡
1/2
= 2⇡ x6 /18 + x2 /6
Z 2
2
1
Z 2
(x3 /3 + x 1 /4)
1/2
p
x4 + 1/2 + x 4 /16 dx =
(x3 /3 + x 1 /4)(x2 + x 2 /4) dx
1/2
Z 2
(x5 /3 + x/3 + x 3 /16) dx
1/2
2
x 2 /32
= 2⇡(32/9 + 2/3
1/128
(1/1152 + 1/24
1/8)) =
1/2
p
p
6.6.34 We can rewrite the p
given curve as 2x+ p
4x2 1 = e2y . Squaring yields 4x2 +4x 4x2
e4y . So e4y + 1 = 8x2 + 4x 4x2 1 = 4x(2x + 4x2 1) = 4x(e2y ). Thus
275⇡
32
1+4x2
e4y + 1
e2y + e 2y
=
.
2y
4e
4
The portion of the curve in question is for 0  y  ln 2.
The surface area is
Z ln 2 2y
Z ln 2 2y
e + e 2y p
e + e 2y p 4y
2y
2y
2
S = 2⇡
1 + (e /2 e
/2) dy = 2⇡
e /4 + 1/2 + e 2y /4 dy
4
4
0
0
p
Z ln 2 2y
Z
⇡ ln 2 4y
⇡ 4y
e + e 2y (e2y + e 2y )2
= 2⇡
dy =
(e + 2 + e 4y ) dy =
e /4 + 2y e 4y /4
4
4
2
4 0
0
✓
◆
⇡
⇡ 255
= (4 + 2 ln 2 1/64 (1/4 + 0 1/4)) =
+ 2 ln 2
4
4 64
1=
x=
Copyright c 2019 Pearson Education, Inc.
ln 2
0
6.6. Surface Area
631
6.6.35
S = 2⇡
Z 4
y 1/2 /12)
(4y 3/2
y 1/2 /12)
(4y 3/2
y 1/2 /12)
(4y 3/2
y 1/2 /12)
(4y 3/2
y 1/2 /12)(6y 1/2 + y 1/2 /24) dy
(24y 2
y/3
1
= 2⇡
Z 4
1
= 2⇡
Z 4
1
= 2⇡
Z 4
1
= 2⇡
Z 4
q
(4y 3/2
p
p
q
1 + (6y 1/2
y 1/2 /24)2 dy
1 + 36y
1/2 + 1/(576y) dy
36y + 1/2 + 1/(576y) dy
(6y 1/2 + y 1/2 /24)2 dy
1
= 2⇡
Z 4
1/288) dy
1
= 2⇡ 8y 3
4
y 2 /6
y/288
p
x2 .
1
r2
✓
= 2⇡ 512
8
3
1
72
✓
8
1
6
1
288
◆◆
=
48143⇡
.
48
6.6.36 Let y = f (x) = a +
Revolving this curve around p
the x-axis will give a portion of the
surface of the torus. The other comes from revolving y = g(x) = a
r2 x2 around the x-axis. We will
compute the surface area for the first portion first.
Note that
s
✓
◆2 r 2
p
x
r
x2 + x 2
r
0
2
1 + f (x) = 1 + p
=
=p
.
2
2
2
2
2
r
x
r
x
r
x2
So (using symmetry) the surface area for this portion of the torus is
◆
✓
Z r
Z r
Z r
p
r
1
p
4⇡
(a + r2 x2 ) p
dx = 4⇡ar
dx + 4⇡r
1 dx
r 2 x2
r 2 x2
0
0
0
r
= 4⇡ar sin 1 (x/r)
p
+ 4⇡r2 = 4⇡ar (⇡/2
0) + 4⇡r2 = 2⇡ 2 ar + 4⇡r2 .
0
If we revolve y = g(x) = a
x2 around the x-axis, the surface area will be 2⇡ 2 ar
similar calculation. Thus, the total surface area of the torus is
r2
4⇡r2 by a very
S = 2⇡ 2 ar + 4⇡r2 + 2⇡ 2 ar 4⇡r2 = 4⇡ 2 ar.
p
6.6.37 By symmetry, we can let y = f (x) = r2 x2 , and imagine the surface obtained by revolving this
curve around the x-axis for a  x  a + h. The surface area is
s
✓
◆2
Z a+h p
x
2
2
S = 2⇡
r
x 1+ p
dx.
r 2 x2
a
This can be written as
Z a+h p
2⇡
r2
a
= 2⇡
a. Let y = f (x) =
r
r2
x2 + x 2
dx
x2
p
Z a+h
r2
2
dx = 2⇡
r dx = 2⇡r(a + h
x p
r 2 x2
a
r2
Z a+h p
r2
bp 2
a
a
x2 . Then f 0 (x) =
a
6.6.38
x2
a) = 2⇡rh.
x
b
·p
, so
2
a
a
x2
r
r
2
2
p
a4 a 2 x 2 + b 2 x 2
b
x
1
1 + f 0 (x)2 = 1 + 2 · 2
=
·p
.
2
a2
x
a a
a2 x2
Copyright c 2019 Pearson Education, Inc.
632
Chapter 6. Applications of Integration
p
1 + f 0 (x)2 is
r
r
bp 2
a4 a2 x 2 + b 2 x 2
1
b
2
a
x
·p
=
a2
a
a2
a
a2 x2
Thus f (x)
Letting c2 = 1
b2
and using symmetry gives
a2
Z
4⇡b a p 2
S=
a
a 0
(a2
b2 )x2
a2
b
=
a
s
a2
✓
b2
a2
1
◆
x2 .
c2 x2 dx.
b. Let u = r
cx, so that du = c dx, Note that when x = 0, u is also 0, and when x = a, we have
b2 p 2 p 2
a = a
b2 . Thus,
u = ca = 1
a2
Z pa 2 b2 p
Z pa 2 b2 p
4⇡b
4⇡b
S=
a2 u2 du = p
a2 u2 du.
ac 0
a 2 b2 0
c. S is
⇣ p
2⇡b
p
u a2
a 2 b2
u2 + a2 sin 1 (u/a)
p
⌘
p
a2 b2
2⇡b
=p
a 2 b2
0
a2
b2 + a2 sin 1
p
a 2 b2
a
!!
.
b2 in the denominator gives
!!
p
a2
a 2 b2
1
S = 2⇡b b + p
sin
.
a
a 2 b2
Multiplying through by the factor of
a2
b
p
d. S would be measured in square meters.
e. Returning to our work in the first
Z part of this problem, if a = b then f (x)
a
So the surface area is S = 4⇡
b dx = 4⇡b(a
2
0) = 4⇡ab = 4⇡a .
p
p
1 + f 0 (x)2 = ab a2
0 = b.
0
6.6.39
a. Using R for the radius of the sphere, The ratio is
b. The ratio is
6
6R2
= .
R
R3
4⇡R2
3
= .
3
R
(4/3)⇡R
p
c. We have an ellipsoid whose axes have lengths 2R, R, and R, where R = ` 3 4. The volume of the
ellipsoid is
4⇡
R R
⇡R3
4⇡`3
V =
·R· ·
=
=
.
3
3
3
2 2
The surface area is
!
p
R2 (R2 /4)
R R
R2
1
+p
sin
S = 2⇡ ·
R
2 2
R2 (R2 /4)
!!
p
R
2R2
3R
= ⇡R
+p
sin 1
2R
2
3R
p
p
p !!
p
` 3 4 2` 3 4
3
3
1
= ⇡` 4
+ p sin
2
2
3
!
p
p
p
` 3 4 2` 3 4 ⇡
3
+ p ·
= ⇡` 4
2
3
3
✓
◆
1
2⇡
+ p
.
= ⇡`2 42/3
2 3 3
Copyright c 2019 Pearson Education, Inc.
6.6. Surface Area
633
The SAV ratio is therefore
3⇡`2 42/3
⇣
1
2⇡
p
2 + 3 3
4⇡`3
⌘
p
p
3(3 3 + 4⇡)
9 + 4 3⇡
p
=
.
=
6 · 41/3 `
41/3 `6 3
d.
SAV
2.5
2.0
1.5
1.0
0.5
1
2
3
4
5
a
The lower curve is for the sphere and the upper for the ellipsoid.
e. More spherical is better for minimizing heat loss.
g(b) g(a)
.
b a
p
The slant height ` can be written as (g(b) g(a))2 + (b a)2 , so the quantity ⇡(g(b) + g(a))l can be
written as
p
p
b a
p
⇡(g(b) + g(a)) (g(b) g(a))2 + (b a)2 = (b a)⇡(g(b) + g(a)) c2 + 1.
(b a)2
6.6.40 Let c =
We will show that the surface area of the frustrum is this quantity.
g(b) g(a)
The line between (a, g(a)) and (b, g(b)) has slope
= c. Then the equation of the line is
b a
y = f (x) = g(a) + c(x a). The surface area is
Z b
Z b
p
p
S = 2⇡
(g(a) + c(x a)) 1 + c2 dx = 2⇡ 1 + c2
(g(a) + c(x a)) dx
a
2⇡
p
=⇡
=⇡
as desired.
1 + c2
p
p
✓
g(a)x +
1 + c2 (2g(a)(b
1 + c2 (b
a)2
c(x
2
◆ b
= 2⇡
a
a) + (g(b)
p
1 + c2
g(a))(b
a)(g(a) + g(b))
✓
a
c(b
a)2
◆
g(a)(b a) +
2
p
2
a)) = ⇡ 1 + c (b a)(2g(a) + g(b)
g(a))
6.6.41
a. Note that g(x) = cf (x), so that g 0 (x) = cf 0 (x).
Z b
Z b
Z b
p
p
p
2
2
0
2
2
2
0
2
cf (x) c + c f (x) dx = 2c ⇡
f (x) 1 + f 0 (x)2 dx = c2 A.
2⇡
g(x) c + g (x) dx = 2⇡
a
a
0
a
0
b. Note that h (x) = cf (cx). Consider
Z b/c
Z b/c
p
p
2
2
0
2
f (cx) 1 + f 0 (cx)2 dx.
2⇡
f (cx) c + c f (cx) dx = 2c⇡
a/c
a/c
Let u = cx so that du = c dx. Then our integral is equal to
Z b
p
2⇡
f (u) 1 + f 0 (u)2 du = A.
a
Copyright c 2019 Pearson Education, Inc.
634
Chapter 6. Applications of Integration
6.6.42 The area obtained by revolving f (x) + C around the x-axis is
2⇡
Z b
(f (x) + C)
a
a
as desired.
6.7
Z b
Z bp
p
p
1 + f 0 (x)2 dx = 2⇡
f (x) 1 + f 0 (x)2 dx + 2⇡C
1 + f 0 (x)2 dx = S + 2⇡CL,
a
Physical Applications
6.7.1 m = ⇢1 · l1 + ⇢2 · l2 = 1 g/cm · 50 cm + 2 g/cm · 50 cm = 150 g.
6.7.2 The mass is given by m =
Z b
⇢(x) dx.
a
6.7.3 The work is the product of 5 Newtons and 5 meters, which is 25 J.
6.7.4 If the force is not constant, the interval must be divided up into pieces, and on each small piece the
work can be approximated by assuming a constant force. These approximations are then added up and then
the sum is refined through a limiting process, which leads to a definite integral.
6.7.5 Di↵erent volumes of water are moved di↵erent distances.
6.7.6 Di↵erent parts of the dam have di↵erent depths and thus di↵erent amount of pressure.
6.7.7 F = ⇢gh = 1000 · 9.8 · 4 = 39, 200 N/m2 .
6.7.8 Along a thin horizontal strip, the pressure is the same, because the depth is constant.
6.7.9 W =
Z 10
25⇡⇢g(15
y) dy.
25⇡⇢g(15
y) dy.
5
6.7.10 W =
Z 5
0
6.7.11 W =
Z 10
25⇡⇢g(10
y) dy.
25⇡⇢g(10
y) dy.
0
6.7.12 W =
Z 3
0
6.7.13 m =
Z ⇡
⇡
(1 + sin x) dx = ( x
cos x)
=⇡
0
6.7.14 m =
Z 1
(1 + x3 ) dx = x + x4 /4
Z 2
(2
Z 4
5e 2x dx = ( 5/2)e 2x
1
0
6.7.15 m =
0
1) = ⇡ + 2.
2
x/2) dx = 2x
x2 /4
= 3.
0
4
0
Z 1 p
6.7.17 m =
x 2
0
(0
5
.
4
=
0
6.7.16 m =
( 1)
0
=
0
x2 dx =
✓
1
(2
3
5
(1
2
2 3/2
x )
e 8 ).
p
p
2 2
= ( 1/3) + (2 2/3) =
3
0
◆ 1
Copyright c 2019 Pearson Education, Inc.
1
.
6.7. Physical Applications
Z 2
6.7.18 m =
1 dx +
0
6.7.19 m =
Z 3
2 dx = 2 + 2 = 4.
Z 4
(1 + x) dx = 2 + x + x2 /2
2
Z 2
1 dx +
Z 1
x2 dx +
Z 3
2x dx = x2
Z 3
2/x2 dx = ( 2/x)
0
6.7.20 m =
4
2
0
6.7.21 W =
Z 2
= 2 + (4 + 8)
(2 + 2) = 10.
2
1
x) dx = x3 /3
x(2
2
+ x2
1
x3 /3
=
0
1
1
+4
3
8
3
✓
1
1
3
3
0
6.7.22 W =
635
= 9 J.
0
3
=
1
1
2
3
( 2) =
4
J.
3
6.7.23
a. Because f (0.2) = 0.2k = 30, we have k = 150.
b. W =
c. W =
Z
0.4
0.4
150x dx = 75x2
0
Z 0.3
0.3
150x dx = 75x2
0
d. W =
= 75 · 0.16 = 12 J.
0
Z 0.4
= 75 · 0.09 = 6.75 J.
0
0.4
150x dx = 75x2
= 75(0.16
0.2
0.04) = 9 J.
0.2
6.7.24
a. f (0.25) = 0.25k = 15, so k = 60.
b. W =
c. W =
Z
0.2
0.2
60x dx = 30x2
0
Z 0.55
= 30 · 0.04 = 1.2 J.
0
0.55
60x dx = 30x2
0.25
= 30(0.3025
0.25
0.0625) = 30 · 0.24 = 7.2 J.
6.7.25
a. f (x) = kx, and f (0.5) = 50, so k(0.5) = 50, so k = 100.
Z 1.5
1.5
Therefore W =
100x dx = 50x2
= 112.5 J.
0
b.
R
0.5
0
0
0.5
100x dx = 50x2
= 12.5 J.
0
6.7.26 f (x) = kx, and f (0.02) = 0.02k = 500 · 9.8 = 4900, so k = 245000.
Z 0.04
0.04
W =
245000x dx = 122500x2
= 196 J.
0
0
6.7.27
a. f (0.2) = 0.2k = 50, so k = 250. W =
Z 0.5
0
0.5
250x dx = 125x2
0
= 125 · 0.25 = 31.25 J.
Copyright c 2019 Pearson Education, Inc.
◆
= 1.
636
b.
Chapter 6. Applications of Integration
R 0.2
0
0.2
2
kx dx = kx /2
= 0.02k = 50, so k = 2500. W =
0.5
2500x dx = 1250x2
0
0
312.5 J.
Z 0.5
= 1250 · 0.25 =
0
6.7.28
a. f (0.1) = 0.1k = 50, so k = 500. W =
Z 0.4
0.4
500x dx = 250x2
0
b.
R 0.1
0
0.1
kx dx = kx2 /2
= 250 · 0.16 = 40 J.
0
= 0.005k = 2, so k = 400. W =
Z 0.4
0.4
400x dx = 200x2
0
0
0
= 200 · 0.16 = 32 J.
6.7.29
a. We have
Z 0.5
0.5
kx dx =
0
k 2
x
= 0.125k = 100, so that k = 800. Then
2 0
W =
Z 1.25
1.25
800x dx = 400x2
0
= 625 J.
0
b. f (0.5) = 0.5k = 250, so k = 500 and thus
W =
Z 1.25
1.25
500x dx = 250x2
0
= 390.625 J.
0
y
W (x) =
Z x
x
25t dt = 25t2 /2
0
6.7.30
100
= 25x2 /2.
80
0
Note that W is an even function, so that
W ( x) = W (x), and thus the work is the
same to compress or stretch the spring a given
distance from its equilibrium position.
60
40
20
-3
-2
-1
1
2
x
6.7.31
a. W1 =
Z 30
30
⇢g(30
y) dy = 5g 30y
y 2 /2
0
0
= 2250g = 2250 · 9.8 = 22050 J.
b. W = W1 + W2 , where W2 is the work to just lift the block. W2 = 50g · 30 = 1500g, so W =
2250g + 1500g = 3750g = 3750 · 9.8 = 36750 J.
6.7.32 W =
Z 60
0
55
g(60
1000
y) dy =
55g
60y
1000
60
y 2 /2
0
= 99g = 99 · 9.8 = 970.2 J.
6.7.33 The bottom half of the chain weighs 25 kg, so this problem is equivalent to determining the sum of
the work done by the winch in winding up a 25-kg, 10-m chain and lifting a 25-kg mass a distance of 10 m.
So
W =
Z 20
2.5(9.8)(20
y) dy + 25(9.8)(10) = 2450 + 24.5
10
= 2450 + 24.5 (400
Z 20
(20
10
200
(200
50)) = 2450 + 1225 = 3675 J.
✓
y) dy = 2450 + 24.5 20y
Copyright c 2019 Pearson Education, Inc.
y2
2
◆ 20
10
6.7. Physical Applications
✓
6.7.34 F (x) = 9.8 4
637
◆
1
y , so
5
W =
Z 10
0
Z 2.5
6.7.35 W =
⇢gA(y)(2.5
0
2
3675000 ·
✓
9.8 4
1
y
5
◆
✓
dy = 9.8 4y
y2
10
y) dy = 1000 · 9.8 · 25 · 15
(2.5)
= 11, 484, 375 J.
2
◆ 10
= 9.8(40
10) = 294 J.
0
Z 2.5
(2.5
y) dy = 3675000
0
✓
2.5y
y2
2
◆ 2.5
=
0
6.7.36
a. W =
Z 8
0
8
⇢g⇡ · 22 (8
y 2 /2
y) dy = 4⇡⇢g 8y
⇡ 3.941 ⇥ 106 J.
0
b. Not true. For pumping half the water from a full tank the work is
Z 8
8
4⇢⇡g(8 y) dy = 4⇡⇢g 8y y 2 /2
= 32⇡⇢g ⇡ 985203 J.
4
4
To empty a half-full tank, the work is
Z 4
4⇢⇡g(8 y) dy = 4⇡⇢g 8y
4
y 2 /2
0
6.7.37 W =
Z 4
2
⇢⇡g2 (10
y) dy = 4⇡⇢g
0
6.7.38 W =
1.47 ⇥ 107 J.
Z 2
0
✓
y2
2
10y
⇢gA(y)(3 y) dy = 1000 · 9.8 · 25 · 15
0
◆ 4
= 96⇡⇢g ⇡ 2.95561 ⇥ 106 J.
= 4⇡⇢g(40
0
Z 2
8) = 128⇡⇢g ⇡ 3.941 ⇥ 106 J.
(3 y) dy = 3675000
0
✓
3y
y2
2
◆ 2
0
= (3675000) · 4 =
6.7.39
a. Let the vertex of the cone be at (0, 0), with the y-axis vertically oriented. Note that the area of a
horizontal slice at height y is ⇡y 2 /16, and it must move 6 y meters to get to the top.
✓
◆ 6
Z
Z 6
y2
⇡⇢g 6 2
⇡⇢g
y4
⇡⇢g
W =
⇢g⇡ (6 y) dy =
(6y
y 3 ) dy =
2y 3
=
(108) = 66, 150⇡ J.
16
4
16
16
16
0
0
0
b. Not true.
Z 3
y2
⇢g⇡ (6
16
0
⇡⇢g
y) dy =
16
Z 3
(6y
2
0
⇡⇢g
y ) dy =
16
3
✓
2y
3
y4
4
◆ 3
0
=
⇡⇢g
· 33.75 ⇡ 20672⇡ J,
16
less than half the amount from part (a). Note that while the water must be raised further than water
in the top half, due to the shape of the tank, there is far less water in the bottom half than in the top.
6.7.40
a. W = ⇡⇢g
Z 8
8
9(10 y) dy = ⇡⇢g 90y
9y 2 /2
4
b. W = ⇡⇢g
Z 4
0
8.867 ⇥ 106 J.
4
= (720 288 360+72)⇡⇢g = 144⇡⇢g ⇡ 4.433⇥106 J.
4
9(10
y) dy = ⇡⇢g 90y
9y 2 /2
= (360
0
72)⇡⇢g = 288⇡⇢g ⇡ 4.43342 ⇥ 106 ⇡
Copyright c 2019 Pearson Education, Inc.
638
Chapter 6. Applications of Integration
c. The water in the lower part of the tank has to travel farther, so it requires more work to pump it out.
6.7.41 A vertical cross section of the tank that passes through the center of the tank is a circle of radius 8.
If we place the origin of the xy-plane at the center of this circle, then we have
W =
Z 8
y 2 )⇡(10 + y) dy = 9800⇡
1000(9.8)(64
8
✓
= 9800⇡ 640y + 32y 2
10y 3
3
y4
4
◆ 8
Z 8
10y 2
(640 + 64y
y 3 ) dy
8
8
✓
✓
5120
5120
= 9800⇡ 5120 + 2048
1024
5120 + 2048 +
3
3
✓
◆
✓
◆
10240
20480
= 9800⇡
⇡ 210, 176, 737 J.
= 9800⇡ 10240
3
3
1024
◆◆
6.7.42
a. Orient the axes so that (0, 0) is the south pole of the semicircle at one end of the trough.pThe equation
of the semicircle is x2 + (y (1/4))2 = (1/4)2 . At a height of y, a slice has area 2x · 3 = 6 (1/2)y y 2 .
The distance it must travel to the top is 1/4 y.
W =
Z 1/4
⇢g6
0
y 2 , so that du = ((1/2)
Let u = (1/2)y
W = 3⇢g
Z 1/16
0
p
(1/2)y
y 2 ((1/4)
2y) dy = 2((1/4)
y) dy.
y) dy. Then we have
⇣
⌘ 1/16
u1/2 du = 2⇢g u3/2
= ⇢g/32 = 306.25 J.
0
b. Yes. If we double the length of the trough, the area of a slice is doubled, and the work integral is
doubled.
c. No. If the radius is doubled, the work is more than doubled. (There are “more slices,” and each must
travel farther to get to the top of the tank.)
6.7.43
a. Orient the axes so that the lower corners of the trough are at ( 0.25, 0) and at (0.25, 0). Then the
upper corners are at ( 0.5, 1) and at (0.5, 1). Note that the line between (0.25, 0) and (0.5, 1) is given
by y = 4x 1. The area of a slice at height y is 2x · 10 = 20 · 14 (y + 1) = 5(y + 1). Thus,
W = ⇢g
Z 1
5(y + 1)(1
y) dy = 5⇢g
0
Z 1
1
(1
y 2 ) dy = 5⇢g y
y 3 /3
0
=
0
10⇢g
⇡ 32, 667 J.
3
b. Yes. If the length is doubled, the area of each slice is doubled, so the work integral is doubled as well.
6.7.44 Note that this tank is full of water, rather than gasoline.
W = 1000 · 9.8
Z 5
5
p
20 25
y 2 (10
Z 5p
y) dy = 196, 000
25
5
The integral can be divided into two integrals as in the example. We have
10
Z 5p
5
25
y 2 dy
Z 5 p
y 25
y 2 dy.
5
Copyright c 2019 Pearson Education, Inc.
y 2 (10
y) dy.
6.7. Physical Applications
639
25⇡
= 125⇡. The
2
2y dy. The resulting integral is
The first represents 10 times the area of a half-circle of radius 5, so the value is (10)
y 2 , du =
integral can be computed with the substitution u = 25
second
Z
1 0p
u du = 0. So the total work is
2 0
W = 196000 · 125⇡ ⇡ 7.697 ⇥ 107 J.
6.7.45 Let the vertex of the cone be at (0, 0), with the y-axis vertically oriented. Note that the area of a
horizontal slice at height y is ⇡y 2 /16, and it must move 3 y meters to get to the point 1 meter above the
top.
Z 2
Z 2
y2
W =
⇢g⇡ (3 y) dy = (⇡⇢g/16)
(3y 2 y 3 ) dy
16
0
0
2
= (⇡⇢g/16) y 3
y 4 /4
= (⇡⇢g/16)(4) =
0
6.7.46 F =
Z 10
⇡⇢g
⇡ 7696.9 J.
4
10
⇢g(10
0
y 2 /2
y) · 40 dy = 40⇢g 10y
= 200⇢g = 1.960 ⇥ 106 N.
0
6.7.47 Orient the axes so that the lower corners of the trapezoid are at (5, 0) and ( 5, 0), and the upper
corners are at (10, 15) and ( 10, 15). Note that the line between the corners for x > 0 is given by y = 3(x 5),
2y
+ 10.
so at level y, we have a width of 2x =
3
F = ⇢g
Z 15
(15
0
2⇢g
=
225y
3
y)
✓
2y + 30
3
15
y 3 /3
0
◆
dy =
2⇢g
3
Z 15
(225
y 2 ) dy
0
2⇢g
=
· 2250 = 1500⇢g = 1.470 ⇥ 107 N.
3
6.7.48 Orient the axes so that (0, 0) is at the “south pole” of the semicircle. The
p center of the circle is
2
2
2
and the radius is 20, so the equation is x + (y 20) = 20 , so x = 40y y 2 . The width is
(0, 20) p
2x = 2 40y y 2 .
F = 2⇢g
Z 20
(20
0
Let u = 40y
p
y)
40y
y 2 dy.
y 2 , so that du = 2(20 y) dy. Then
Z 400
⇣
⌘ 400
p
16000
u du = ⇢g 2u3/2 /3
=
⇢g = 5.227 ⇥ 107 N.
F = ⇢g
3
0
0
6.7.49 Orient the axes so that the bottom vertex is at (0, 0). The other vertices are at (±10, 30), and the
line between (0, 0) and (10, 30) is given by y = 3x. Thus, a slice at height y has width 2x = 2y/3.
F =
Z 30
0
⇢g(30
y)
2y
2⇢g
dy =
15y 2
3
3
30
y 3 /3
=
0
2⇢g
(4500) = 3000⇢g = 2.940 ⇥ 107 N.
3
p
6.7.50 At a height of y, the width of a slice is 2x = 2(4 y) = 8y 1/2 . This gives
✓
◆
Z 25
⇣
⌘ 25
2500
1/2
3/2
5/2
(25 y)y dy = 8⇢g 50y /3 2y /5
= 8⇢g
= 6.533 ⇥ 107 N.
F = 8⇢g
3
0
0
6.7.51 The width of the plate at depth y is 2 y, so the force on the plate is
✓
◆ 2
Z 2
y3
2
2
⇢g(2 y)y dy = ⇢g y
= ⇢g · ⇡ 6533 N.
F =
3
3
1
1
Copyright c 2019 Pearson Education, Inc.
640
Chapter 6. Applications of Integration
6.7.52 F =
Z 1/2
⇢g(4
Z 1.5
⇢g(4
1/2
0
6.7.53 F =
y) · 0.5 dy = ⇢g 2y
y 2 /4
y) · 0.5 dy = ⇢g 2y
y 2 /4
=
15⇢g
= 9187.5 N.
16
=
11⇢g
= 6737.5 N.
16
0
1.5
1
1
6.7.54 Orient the axes so that (0, 0) is at the bottom of the circle at the bottom of the pool. Then
the
p
2
2
2
equation of the circle is x + (y 1/2) = (1/2) , so the width of a slice at a height of y is 2x = 2 y y 2 .
Thus,
F = 2⇢g
Z 1
(4
y)(
0
6.7.55 F =
Z 50
0
p
Z 1 p
Z 1
p
7
1
2
y y dy + 2⇢g
(
y) y y 2 dy
0 2
0 2
✓
◆ 1
⇣⇡ ⌘
1
7⇡⇢g
+ 2⇢g
(y y 2 )3/2
=
+ 0 ⇡ 2.694 ⇥ 104 N.
= 7⇢g
8
3
8
0
y 2 ) dy = 2⇢g
y
50
(150 + 2y) · 80 dy = 80 150y + y 2
= 8 ⇥ 105 N.
0
6.7.56 Arrange the axes so that the origin is in the center of the circular
end of the tank. The equation of
p
the circle is x2 + y 2 = 25, so the width of the tank at height y is 2 25 y 2 , and the depth is 5 y. Thus
the force is given by
⇢g
Z 5
5
(4
p
y)2 25
y 2 dy = 14445.2
= 14445.
Z 5
(5
5
✓Z 5
5
y)
p
5 25
p
25
y 2 dy
y 2 dy +
Z 5 p
y 25
5
y 2 dy
◆
.
The first integral is five times the area of a semicircle of radius 5, while the second integral is zero since it
125⇡
is a symmetric integral of an odd function. So the total force is 14445.2 ·
⇡ 2.836 ⇥ 106 N.
2
6.7.57
Z b
1
Z b
⇢(x) dx · (b
b a a
Z L
Z L
kL2
b. True.
kx dx =
=
kx dx.
2
0
0
a. True. m =
⇢(x) dx =
a
a) = ⇢ · L.
c. True. This follows because work is force times distance.
d. False. Although they have the same geometry, they are placed at di↵erent depths of the water, so the
force is di↵erent.
6.7.58
a. m1 =
Z L
L
4e x dx =
4e x
0
m2 =
Z L
= 4(1
e L ).
0
L
6e 2x dx =
3e 2x
0
= 3(1
0
L 2
e 2L ).
These are the same when 3(e )
4e L + 1 = 0, or (e L
bigger on (0, ln 3) and m1 is bigger for L > ln 3.
1)(3e L
b. No. lim m1 = 4 and lim m2 = 3.
L!1
L!1
Copyright c 2019 Pearson Education, Inc.
1) = 0, so L = ln 3. m2 is
6.7. Physical Applications
641
6.7.59
y
100
50
Compared to the linear spring F (x) = 16x,
the restoring force is less for large displacements.
a.
-6
-4
2
-2
4
6
x
-50
-100
b. W =
Z 1.5
1.5
(16x
0.1x3 ) dx = 8x2
.025x4
= 17.87 J.
0
c. W =
Z
0
2
2
(16x
0.1x3 ) dx = 8x2
.025x4
= 31.6 J.
0
0
6.7.60
a. f (x) = kx, and f (2) = 2k = 10g, so k = 5g. To compress the spring:
W =g
Z 2
2
5x dx = g 5x2 /2
1.5
= g(10
45/8) = 35g/8 = 42.875 J.
1.5
To move the mass: 10 · 0.5 · 9.8 = 49 J. So the total work is 42.875 + 49 = 91.875 J.
b. To stretch the spring:
W =g
Z 2.5
2.5
5x dx = g 5x2 /2
2
= g(125/8
10) = 45g/8 = 55.125 J.
2
To move the mass: 10 · 0.5 · 9.8 = 49 J. So the total work is 55.125 + 49 = 104.125 J.
6.7.61 Dividing the leak rate by the lifting rate, we find that the bucket is leaking at 20 kg/m. So the
force exerted by the bucket after it has been raised y meters is F (y) = 9.8(5000 20y). Therefore, the work
required to lift the bucket 30 m is
W =
Z 30
30
9.8(5000
20y) dy = 9.8 5000y
0
10y 2
= 9.8 (150, 000
9000) = 1, 381, 800 J.
0
6.7.62 Orient the y-axis vertically, with the point (10, 0) representing a point on the bottom of the pool in
one corner of the deep end, and ( 10, 1) representing a point on the bottom of the pool in the shallow end
1
(10 x).
on the same side of the pool. Note that the straight line between these two points is given by y =
20
So in the first 1 meter of depth, at a height of y the length of the pool is 10 x = 20y, so the area of a slice
is 20y · 10. In the 2nd meter of depth, the slices are uniformly of area 200 square meters.
W =
Z 1
⇢g200y(2.2
y) dy +
0
= 200⇢g (1.1
Z 2
⇢g200(2.2
2
(2.2
1
y) dy = 200⇢g
1.1y
2
3
y /3
1
(1/3) + (4.4
+ 2.2y
0
880
1/2))) =
⇢g ⇡ 2.87467 ⇥ 106 J.
3
Copyright c 2019 Pearson Education, Inc.
2
2
y /2
1
!
642
Chapter 6. Applications of Integration
6.7.63 Orient the axes so that (0, 0) is in the middle of the bottom of the cup. Note that the line between
y
1
+ . The area of a cross section at height y
(0.02, 0) and (0.025, 0.15) is given by y = 30x 3/5, so x =
30 50
✓
◆2
5y + 3
2
. Note that the distance the slice must travel is 0.2 y, because it must go
is given by ⇡x = ⇡
150
0.05 above the top of the glass. Thus,
W = ⇡⇢g
Z 0.15 ✓
0
⇡⇢g
5 · 1502
5y + 3
150
Z 0.15
◆2
(.2
⇡⇢g
5 · 1502
Z 0.15
(5y + 3)2 (1
5y) dy
0
125y 2
15y + 9) dy
⇡⇢g
125y 4 /4 125y 3 /3
5 · 1502
⇡⇢g
=
· 1.0248 ⇡ 0.280 J.
5 · 1502
15y 2 /2 + 9y
=
( 125y 3
y) dy =
0
=
0.15
0
6.7.64
a. The acceleration due to gravity is F = mg, and is in the vertical direction. The tangent direction to
the curve is perpendicular to the normal, which makes an angle of ✓ with the the vertical. If we form a
right triangle with hypotenuse of length mg, and angle ✓, then the two legs must have lengths mg sin ✓
(parallel to the curve) and mg cos ✓ (normal to the curve).
b. Note that the angle t is t = S/L, where S is arc length. Then S = Lt and dS = L dt. So
Z ✓
W =
F ds =
0
Z ✓
0
✓
mg sin ✓ · L d✓ = mgL (
cos ✓))
= mg(L
L cos ✓) = mgh.
0
6.7.65
a. F =
Z 3
3
⇢g(4
1
y2
y) · 2 dy = ⇢g 8y
can withstand the force.
Z 3
b. F (h) =
⇢g(h y) · 2 dy = ⇢g 2hy
= 8⇢g = 78, 400 N. This is less than 90,000 N., so the window
1
3
y2
1
= ⇢g(6h
9
(2h
1)) = 4⇢g(h
2). This is less than
1
or equal to 90,000 Newtons when h  4.296 meters.
6.7.66 The plate pictured on the left should have more force, because it has its wider part lower in the pool.
For each plate, orient the axes so that the vertical line through the vertex pointing up (or down) is the
y-axis, and the x-axis lies along the horizontal base of the triangle. For the plate on the left, the line through
p
p
p
1
(1/2, 0) and (0, 3/2) is given by y = 3((1/2) x), so x =
( 3/3)y, and the width at height y is given
2
p
2 3
by 1
y.
3
For the plate on the left,
F =
Z p3/2
0
= ⇢g
p
⇢g(1 + ( 3/2)
Z p3/2
0
= ⇢g
✓
1
6
✓
2y 2
p
3
4y 3
p
3
2y
p
3
2
p y) dy
3
!
p
3
2y +
+ 1 dy
2
y)(1
⇣
p
p ⌘
2(3 + 3)y 2 + 3 2 + 3 y
◆◆ p3/2
0
p
1
3
= +
⇡ 0.683 N.
4
4
Copyright c 2019 Pearson Education, Inc.
6.7. Physical Applications
643
p
p
y
3/2) is given by y = 3x, so x = p , and
3
For the plate on the right, the line through (0, 0) and (1/2,
2y
the width at height y is p .
3
Thus,
F =
Z p3/2
0
= ⇢g
✓
◆
Z p3/2 ✓
2y
2y 2
2y
p + p + y dy
⇢g(1 + 3/2 y) p dy = ⇢g
3
3
3
0
p
◆ p3/2
3
2
2
2y
y
y
1+2 3
p +p +
=
⇡ 0.558 N.
2
8
3 3
3
0
p
6.7.67 The plate on the left has more than half of its area below the horizontal line which is 3/2 below the
surface, while the plate on the right has exactly half its area below that line, so the plate on the left should
have more force than the plate on the right.
p
p
first quadrant forom (0, 0)pto ( 2/2, 2/2) is given by y = x,while the line
a. Note that
p the
p line in the p
from ( 2/2, 2/2) to (0, 2) is given by y = x + 2. So the width
p of a slice at height y in the lower
part of the region is 2y, and in the upper part of the region is 2( 2 y).
Z p2/2 p
F = ⇢g
( 2+1
0
= ⇢g
Z p2/2
0
= ⇢g
✓
= ⇢g
Z p2 p
y)(2y) dy + ⇢g p ( 2 + 1
2/2
p
y)2( 2
p
Z 2
p
( 2y + 2 2y + 2y) dy + ⇢g p (2y 2
p
4 2y
2
y) dy
p
2y + 2 2 + 4) dy
2/2
◆ p2/2
✓ ✓ 3
◆◆ p2
⇣
⌘
p
p
2y
y
1
+ 2y 2 + y 2
+ ⇢g 2
1 + 2 2 y 2 + 2y + 2y
p
3
3
2
0
2/2
!
p
p !
1
2 1
1
2
+
+ + p
= ⇢g 1 +
N.
2
3
2 3 2
2
p
3
This is 16,730 N.
Z 1
b. F =
⇢g(2 y) · 1 dy = ⇢g 2y
1
y 2 /2
=
0
0
3⇢g
N. This is 14,700 N.
2
6.7.68
a. W =
Z xf
00
mx dx =
x0
b. W =
Z xf
x0
Z x(5)
0
2 · 8 dx = 16(x(5)) = 1600 J.
Z 5
dx
mx dx =
2 · 8 dt =
dt
0
00
Z 5
0
5
16 · 8t dt = 64t2
0
= 64 · 25 = 1600 J.
6.7.69
a.
✓
◆ 2500000
GM m
1
dx
=
GM
m
(x + R)2
x+R 0
0
✓
◆
1
1
GM m2500000
=
⇡ 8.87435 ⇥ 109 J.
= GM m
R(R + 2500000)
R R + 2500000
W =
b. W (x) =
Z x
0
Z 2500000
GM m
dt = GM m
(t + R)2
✓
1
t+R
◆ x
0
= GM m
✓
1
R
1
R+x
◆
Copyright c 2019 Pearson Education, Inc.
=
GM mx
500GM x
=
.
R(R + x)
R(R + x)
644
Chapter 6. Applications of Integration
c. lim
GM mx
x!1 R(R + x)
= lim
x!1 R
GM m
GM m
=
.
R
R
x +1
GM mx
1
2GM x
2GM
d. Suppose
= mv 2 , then v 2 =
, and as x ! 1 we have v 2 =
, so v =
2
R(R + x)
R
R(R + x)
r
2GM
.
R
6.7.70 When the box is fully submerged in the water, the buoyant force is 8g⇢w . The force required is
F = 8g⇢w 8g⇢ = 8g⇢w 8g(⇢w /2) = 4g⇢w = 39, 200 N.
Chapter Six Review
1
a. True. A vertical slice would lead to shells, while a horizontal slice would lead to either disks or washers.
b. True. In order to find position, you would also need to know either its initial position, or at least its
position at some time.
c. True. If dV /dt is constant, then V is a linear function of time.
2
✓ ◆
1
⇡t
, which occurs on the given interval for t = 1. For 0  t < 1, we have
a. v(t) = 0 when = cos
2
3
v(t) < 0, so the motion is in the negative direction, and for 1 < t  3, v(t) > 0, so the motion is in the
positive direction.
✓ ◆◆
✓
◆ 3
Z 3✓
⇡t
6 sin(⇡t/3)
1 2 cos
dt = t
= 3 m.
b. The displacement is given by
3
⇡
0
0
c. The distance traveled is
Z 3✓
✓
◆◆
⇡t
1 2 cos
dt
2 cos
dt
3
1
0
✓
◆ 3 ✓
◆ 1
6 sin(⇡t/3)
6 sin(⇡t/3)
= t
t
⇡
⇡
0
p 1
p
= (3 0 (1 3 3/⇡)) (1 3 3/⇡ (0 0))
p
p
6 3
6 3+⇡
.
=1+
=
⇡
⇡
⇡t
3
◆◆
Z 1✓
1
✓
◆ t
6 sin(⇡x/3)
6
⇡t
(1 2 cos ⇡x/3) dx = x
=t
sin .
d. s(t) = s(0) +
⇡
⇡
3
0
0
With the antiderivative method, we would have
✓ ◆◆
✓
◆
Z
Z ✓
6 sin(⇡t/3)
⇡t
1 2 cos
dt = t
+ C,
s(t) = v(t) dt =
3
⇡
Z t
✓
and because s(0) = 0, we would have C = 0. Therefore s(t) = t
6
⇡t
sin .
⇡
3
3
1
30t + 12 = 6(2t2 5t + 2) = 6(2t 1)(t 2), which is zero for t = 2 and t = . v(t) > 0 for
2
1
0  t < and for 2 < t  3, so the motion is in the positive direction on those intervals, while v < 0
2
1
for < t < 2, so the motion is in the negative direction on that interval.
2
a. 12t2
Copyright c 2019 Pearson Education, Inc.
Chapter Six Review
645
b. The displacement is given by
Z 3
3
(12t2
30t + 12) dt = 4t3
15t2 + 12t
= 108
0
c. The distance traveled is
Z 3
Z 1/2
2
(12t
30t + 12) dt +
(12t2
0
30t + 12) dt
2
15t2 + 12t
3
15t2 + 12t
+ 4t3
0
= (1/2 15/4 + 6) + (108
= 22.5 m.
d. s(t) = s(0) +
Z t
Z 2
(12t2
4t3
15t2 + 12t
30t + 12) dt
1/2
1/2
= 4t3
2
2
135 + 36)
(32
1/2
60 + 24)
((32
60 + 24)
(1/2
15/4 + 6))
t
(12x2
30x + 12) dx = 1 + 4x3
= 4t3
15x2 + 12x
0
15t2 + 12t + 1.
0
With the antiderivative method, we would have
Z
Z
s(t) = v(t) dt = (12t2 30t + 12) dt = 4t3
15t2 + 12t + C,
and because s(0) = 1, we would have C = 1. Therefore, s(t) = 4t3
4 The displacement is
135 + 36 = 9 m.
0
Z 1.5
20 cos ⇡t dt =
0
✓
20
sin ⇡t
⇡
◆ 1.5
0
=
15t2 + 12t + 1.
20
.
⇡
5 The position s(t) and the displacement are the same, because the projective started on the ground at
position 0.
Z t
Z t
t
s(t) =
v(x) dx =
(20 10x) dx = 20x 5x2
= 20t 5t2 .
0
0
0
Note that the projectile is moving up for 0 < t < 2 and down for 2 < t < 4. Thus the distance traveled is equal
to the position for 0 < t < 2, but for 2 < t < 4 the distance traveled is 20 + (20 (20t 5t2 )) = 40 20t + 5t2 .
6 a(t) =Z 5, so v(t) Z=
s(t) =
v(t) dt =
s(t) = 80t
7
a. v(t) =
5t + C, and because v(0) = 80, we have v(t) = 80
5t. The position function
2
(80
5t) dt = 80t
5t /2 + D, and because s(0) = 0 we have D = 0. Thus,
Z
2 sin(⇡t/4) dt =
8
cos(⇡t/4) + C, and because v(0) =
⇡
2
5t /2.
Z
a(t) dt =
8
cos(⇡t/4).
Thus, v(t) =
⇡ Z
Z
8
v(t) dt =
cos(⇡t/4) dt =
s(t) =
⇡
32
sin(⇡t/4).
Thus, s(t) =
⇡2
8
, we have C = 0.
⇡
32
sin(⇡t/4) + D, and because s(0) = 0 we have D = 0.
⇡2
b. s is periodic with period 8, so we only consider 0  t  8. There are critical numbers for s at t = 2
32
32
and t = 6. There is a maximum for s of 2 at t = 6 and a minimum of
at t = 2.
⇡2
⇡
Z
8
1 88
4
cos(⇡t/4) dt =
(
sin(⇡t/4))
= 0.
c. The average velocity is
8 0 ⇡
⇡2
0
Z
8
1 8 32
16
sin(⇡t/4)
dt
=
(
cos(⇡t/4))
= 0.
The average position is
⇡3
8 0
⇡2
0
Copyright c 2019 Pearson Education, Inc.
646
Chapter 6. Applications of Integration
8
a.
v
Z 1
1
(2t + 1) dt = t2 + t
= 2
0
0
Z 1
miles, while Benny runs
(4
t) dt =
b. Anna runs
6
5
0
1
4
4t
3
t2 /2
= (4
1/2) = 3.5 miles. Note
0
2
1
0.5
1.0
1.5
2.0
t
2.5
that the area under Anna’s graph from 0 to 1
is smaller than the corresponding area under
Benny’s graph.
c. sA (t) = t2 + t, which is 6 for t = 2. sB (t) = 4t t2 /2 which is 6 for t = 2, so they both take exactly
two hours to run 6 miles. At t = 2, both velocity function have the same area under them, namely 6.
9
a. For 0  t  8 we have R0 (t) = 4t1/3 , so R(t) = 3t4/3 + C, but C = 0, so R(t) = 3t4/3 .
Rt
b. Note that R(8) = 48, so for t > 8, R(t) = 48 + 8 2 dx = 48 + 2(t
We have R(t) =
8
<3t4/3
:
2t + 32
8).
if 0  t  8;
if t > 8.
c. The fuel runs out when 150 = 48 + 2(t
8), which occurs for t = 59.
15
. Then V (t) = 15 ln(t + 1) + C, and because V (0) = 75, we have C = 75. Thus,
t+1
V (t) = 75 15 ln(t + 1). This is 0 when ln(t + 1) = 5, which occurs when t = e5 1 ⇡ 147.413 hours, so
don’t try to hold your breath while the tank is emptying.
10 Let V 0 (t) =
11
a.
v
150
b. The velocity is 50 when 200e t/10 = 50, which
occurs when et/10 = 4, so when t = 10 ln 4.
100
50
10
20
30
t
Copyright c 2019 Pearson Education, Inc.
Chapter Six Review
647
c. .
s
Z t
The position is given by
1500
t
1000
2000e x/10
200e x/10 dx =
0
= 2000(1
e t/10 ).
0
500
10
20
t
30
d. No. lim s(t) = 2000 < 2500.
t!1
12
a.
v
60
40
20
20
40
60
80
t
b.
s
3000
The position is given by
p
2000
t
400 x + 1
Z t
p
200/ x + 1 dx =
0
p
= 400( t + 1
1).
0
1000
20
40
60
80
t
p
c. Yes, the position is 2500 when 400 t + 1 = 2900, which occurs when t + 1 =
✓ ◆2
29
t=
1 = 51.5625.
4
Copyright c 2019 Pearson Education, Inc.
✓
29
4
◆2
, so
648
Chapter 6. Applications of Integration
13
s
a.
Tom’s position
Z t
20e 2x dx =
function
given
12
by
10
= 10(1 e 2t ).
Sue’s position is given by
t
15e
is
10e 2x
0
x
t
0Z
8
t
6
15e x dx =
0
4
t
= 15(1
e ).
2
0
0.5
1.0
t
1.5
b. We are looking for t so that 10(1 e 2t ) = 15(1 e t ), which occurs when 10(e t )2 15e t + 5 = 0, or
2u2 3u + 1 = 0 where u = e t . This quadratic factors as (2u 1)(u 1), so if e t = 1 or e t = 1/2,
so t = 0 or t = ln 2.
c. Sue takes the lead and doesn’t relinquish it at t = ln 2.
14 The area is given by
Z 1
Z 1
x
((x + 1) (2 + x 1)) dx =
(2
0
x
2 ) dx =
0
15 The area is given by
✓
Z ⇡/4
cos 2x
(sin 2x 1 ( cos 2x)) dx =
2
0
16 The area is given by
Z ⇡/3
x+
✓
2x
ln 2
2x
sin 2x
2
◆ 1
=
0
⇡/3
0
✓
0
0
◆ ⇡/4
cos y dy = sin y
=
✓
=
0
2
ln 2
◆
✓
⇡ 1
+
4
2
◆
✓
2
1
ln 2
0
1
2
◆
=2
1
.
ln 2
◆
0+0 =1
⇡
.
4
p
3
.
2
17 The area is given by
Z 1
1
(ey
1) dy = (ey
0
y)
= (e
1)
(1
0) = e
2.
0
⇡
⇡
⇡
= cos , the curves intersect at x = . By symmetry, the area is given by
4
4
2
p
p !
p
p !!
Z 5⇡/2 ⇣
5⇡/2
⇣
p
x⌘
2
2
2
2
x
x⌘
x
cos
dx = 2
2 cos
2 sin
= 4
+
= 8 2.
2
sin
2 ⇡/2
2
2
2
2
2
2
2
⇡/2
18 Because sin
p
p
19 These two curves meet p
when 4x = x 25 x2 , so at x = 0 and when 4 = 25 x2 , which occurs for
x = 3. On [0, 3], we have x 25 x2 > 4x (for example check at x = 1), so the area is
Z 3⇣ p
⌘
A=
x 25 x2 4x dx.
0
Note that
du =
Z 3
0
3
4x dx =
2x2
=
18. For the first term, use the substitution u = 25
0
2x dx and the integration bounds become u = 25 to u = 16. Then we have
✓
◆ 16
Z
1 16 1/2
1 2 3/2
64 125
7
A=
u du 18 =
u
18 =
+
18 = .
3
3
3
2 25
2 3
25
Copyright c 2019 Pearson Education, Inc.
x2 , so that
Chapter Six Review
649
20 The curves meet where 6 3x2 = 6x 3, or 3x2 + 6x
occurs at x = 1 and x = 3. The area is given by
Z 1
(6 3x2 (6x 3)) dx =
3
Z 1
9 = 3(x2 + 2x
3) = 3(x + 3)(x
1) = 0, which
1
( 3x2 6x+9) dx =
3x2 + 9x
x3
= ( 1 3+9) (27 27 27) = 32.
3
3
21 The curves intersect for x2 = 2x2 4x, or x2 4x = x(x 4) = 0, so for x = 0 and x = 4. For 0  x  2
the region is bounded above by x2 and below by the x-axis, and for 2  x  4 the region is bounded above
by x2 and below by 2x2 4x. The area is given by
Z 2
2
x dx +
0
Z 4
2
4
2x2 + 4x) dx = x3 /3
(x2
+
2
x3 /3 + 2x2
= (8/3) +
64/3 + 32
0
2
⇣
p
2 x + x) dx = x
4x3/2 /3 + x2 /2
( 8/3 + 8) = 8.
22 The area is given by
Z 1
p
(1
2
x) dx =
0
Z 1
(1
0
23 Note that we can write 1
x
2
8
>x
<
1 as 2
>
:
2
⌘ 1
0
=
1
.
6
if x  2;
x
if x 2.
2
x
x
For 0  x  2, the region is bounded above by and below by . For 2  x  3, the region is bounded
2
6
x
x
above by 2
and below by . The area is then given by
2
6
Z 2
(x/2
x/6) dx +
0
24
Z 1
Z 3
2
(2
x/2
x/6) dx = x2 /6
2
(x1/p
3
+ 2x
x2 /3
0
=
2
2
+ (6
3
3)
(4
4/3) = 1.
xp ) dx is evaluated as
0
✓
p
x(p+1)/p
p+1
1
xp+1
p+1
◆ 1
0
=
p
p+1
1
p 1
=
.
p+1
p+1
For p = 100 the area is 99/101, and for p = 1000 the area is 999/1001.
25 The curve and the line intersect where 2x2 6x+5 = 1, or 2x2 6x+4 = 2(x2 3x+2) = 2(x 1)(x 2) = 0,
which is for x = 1 and x = 2. The area is given by
Z 2
(1
(2x
2
6x + 5)) dx =
1
Z 2
( 4 + 6x
1
=
8 + 12
2
2x ) dx =
✓
16
4+3
3
✓
4x + 3x
◆
1
2
= .
3
3
2
2x3
3
◆ 2
1
26 Substituting x for y 2 in the equation x = (2 y 2 )2 , we obtain x = (2 x)2 , which can be written as
x = 4 4x + x2 or x2 5x + 4 = (x 4)(x 1) = 0, so the curves intersect at x = 4 and x = 1. The points
of intersection are (1, ±1) and (4, ±2). It will be easier to integrate with respect to y. Exploiting symmetry,
we can integrate with respect to y from 0 to 1 and then again from 1 to 2 and double our result.
Note that between y = 1 and y = 2 we have y 2 > (2 y 2 )2 , while between y = 0 and y = 1 the reverse
Copyright c 2019 Pearson Education, Inc.
650
Chapter 6. Applications of Integration
is true. Thus the area is given by
A=2
Z 1
y 2 )2 )
((2
y 2 ) dy + 2
0
=2
Z 1
(y 4
5y 2 + 4) dy + 2
0
Z 2
Z 2
(y 2
y 2 )2 ) dy
(2
1
( y 4 + 5y 2
4) dy
1
✓ 5
◆ 1
✓ 5
y
5y 3
y
5y 3
=2
+ 4y
+2
+
3
5
3
5
0
✓ ◆
✓
◆
38
16 38
120
+
=
= 8.
=2
+2
15
15
15 15
4y
◆ 2
1
27 The area of R1 is
Z 1
(3
p
2 x) dx =
x
0
✓
x2
2
3x
4x3/2
3
◆ 1
1
2
=3
0
4
7
= .
3
6
The area of R2 can be computed by subtracting the area of R1 from the area of a triangle with base 3
and height 3, so it is
9 7
20
10
=
=
.
2 6
6
3
Z 3
p
The area of R3 is the di↵erence of
2 x dx and the area of a triangle with base 2 and height 2, so it is
1
Z 3
p
2 x dx
✓
2=
1
28 V = ⇡
Z 1
((3
x)
2
4x) dx = ⇡
0
Z 1
4x3/2
3
(x
2
0
◆ 3
2=
1
p
12 3
3
p
2=4 3
4
3
✓ 3
x
10x + 9) dx = ⇡
3
2
5x + 9x
10
.
3
◆ 1
=⇡
0
✓
29
V = 2⇡
Z 1
x(3
x
p
2 x) dx = 2⇡
0
= 2⇡
✓
3x2
2
x3
3
Z 1
◆ 1
= 2⇡
dy = ⇡
Z 2✓
4x5/2
5
x2
(3x
2x3/2 ) dx
0
0
✓
3
2
1
3
4
5
◆
=
11⇡
.
15
30
V =⇡
Z 2✓
(3
y)
y4
16
2
0
✓
= ⇡ 9y
3y 2 +
y3
3
◆
y5
80
◆ 2
0
9
6y + y
0
✓
= ⇡ 18
12 +
8
3
2
◆
y4
dy
16
◆
32
124⇡
=
.
80
15
31
Z 2 ✓
V = 2⇡
y 3
0
= 2⇡
32 V = ⇡
Z 2
0
(3
2
y) dy + ⇡
✓
3y 2
2
Z 2p 3 ✓
2
3
◆
✓
◆
y2
y3
2
y
dy = 2⇡ 3y y
dy
4
4
◆ 2
✓
◆
y3
y4
8
14⇡
1 =
.
= 2⇡ 6
3
3
16 0
3
y2
4
◆2
dy.
Copyright c 2019 Pearson Education, Inc.
1
3
◆
13⇡
5+9 =
.
3
Chapter Six Review
33 V = 2⇡
Z 3
(3
651
p
x) 2 x
3 + x dx.
1
34
a. A =
Z 2
((4 + y)
(y 2 + 2)) dy.
0
b. V = 2⇡
Z 2
0
c. V = ⇡
Z 2
0
⇥
⇥
y (4 + y)
(4 + y)2
⇤
(y 2 + 2) dy.
⇤
(y 2 + 2)2 dy.
◆2
Z ✓
⇡ 2 (4 + y) (y 2 + 2)
d. V =
A(y) dy =
dy.
2 0
2
0
✓
◆ 4
✓
✓
◆◆
Z 4
Z 1
p
2x3/2
16
2
7
35 A =
1 dx +
(2
x) dx = 1 + 2x
=1+ 8
2
= .
3
3
3
3
0
1
1
✓
◆ 1
✓
◆
Z 1
Z 1
11⇡
4y 3
y4
4 1
36 V = 2⇡
y(2 y)2 dy = 2⇡
(4y 4y 2 + y 3 ) dy = 2⇡ 2y 2
+
= 2⇡ 2
+
=
.
4
6
3
4
3
0
0
0
✓
◆ 1
✓
◆
Z 1
1 32
31⇡
(2 y)5
37 V = ⇡
(2 y)4 dy = ⇡
=⇡
+
=
.
5
5
5
5
0
0
Z 2
38 For the portion generated over 0  x  1, we have
Z 1 p
S1 = 2⇡
1 1 + 02 dx = 2⇡.
0
For the portion over 1  x  4, we have
S2 = 2⇡
Z 4
1
(2
p
x)
r
1+
1
dx ⇡ 8.96177.
4x
This calculation was done using a computer algebra system. The surface area for the whole surface is
approximately S1 + S2 = 2⇡ + 8.96177 ⇡ 15.24.
Z ⇡/3
⇡/3
p
39 The area of region R1 is
sec2 x dx = tan x
= 3. The area of region R2 can be calculated by
0
0
⇡
and height 4. Thus the area of R2 is
subtracting the area of R1 from the area of a rectangle with base
3
4⇡ p
3.
3
40
a.
y
x
Copyright c 2019 Pearson Education, Inc.
652
Chapter 6. Applications of Integration
Z x
b. R(x) =
g(t)) dt, and R0 (x) = f (x)
(f (t)
g(x).
a
41 A(a) =
Z p
3 a
⇣p
x/a
✓
⌘
2
x /a dx =
0
2 x3/2
p
3 a
x3
3a
p
3
◆
a
=
0
1
.
3
42
a. V = ⇡
Z 4 p
( 4
y)2 dy = ⇡
0
b. V = 2⇡
Z 2
Z 4
4
(4
y 2 /2
y) dy = ⇡ 4y
= ⇡(16
0
2
x(4
x ) dx = 2⇡
0
8
(0
0)) = 8⇡.
0
Z 2
2
x3 ) dx = 2⇡ 2x2
(4x
x4 /4
= 2⇡ (8
0
4
(0
0)) = 8⇡.
0
p
43 From x = 0 to x = 1 the area function is A(x) = ( x)2 = x, while from x = 1 to x = 2 it is
A(x) = (2 x)2 .
Z 2
Z 1
1
2
1
1
1 1
5
x dx +
(2 x)2 dx = x2
(2 x)3 = + = .
2 0 3
2 3
6
0
1
1
44 The region is the same as in pthe previous exercise. From x = 0 to x = 1 the slices are semicircles with
2
p
diameter x, so their area is ⇡( 8x) = ⇡8 x. From x = 1 to x = 2, the slices are semicircles with diameter
2 x, so their area is ⇡8 (2 x)2 . Thus by the general slicing method, the volume is
⇡
8
✓Z 1
x dx +
0
Z 2
2
(2
x) dx
1
◆
1
⇡
=
8
1 2
x
2 0
2
1
(2
3
x)
3
1
!
⇡
8
=
✓
1 1
+
2 3
◆
=
5⇡
.
48
45 Because the cross sections are parallel to the x-axis, solve for x to get x = y 2 and x = 2 y. Then from
y = 0 to y = 1 the area function is A(y) = (y 2 )2 = y 4 and from y = 1 to y = 2 it is A(y) = (2 y)2 . The by
the general slicing method, the volume is
Z 1
y 4 dy +
0
Z 2
1
(2
y)2 dy =
1
y5
5 0
2
1
(2
3
=
1 1
8
+ =
.
5 3
15
Z 2
(24x
24x2 + 12x3
2x5/2
5
1
x3/2 dx = 4⇡ ·
y)3
1
46
V =⇡
Z 2
2
(( x + 2x + 2)
2
(2x
2
2
4x + 2) ) dx = ⇡
0
= ⇡ 12x2
3x4 ) dx
0
2
8x3 + 3x4
3x5 /5
64⇡
=
.
5
0
47
a. V = 2⇡
Z 1
x((1 +
p
x)
(1
p
x)) dx = 4⇡
0
Z 1
0
=
0
8⇡
.
5
Z 2
Z 2
Z 2
b. V = ⇡
(12 ((1 y)2 )2 ) dy = ⇡
(1 (y 4 4y 3 +6y 2 4y +1)) dy = ⇡
( y 4 +4y 3 6y 2 +4y) dy =
0
0
0
✓
◆ 2
y5
8⇡
4
3
2
⇡
+y
2y + 2y
=
.
5
5
0
48 V = ⇡
Z (ln 2)/2
0
((2e
x 2
)
x 2
(e ) ) dx = ⇡
✓
2e
2x
1 2x
e
2
◆ (ln 2)/2
0
=
⇡
.
2
Copyright c 2019 Pearson Education, Inc.
Chapter Six Review
653
49 It appears to be easiest to use shells, because then we won’t have to split the integral up, and also the
2
2
functions will integrate more easily. Solving for y we have y = ex and y = e2 x . The height of each shell
2
2
is then e2 x
ex and the volume is
2⇡
Z 1
x(e2 x
2
✓
2
ex ) dx = 2⇡
0
1 2 x2
e
2
1 x2
e
2
◆ 1
= ⇡( e
e + e2 + 1) = ⇡(e2
2e + 1) = ⇡(e
1)2 .
0
50 Using washers, we have
⇡
Z ⇡/3
⇡/3
(4
sec2 x) dx = ⇡ (4x
tan x)
0
0
⇣
= ⇡ 4⇡/3
p
Z p
3/2
x
p
dx. Let u = 1
51 We use the shell method. V = 2⇡
1
x2
0
✓
Z 1/4
Z 1
p 1
1
1/2
p du = ⇡
gives V = 2⇡
u
du = 2⇡ u
= 2⇡ 1
2 u
1
1/4
1/4
3
⌘
0) = 4⇡ 2 /3
(0
x2 so that du =
◆
1
= ⇡.
2
p
⇡ 3.
2x dx. Substituting
52 Use the shell method. Then each shell has height 4 x2 and the radius of the shell at x is x ( 2) = x+2.
The volume is then
Z 2
Z 2
2
V = 2⇡
(x + 2)(4 x ) dx = 2⇡
( x3 2x2 + 4x + 8) dx
= 2⇡
✓
2
x4
4
2 3
x + 2x2 + 8x
3
◆ 2
53 We use the disk method. The radius of each disk is 4
V =⇡
Z 4
x2 )2 dx = ⇡
(4x
0
=⇡
✓ 5
x
5
2x4 +
16 3
x
3
Z 4
◆ 4
(x4
0
=⇡
0
✓
2
= 2⇡
✓
(x
2)2 = 4x
2
◆
32
128⇡
+ 32 =
.
3
3
x2 , so the volume is
8x3 + 16x2 ) dx
1024
5
512 +
1024
3
◆
=
512⇡
.
15
54 The curves intersect when 6x = x2 + 5, or when x2 6x + 5 = (x 1)(x 5) = 0, thus at x = 1 and
x = 5. To revolve about the line y = 1, we use the washer method; the outer radius of each washer is
6x ( 1) = 6x + 1 and the inner radius is x2 + 5 ( 1) = x2 + 6. The volume is thus
V =⇡
Z 5
((6x + 1)
2
2
2
(x + 6) ) dx = ⇡
1
✓
x5
+ 8x3 + 6x2 35x
5
✓
◆
1
1856⇡
=
.
= ⇡ 371 +
5
5
=⇡
◆ 5
1
Z 5
( x4 + 24x2 + 12x
35) dx
625 + 1000 + 150
175 +
1
=⇡
✓
1
5
8
6 + 35
◆
To revolve about the line x = 1, we use the shell method. The height of each shell is 6x
6x x2 5, and the radius at x is x ( 1) = x + 1. The volume is thus
V = 2⇡
Z 5
(x + 1)(6x
x2
5) dx = 2⇡
1
= 2⇡
✓
625 625 25
+
+
2
4
3
1
25 +
4
Z 5
( x3 + 5x2 + x 5) dx = 2⇡
1
◆
5 1
256⇡
+5 =
.
3 2
3
Copyright c 2019 Pearson Education, Inc.
✓
x4
5
1
+ x2 + x2
4
3
2
(x2 + 5) =
5x
◆ 5
1
654
Chapter 6. Applications of Integration
55 The two non-horizontal lines intersect at (2, 4). To revolve about y = 2, we use the shell method. First
solve for x to obtain x = y2 and x = 6 y; then the height of each shell is 6 y y2 = 6 23 y, and the radius
at y is y ( 2) = y + 2. The volume is
✓
◆
◆
✓
◆ 4
Z 4
Z 4✓
3
3 2
3
1 3
2⇡
(y + 2) 6
y dy = 2⇡
12 + 3y
y
dy = 2⇡ 12y + y 3
y
= 80⇡.
2
2
2
2
0
0
0
To revolve about x = 2, we use the washer method. With the equations above, the outer radius of each
asher is 6 y ( 2) = 8 y and the inner radius is y2 ( 2) = 2 + y2 . The volume is
◆
◆
Z 4✓
Z 4✓
⇣
y ⌘2
3 2
2
V =⇡
(8 y)
2+
dy = ⇡
y
18y + 60 dy
2
4
0
0
◆ 4
✓
1 3
=⇡
y
9y 2 + 60y
= ⇡(16 144 + 240) = 112⇡.
4
0
56
a. Vx = ⇡
Z 2
1
Vy = 2⇡
2
1
⇡
dx = ⇡ ( 1/x) = .
2
x
2
1
Z 2
1
b. Vx = ⇡
Z 4
1
Vy = 2⇡
1
dx = ⇡
x6
Z 4
1
x · x 1 dx = 2⇡, so Vy > Vx .
4
x 5 /5
⇡
(1
5
=
1
4
x · x 3 dx = 2⇡ ( 1/x)
8
>
<
=
1
1
).
1024
3⇡
, so Vy > Vx .
2
⇡
( 1 + a1 2p ) if p 6= 1/2,
1
1
2p
c. Vx = ⇡
dx =
2p
>
1 x
:
⇡ ln a
if p = 1/2.
8
2⇡(a2 p 1)
>
Z a
<
if p 6= 2,
1
2 p
dx =
d. Vy = 2⇡
p
1
>
1 x
:
2⇡ ln a
if p = 2.
Z a
e. Using part d, let h = 2
2⇡(ah
lim
h!0
h
part c.
1)
2⇡(a2 p 1)
=
p!2
2 p
p, and note that Vy is continuous. So lim Vy = lim
h!0
= 2⇡ ln a, so lim
h!0
(ah
1)
h
= ln a. A similar calculation can be done with the result of
f. No, Vy > Vx for all values of a and p, because
1
1
< p 1 for x > 1 and p > 0.
x2p
x
57 We use disks when revolving about the x-axis. The volume is
✓
◆ 1
✓
Z 1
Z 1
1 3 1 4 1 5
1
2 2
2
2
2
3
4
2
2
⇡
c x (1 x) dx = ⇡c
(x
2x + x ) dx = ⇡c
x
x + x
= ⇡c
3
2
5
3
0
0
0
We use shells when revolving about the y-axis. The volume is
✓
Z 1
Z 1
1 3
2⇡
x · cx(1 x) dx = 2⇡c
(x2 x3 ) dx = 2⇡c
x
3
0
0
2
These are equal when
⇡c
⇡c
=
, which for c =
6 0 occurs when c = 5.
30
6
Copyright c 2019 Pearson Education, Inc.
1 4
x
4
◆ 1
0
=
⇡c
.
6
1 1
+
2 5
◆
=
⇡c2
.
30
Chapter Six Review
58 L =
Z 2
p
655
1 + 4 dx =
p
5(2
p
( 2)) = 4 5.
dx =
Z p5
2
Z p
5
r
1+ 2
59 L = p
x
2
Z 4
p
1
u
u 1/2 du =
2 1
1
1
p
4
=2
2
dx. Let u = x2
1 so that du = 2x dx. Substituting gives
Z 2p
1/(2x2 ), so 1 + y 02 = (x2 /2 + 1/(2x2 ))2 . L =
1/(2x)
p
61 Note that y 0 = 1/(2 x)
17
.
12
=
x + x3/2 /3
p
⌘ 3
1 + y 02 dy =
1
1
x/2 dx =
1
1 = 1.
2
1/(2x2 )) dx = x3 /6
⇣p
x
x2
1
60 Note that y 0 = x2 /2
p
p
Z 3p
p
p
x/2, so 1 + y 02 = (1/(2 x) + x/2)2 . L =
4/3.
(x2 /2 +
1
1 + y 02 dy =
1
p
=2 3
Z 2
Z 3
p
1/(2 x) +
1
1
62 f 0 (x) = ⇡ cos ⇡x. Therefore the arc length is L =
Z 1p
0
1 + ⇡ 2 cos2 ⇡x dx ⇡ 2.305.
p
p
p
1 + f 0 (x)2 = 1 + (2x + 2)2 = 4x2 + 8x + 5.
Z 4p
The arc length is
4x2 + 8x + 5 dx ⇡ 16.127.
63 y 0 = 2(x + 1), so
2
p
p
p
64 y 0 = x2 + x, so
1 + f 0 (x)2 =
1 + (x2 + x)2 = x4 + 2x3 + x2 + 1. So the arc length is L =
Z 2p
x4 + 2x3 + x2 + 1 dx ⇡ 5.351.
0
p
x2 + 1
, and 1 + y 02 =
2
x
65 Note that y 0 = 1/x, so 1 + y 02 =
So
Z bp 2
x +1
dx =
L=
x
1
p
x2 + 1
ln
1+
p
x2 + 1
x
!!
p
b
=
1
x2 + 1
.
x
p
b2 + 1
p
2+ln
p
( b2 + 1
1)(1 +
b
p !
2)
.
Using a computer algebra system, we see that this has value 2 for b ⇡ 2.715.
66
a. S = 2⇡
Z 2
0
3
(x /3)
p
⇡
1 + x4 dx. Let u = 1 + x4 , so that du = 4x3 dx. Substituting gives
⇡ ⇣ 3/2
⇡ ⇣ 3/2 ⌘
17
u
=
9
9
1
b. V = 2⇡
Z 2
0
c. V = ⇡
17
Z 2
0
3
6
⌘
1 .
x(x /3) dx = 2⇡
Z 2
2
x4 /3 dx = 2⇡ x5 /15
0
= 2⇡(32/15
0
2
x6 /9 dx = ⇡ x7 /63
= ⇡(128/63) = 128⇡/63.
0
67
Copyright c 2019 Pearson Education, Inc.
0) = 64⇡/15.
Z 17
1
u1/2 du =
656
Chapter 6. Applications of Integration
Z 3p
a. S = 2⇡
3x
0
Z 3
⇡
3 dx = 3⇡(3
x2
s
(3 2x)2
1+
dx = 2⇡
4(3x x2 )
Z 3p
3x
x2
0
s
12x
4x2 + 9 12x + 4x2
dx =
4(3x x2 )
0) = 9⇡.
0
b. V = ⇡
Z 3
3
x2 ) dx = ⇡ 3x2 /2
(3x
x3 /3
0
= ⇡ (27/2
9) = 9⇡/2.
0
68 Consider the line y = x/2 over the interval [0, 8]. If we revolve this around the x-axis, we generate a cone
Z 8
8
p
p
p
(x/2) 1 + 1/4 dx = 5⇡/2 x2 /2
= 16 5⇡.
with radius 4 and height 8. The surface of this cone is 2⇡
0
69
0
◆
Z 2✓ 4
p
x
1
a. S = 2⇡
+
1 + (2x3 (1/8x3 ))2 dx =
2
16x2
1
◆
◆
Z 2✓ 4
Z 2✓ 4
p
p
x
1
x
1
6
6 ) dx = 2⇡
2⇡
+
1
+
(4x
1/2
+
(64/x
+
4x6 + 1/2 + 64/x6 dx =
2
2
2
16x
2
16x
1
1
◆
◆✓
◆
Z 2✓ 4
Z 2✓ 4
Z 2
p
x
1
x
1
1
3
3 + (1/8x3 ))2 dx = 2⇡
2⇡
+
(2x
+
2x
+
dx
=
2⇡
(x7 +
2
16x2
2
16x2
8x3
1
1
1
Z 2
2
x/16 + x/8 + x 5 /128) dx = 2⇡
(x7 + 3x/16 + x 5 /128) dx = 2⇡ x8 /8 + 3x2 /32 x 4 /512
=
1
2⇡((32 + 3/8
1/8192)
1
263439⇡
1/512)) =
.
4096
(1/8 + 3/32
p
b. Using the fact that 1 + f 0 (x)2 = 2x3 + 1/(8x3 ) which was discovered during the previous calculation,
Z 2
2
483
we have L =
(2x3 + x 3 /8) dx = x4 /2 x 2 /16
= (8 1/64) (1/2 1/16) =
.
64
1
1
Z 2
2
c. V = 2⇡
(x5 /2 + x 1 /16) dx = 2⇡ x6 /12 + (1/16) ln x
= 2⇡(16/3 + (ln 2)/16 (1/12 + 0)) =
1
1
21⇡ ⇡ ln 2
+
.
2
8
d.
V =⇡
Z 2
(x4 /2 + x 2 /16)2 dx = ⇡
1
70 m =
0
71 m =
Z 3
72 m =
0
1/6144
(1/36 + 1/48
0
150e x/3 dx =
dx +
Z 4
⇣
2 dx +
2
450e x/3
Z 6
⌘
3
= 450(1
e 1 ) gm.
0
4 dx = 2 + 4 + 8 = 14.
4
73
a. Because 50 =
Z .2
0
kx dx =
kx2
2
0.2
=
0
W =
k
, we must have k = 2500.
50
Z 0.7
0.2
x 3 /768
1
264341⇡
1/768)) =
.
18432
⇣
⌘ 9
p
(3 + 2 x) dx = 3x + 4x3/2 /3
= 63 gm.
0
Z 2
2
(x8 /4 + x2 /16 + x 4 /256) dx = ⇡ x9 /36 + x3 /48
1
= ⇡ (512/36 + 8/48
Z 9
Z 2
0.7
2500x dx = 1250x2
= 562.5 J.
0.2
Copyright c 2019 Pearson Education, Inc.
Chapter Six Review
657
b. f (0.2) = 0.2k = 50, so k = 250, and thus
W =
Z 0.7
0.7
250x dx = 125x2
0.2
⇣
74 The force function is given by f (y) = 9.8 7
W = 9.8
Z 8⇣
0
7
✓
y⌘
dy = 9.8 7y
4
= 56.25 J.
0.2
y⌘
. So the work required is
4
y2
8
◆ 8
= 9.8(56
8) = 9.8(48) = 470.4 J.
0
Copyright c 2019 Pearson Education, Inc.
658
Chapter 6. Applications of Integration
75
a. Note that the density of the chain is 2 kg/m.
W =
Z 10
✓
2(9.8)(10
y2
2
y) dy = 19.6 10y
0
◆ 10
= 19.6 (100
50) = 980 J.
0
b. The lower 6 meters of the chain are lifted 4 meters. The work required for this is 6(2)(9.8)(4) = 470.4 J
The work required to lift the upper four meters of the chain is
✓
◆ 10
Z 10
y2
2(9.8)(10 y) dy = 19.6 10y
= 19.6(100 50 (60 18) = 156.8.
2
6
6
So the total work required is 470.4 + 156.8 = 672.2 J.
76 The density of the chain is 3 kg/m. The work to lift the mass 5 meters is 4(9.8)(5) = 196 J. The work
✓
◆ 5
Z 5
y2
3(9.8)(5 y) dy = 29.4 5y
= 29.4(25 12.5) = 29.4(12.5) = 367.5 J.
to lift the chain is W =
2
0
0
Therefore the total work is 196 + 367.5 = 563.5 J.
77
a. W =
Z 6
1000(9.8)(8)(6
0
b. W =
Z 2
✓
y) dy = 78, 400 6y
y2
2
y) dy = 78, 400 7y
y2
2
◆ 2
6y
y2
2
✓
1000(9.8)(8)(7
0
78 W =
Z 6
0
⇡1000(9.8) · 16(6
y) dy = 16⇡(9800)
✓
◆ 6
= 78, 400(36
18) = 1, 411, 200 J.
= 78, 400(14
2) = 940, 800 J.
0
0
◆ 6
0
= 288⇡(9800) ⇡ 8, 866, 830 J.
79
a.
Z 6
Z 6
4 2
39200
W =
1000(9.8) ⇡y (6 y) dy =
⇡
(6y 2 y 3 ) dy
9
9
0
0
✓
◆ 6
39200
39200
y4
3
=
⇡ (432 324) ⇡ 1, 477, 805 J.
=
⇡ 2y
9
9
4
0
b. The work to pump out the top 3 feet is
✓
Z 6
39200
39200
2
3
⇡(6y
y ) dy =
⇡ 2y 3
9
9
3
y4
4
◆ 6
=
3
39200
⇡(432
9
324
20.25)) ⇡ 1, 015, 592 J.
(54
Therefore, the work to pump out the bottom 3 feet is about 1, 477, 805
1, 015, 592 = 461, 814 J.
80
W =
Z 0
9.8(1000)(4
2
y )⇡(2
y) dy = 9800⇡
2
✓
= 9800⇡ 8y
81 W =
Z 9
0
4, 987, 592 J.
2y 2
1000(9.8)⇡y(10
2y 3
y4
+
3
4
◆ 0
y) dy = 9800⇡
Z 0
(8
4y
2y 2 + y 3 ) dy
2
2
✓
= 9800⇡ 16 + 8
Z 9
0
(10y
16
3
4
◆
= 9800⇡ ·
✓
y ) dy = 9800⇡ 5y 2
2
Copyright c 2019 Pearson Education, Inc.
y3
3
◆ 9
0
44
⇡ 451, 552 J.
3
= 9800⇡(405
243) ⇡
Chapter Six Review
82 F =
Z 1
659
✓
1000(9.8)(3
y) dy = 9800 3y
0
Z 1
y
1000(9.8)(3 y) dy = 4900
83 F =
2
0
84 F =
Z 2
y2
2
p
y)2 4
1000(9.8)(4
2
Z 1
◆ 1
= 9800(2.5) = 24, 500 N.
0
2
y 2 dy = 19600
Z 2 p
4 4
◆ 1
3y 2
2
y3
3
y 2 dy
19600
(3y y ) dy = 4900
0
✓
2
= 4900
0
Z 2 p
y 4
✓
3
2
1
3
◆
⇡ 5716.7 N.
y 2 dy. The first term
2
in this sum is 19, 600· 4 times the area of the upper half of a circle of radius 2 so it is 19600· 4· 2⇡ = 156, 800⇡.
The second term is zero because it is the integral of an odd function over a symmetric interval. Therefore
F = 156, 800⇡ ⇡ 492, 602 N.
85 Orient the semicircle so that the center is at the point (0, 20). Then the force is given by
Z 20
⇢g(20
y)(2)
0
. Let u = 40y
y 2 so that du = (40
⇢g
Z 400
p
y 2 dy.
40y
2y) dy. Then we have
u
1/2
du = ⇢g
0
✓
2 3/2
u
3
◆ 400
0
⇡ 5.2 ⇥ 107 .
86 The equation of L1 is y = (2ap + b)(x p) + f (p), which can be written y = (2ap + b)x ap2 + c. The
equation of L2 is y = (2aq + b)(x q) + f (q), which can be written y = (2aq + b)x aq 2 + c. If we set these
p+q
. Thus, the area
equal to each other and solve to find the point of intersection, we find s =
2
Z s
Z s
R1 =
f (x) ((2ap + b)x ap2 + c) dx =
ax2 2apx + ap2 dx
p
p
s
= ax3 /3
=
as
3
Z q
ax2
apx2 + ap2 x
p
Similarly, the area
Z q
R2 =
f (x) ((2aq + b)x
aq 2 + c) dx =
s
3
= ax /3
2
2
aqx + aq x
3
aps2 + ap2 s
ap3
a
=
(q
24
3
p)3 .
2aqx + aq 2 dx
s
q
aq 3
=
3
s
as3
+ aqs2
3
aq 2 s =
aq 3
3
as 2
(s
3
Copyright c 2019 Pearson Education, Inc.
3qs + 3q 2 ) =
a
(q
24
p)3 .
660
Chapter 6. Applications of Integration
Copyright c 2019 Pearson Education, Inc.