C H A P T E R 7 Applications of Integration
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Transcript of C H A P T E R 7 Applications of Integration
C H A P T E R 7Applications of Integration
Section 7.1 Area of a Region Between Two Curves . . . . . . . . . . . 2
Section 7.2 Volume: The Disk Method . . . . . . . . . . . . . . . . . 18
Section 7.3 Volume: The Shell Method . . . . . . . . . . . . . . . . . 33
Section 7.4 Arc Length and Surfaces of Revolution . . . . . . . . . . . 44
Section 7.5 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Section 7.6 Moments, Centers of Mass, and Centroids . . . . . . . . . 61
Section 7.7 Fluid Pressure and Fluid Force . . . . . . . . . . . . . . . 73
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
C H A P T E R 7Applications of Integration
Section 7.1 Area of a Region Between Two Curves
2
1. A � �6
0�0 � �x2 � 6x�� dx � ��6
0�x2 � 6x� dx 2.
� �2
�2��x2 � 4� dx
A � �2
�2 ��2x � 5� � �x2 � 2x � 1�� dx
3.
� �3
0��2x2 � 6x� dx
A � �3
0 ���x2 � 2x � 3� � �x2 � 4x � 3�� dx 4. A � �1
0 �x2 � x3� dx
5.
or �6�1
0 �x3 � x� dx
A � 2�0
�1 3�x3 � x� dx � 6�0
�1 �x3 � x� dx 6. A � 2�1
0 ��x � 1�3 � �x � 1�� dx
7.
5
4
3
2
1
542 31x
y
�4
0 ��x � 1� �
x2� dx 8.
x−2 2
−2
2
y
�1
�1 ��1 � x2� � �x2 � 1�� dx 9.
6
5
3
2
654321
1
x
y
�6
0 �4�2�x�3� �
x6� dx
10.
1
−1
2
2 4 5 6 7
3
4
5
6
7(3, 6)
(3, 1)
23
2,( )
y
x
�3
2 �x3
3� x �
x
3� dx 11.
x
y
3
−13 3
π3
2π3
2π π
���3
���3 �2 � sec x� dx 12.
π4
π4−
π4− 2
2,( )
π4− , 2( ) π
4 , 2( )
π4
22,( )
y
x
2
3
���4
���4�sec2 x � cos x� dx
Section 7.1 Area of a Region Between Two Curves 3
13. (a)
Intersection points: and
(b) A � �2
�3 ��4 � y2� � �y � 2�� dy �
1256
A � �0
�5 ��x � 2� � �4 � x dx � �4
02�4 � x dx �
616
�323
�1256
��5, �3��0, 2�
�y � 3��y � 2� � 0
y2 � y � 6 � 0
4 � y2 � y � 2
x � y � 2
x−6 −4 6
−6
4
6
(−5, −3)
(0, 2)
y x � 4 � y2
14. (a) and
Intersection points: and
A � �2
�3 ��6 � x� � x2� dx �
1256
x
(−3, 9)
(2, 4)
y
−2−6 −4 2 4 6−2
4
6
8
10
��3, 9��2, 4�
x2 � 6 � x ⇒ x2 � x � 6 � 0 ⇒ �x � 3��x � 2� � 0
y � 6 � xy � x2 (b)
�323
�616
�1256
A � �4
0 2�y dy � �9
4 ��6 � y� � �y dy
15.
Matches (d)
A � 4
g�x� � �x � 1�2
x1 2 3
2
3
(0, 1)
(3, 4)
yf �x� � x � 1 16.
Matches (a)
A � 1
g�x� � 2 � �x
x
1
3
4
1 2 3
(0, 2)
(4, 0)
yf �x� � 2 �12x
17.
� 168
�42
� 2 � 0 � 2
� �x4
8�
x2
2� x�
2
0
� �2
0 1
2x3 � x � 1 dx
6
5
4
3
43−2 1
1
x
y
(0, 1)
(0, 2)
(2, 6)
(2, 3)
A � �2
0 �1
2x3 � 2 � �x � 1�� dx
4 Chapter 7 Applications of Integration
18.
� �64 � 112 � 80� � �1 � 7 � 20� � 18
� �x3
8�
7x2
4� 10x�
8
2
� �8
23
8x2 �
72
x � 10 dx
2
2
4
6
8
4 6 10
y
x
(2, 9)
2,
(8, 6)
(8, 0)
92( )
A � �8
2 �10 �
12
x � �38
x�x � 8�� dx
19. The points of intersection are given by:
�323
� ��x3
3� 2x2�
4
0
� ��4
0 �x2 � 4x� dx
x(4, 0)(0, 0)
1 2 3 5−1
−2
−3
−4
−5
y A � �4
0 �g�x� � f �x�� dx
x�x � 4� � 0 when x � 0, 4
x2 � 4x � 0
20. The points of intersection are given by:
� �9 �272
�92
� ��x3
3�
3x2
2 �3
0(3, 4)
(0, 1)
1 2
2
3
4
5
6
3 4 5 6
y
x
� �3
0��x2 � 3x� dx
A � �3
0���x2 � 4x � 1� � �x � 1�� dx
x2 � 3x when x � 0, 3
�x2 � 3x � 0
�x2 � 4x � 1 � x � 1
21. The points of intersection are given by:
� �2x �x2
2�
x3
3 �2
�1�
92
10
6
4
21234
8), 9(2
x
)01,(
y � �2
�1 �2 � x � x2� dx
� �2
�1 ��3x � 3� � �x2 � 2x � 1�� dx
A � �2
�1 �g�x� � f �x�� dx
�x � 2��x � 1� � 0 when x � �1, 2
x2 � 2x � 1 � 3x � 3
22. The points of intersection are given by:
� ��x3
3�
32
x2�3
0�
92
x−1 1 2 3 4 5
3
4
5
6
(0, 2)
(3, 5)
y � �3
0 ��x2 � 3x� dx
� �3
0 ���x2 � 4x � 2� � �x � 2�� dx
A � �3
0 � f �x� � g�x�� dx
x�3 � x� � 0 when x � 0, 3
�x2 � 4x � 2 � x � 2
23. The points of intersection are given by:
Note that if we integrate with respect to we need two integrals. Also, note that the region is a triangle.
x,
A � �1
0 ��2 � y� � �y�� dy � �2y � y2�
1
0� 1
x � 1 x � 0 x � 2
x � 2 � x and x � 0 and 2 � x � 0
32
3
1
2
)1,1
x)0,2(
0),0(
(
y
Section 7.1 Area of a Region Between Two Curves 5
24.
x1 2 3 4 5
0.2
0.4
0.6
0.8
1.0 (1, 1)
(5, 0.04)
y
A � �5
1 1
x2 � 0 dx � ��1x�
5
1�
45
25. The points of intersection are given by:
� �29
�3x�3�2 �x2
2 �3
0�
32
5
3
2
4
1 2−2 3 4x
), 4(3
0( 1, )
y � �3
0 ��3x�1�2 � x� dx
� �3
0 ���3x � 1� � �x � 1�� dx
A � �3
0 � f �x� � g�x�� dx
�3x � x when x � 0, 3
�3x � 1 � x � 1
26. The points of intersection are given by:
� 2�12
� 1 � 0 � �34� �
12
� 2�x2
2� x �
34
�x � 1�4�3�1
0
A � 2�1
0 ��x � 1� � 3�x � 1 � dx
x�x � 2��x � 1� � 0 ⇒ x � 0, 1, 2
x�x2 � 3x � 2� � 0
x3 � 3x2 � 2x � 0
x � 1 � �x � 1�3 � x3 � 3x2 � 3x � 1 1
2
(0, −1)
(1, 0)
(2, 1)
y
x
3�x � 1 � x � 1
27. The points of intersection are given by:
� �2y �y2
2�
y3
3 �2
�1�
92
� �2
�1 ��y � 2� � y2� dy
3
2
1
542 31
1
3
, 2)(4
x
1, )(1
y A � �2
�1 �g�y� � f �y�� dy
�y � 2��y � 1� � 0 when y � �1, 2
y2 � y � 2
28. The points of intersection are given by:
� �32
y2 �13
y3�3
0�
92
x
−1
1
3
−3 −2 1
( 3, 3)−
(0, 0)
y � �3
0 �3y � y2� dy
� �3
0 ��2y � y2� � ��y�� dy
A � �3
0 � f �y� � g�y�� dy
y�y � 3� � 0 when y � 0, 3
2y � y2 � �y
6 Chapter 7 Applications of Integration
29.
� �y3
3� y�
2
�1� 6
� �2
�1 ��y2 � 1� � 0� dy
64 5
3
1
32
2
x
y
(0, −1)
(0, 2) (5, 2)
(2, −1)
A � �2
�1 � f �y� � g�y�� dy
30.
� ���16 � y2�3
0� 4 � �7 � 1.354
� �12
�3
0 �16 � y2��1�2��2y� dy
� �3
0 � y�16 � y2
� 0� dy
x
1
2
3
4
−1 1 2 3
( ), 37
3
y A � �3
0 � f �y� � g�y�� dy
31.
� 10 ln 5 � 16.0944
� 10�ln 10 � ln 2�
� �10 ln y�10
2
A � �10
2 10y
dy
12
8
6
864−4 −2 2
4
x
y
(1, 10)(0, 10)
(5, 2)(0, 2)
y �10x
⇒ x �10y
32.
� 1.227
� 4 � 4 ln 2
� �4x � 4 ln 2 � x �1
0
x
1
3
−1 1 3
(0, 2)
(1, 4)
y
A � �1
0 4 �
42 � x dx
33. (a)
(c) Numerical approximation:0.417 � 2.667 � 3.083
−6 12
−1
(3, 9)
(1, 1)(0, 0)
11 (b) The points of intersection are given by:
� �x4
4�
43
x3 �32
x2�1
0� ��x4
4�
43
x3 �32
x2�3
1�
512
�83
�3712
� �1
0 �x3 � 4x2 � 3x� dx � �3
1 ��x3 � 4x2 � 3x� dx
� �1
0 ��x3 � 3x2 � 3x� � x2� dx � �3
1 �x2 � �x3 � 3x2 � 3x�� dx
A � �1
0 � f �x� � g�x�� dx � �3
1 �g�x� � f �x�� dx
x�x � 1��x � 3� � 0 when x � 0, 1, 3
x3 � 3x2 � 3x � x2
34. (a)
(c) Numerical approximation: 2.0
x−2 −1 2
−2
−1
( 1, 2)−
(1, 0)
(1, 2)−
y (b) The point of intersection is given by:
� �1
�1 �x3 � 1� dx � �x4
4� x�
1
�1� 2
� �1
�1 ��x3 � 2x � 1� � ��2x�� dx
A � �1
�1 � f �x� � g�x�� dx
x3 � 1 � 0 when x � �1
x3 � 2x � 1 � �2x
Section 7.1 Area of a Region Between Two Curves 7
35. (a)
(b) The points of intersection are given by:
(c) Numerical approximation: 21.333
� ��2x3
3� 4x2�
4
0�
643
� �4
0 ��2x2 � 8x� dx
A � �4
0 ��3 � 4x � x2� � �x2 � 4x � 3�� dx
2x�x � 4� � 0 when x � 0, 4
x2 � 4x � 3 � 3 � 4x � x2
(4, 3)(0, 3)
−6 12
−3
9 36. (a)
(b) The points of intersection are given by:
(c) Numerical approximation: 8.533
� 2 �4x3
3�
x5
5 �2
0�
12815
� 2 �2
0 �4x2 � x4� dx
A � 2 �2
0 �2x2 � �x4 � 2x2�� dx
x2�x2 � 4� � 0 when x � 0, ±2
x4 � 2x2 � 2x2
( 2, 8)−
(0, 0)
(2, 8)
−4 4
−2
10
37. (a)
(c) Numerical approximation:5.067 � 2.933 � 8.0
−4 4
−5
2
(−2, 0) (2, 0)
(−1, −3) (1, −3)
f �x� � x4 � 4x2, g�x� � x2 � 4 (b) The points of intersection are given by:
By symmetry:
� 2� 15
�53
� 4� � 2��325
�403
� 8 � �15
�53
� 4� � 8
� 2�x5
5�
5x3
3� 4x�
1
0� 2��x5
5�
5x3
3� 4x�
2
1
� 2 �1
0 �x4 � 5x2 � 4� dx � 2 �2
1 ��x4 � 5x2 � 4� dx
A � 2 �1
0 ��x4 � 4x2� � �x2 � 4�� dx � 2 �2
1 ��x2 � 4� � �x4 � 4x2�� dx
�x2 � 4��x2 � 1� � 0 when x � ±2, ±1
x4 � 5x2 � 4 � 0
x4 � 4x2 � x2 � 4
38. (a)
(c) Numerical approximation:8.267 � 0.617 � 0.883 � 9.767
x−1−3−4 1 3 4
−3
−4
3
4
f
g
y
f �x� � x4 � 4x2, g�x� � x3 � 4x (b) The points of intersection are given by:
�24830
�3760
�5360
�29330
� �2
1 ��x3 � 4x� � �x 4 � 4x2�� dx
A � �0
�2 ��x3 � 4x� � �x4 � 4x2�� dx � �1
0 ��x4 � 4x2� � �x3 � 4x�� dx
x�x � 1��x � 2��x � 2� � 0 when x � �2, 0, 1, 2
x4 � x3 � 4x2 � 4x � 0
x4 � 4x2 � x3 � 4x
8 Chapter 7 Applications of Integration
39. (a)
(b) The points of intersection are given by:
(c) Numerical approximation: 1.237
� 2�
4�
16 �
�
2�
13
� 1.237
� 2 �arctan x �x3
6 �1
0
� 2 �1
0 � 1
1 � x2 �x2
2 � dx
A � 2�1
0 � f �x� � g�x�� dx
x � ±1
�x2 � 2��x2 � 1� � 0
x4 � x2 � 2 � 0
1
1 � x2 �x2
2
−3 3
−1
−1, 12 1, 1
2( (( (
3 40. (a)
(b)
(c) Numerical approximation: 6.908
� 6.908
� 3 ln 10
� �3 ln�x2 � 1��3
0
A � �3
0 � 6x
x2 � 1� 0� dx
(0, 0)
3, 95( )
−1 5
−1
3
41. (a)
(b) and (c) on
You must use numerical integration becausedoes not have an elementary
antiderivative.
A � �2
0 �1
2x � 2 � �1 � x3� dx � 1.759
y � �1 � x3
�0, 2��1 � x3 ≤12
x � 2
−4 5
−1
(0, 2)
(0, 1)
(2, 3)
5 42. (a)
(b) and (c) You must use numerical integration:
A � �4
0x�4 � x
4 � x dx � 3.434
−1 5
−1
2
43.
� 2�1 � ln 2� � 0.614
� 2��2 cos x � ln cos x ���3
0
� 2 ���3
0 �2 sin x � tan x� dx
2
3
4
1
2
2
3
4
3,3
x
y
0)0,(
3,3
g
f
π
π
ππ
A � 2 ���3
0 � f �x� � g�x�� dx 44.
�3�3
4� 1.299
� �34
��32 � �0�
� �12
sin 2x � cos x���6
���2
π2
π6
−
π2
−
−1, −1( )
π6 2
1, ( )
y
x
A � ���6
���2�cos 2x � sin x� dx
Section 7.1 Area of a Region Between Two Curves 9
45.
x
y
2
3
−1
ππ
π
π2
2
g
f
(2 , 1)
(0, 1)
� 2�x � sin x�2�
0� 4� � 12.566
� 2�2�
0�1 � cos x� dx
A � �2�
0��2 � cos x� � cos x� dx 46.
1, 2 )(
1
1
2
3
4
y
x
��22
� 2 �4�
�1 � �2� � 2.1797
� �2 � 42
� 4 �4��2 � �
4�
� ��2 � 42
x2 � 4x �4�
sec �x4 �
1
0
A � �1
0 ���2 � 4�x � 4 � sec
�x4
tan �x4 � dx
47.
1
x
,,
1
)0( ,0
1( )
y
1e
� ��12
e�x2�1
0�
12 1 �
1e � 0.316
A � �1
0 �xe�x2
� 0� dx 48. From the graph we see that f and g intersect twice atand
� 21 �1
ln 3 � 0.180
� �x2 � x �1
ln 3�3x��
1
0
� �1
0 ��2x � 1� � 3x� dx
x
2
3
−1 1 2
(0, 1)
(1, 3)
y A � �1
0 �g�x� � f �x�� dx
x � 1.x � 0
49. (a)
00
3
�
(b)
� 2 �12 � �2 �
12 � 4
� ��2 cos x �12
cos 2x��
0
A � ��
0�2 sin x � sin 2x� dx (c) Numerical approximation: 4.0
50. (a)
−2
(0, 1)
−
(π, 1)2
�4
�45
(b)
� ��2 cos x �12
sin 2x��
0� 4
A � ��
0�2 sin x � cos 2x� dx (c) Numerical approximation: 4
51. (a)
060
4
(1, e)
(3, 0.155)
(b)
� e � e1�3
� ��e�1�x�3
1
A � �3
1 1x2 e1�x dx (c) Numerical approximation: 1.323
10 Chapter 7 Applications of Integration
52. (a)
(1, 0)
(5, 1.29)
−2
0 6
2 (b)
� 2�ln 5�2
� �2�ln x�2�5
1
A � �5
1 4 ln x
x dx (c) Numerical approximation: 5.181
53. (a)
−1
−1
4
6 (b) The integral
does not have an elementaryantiderivative.
A � �3
0� x3
4 � x dx
(c) A � 4.7721
54. (a)
−1
−1
2
4
(1, e)
(b) The integral
does not have an elementaryantiderivative.
A � �1
0
�xe x dx
(c) 1.2556
55. (a)
(b) The intersection points are difficult to determine by hand.
(c) where
c � 1.201538.
Area � �c
�c
�4 cos x � x2� dx � 6.3043
5
3
−1
−3
56. (a)
(b) The intersection points are difficult to determine.
(c) Intersection points: and
A � �1.4526269
�1.164035 ��3 � x � x2� dx � 3.0578
�1.4526269, 2.1101248���1.164035, 1.3549778�
4
3
−1
−4
57. F�x� � �x
0 1
2t � 1 dt � �t 2
4� t�
x
0�
x2
4� x
(a)
t
4
y
5
6
2
3
−1−1 1 2 3 4 5 6
F�0� � 0 (b)
t
4
y
5
6
2
3
−1−1 1 2 3 4 5 6
F�2� �22
4� 2 � 3 (c)
t
4
y
5
6
2
3
−1−1 1 2 3 4 5 6
F�6� �6 2
4� 6 � 15
Section 7.1 Area of a Region Between Two Curves 11
58. F �x� � �x
0 1
2t 2 � 2 dt � �1
6t 3 � 2t�
x
0�
x3
6� 2x
(a)
1 2 3 4
4
8
12
16
20
5 6
y
x
F �0� � 0 (b)
1 2 3 4
4
8
12
16
20
5 6
y
x
F �4� �43
6� 2�4� �
563
(c)
1 2 3 4
4
8
12
16
20
5 6
y
x
F �6� � 36 � 12 � 48
59. F ��� � ��
�1cos
��
2 d� � � 2
� sin
��
2 ��
�1�
2�
sin ��
2�
2�
(a)
y
1
32
12
12
−
12
12
−θ
F ��1� � 0 (b)
y
1
32
12
12
−
12
12
−θ
F �0� �2�
� 0.6366 (c)
y
1
32
12
12
−
12
12
−θ
F12 �
2 � �2�
� 1.0868
60. F �y� � �y
�14ex�2 dx � �8ex�2�
y
�1� 8ey�2 � 8e�1�2
(a)
10
5
15
20
25
30
−1 1 2 3 4
y
x
F ��1� � 0 (b)
10
5
15
20
25
30
−1 1 2 3 4
y
x
F �0� � 8 � 8e�1�2 � 3.1478 (c)
10
5
15
20
25
30
−1 1 2 3 4
y
x
F �4� � 8e2 � 8e�1�2 � 54.2602
61.
� � 74
x2 � 7x�4
2� ��7
4x2 � 21x�
6
4� 7 � 7 � 14
� �4
2 7
2x � 7 dx � �6
4 �7
2x � 21 dx
(4, 6)
(6, 1)
(2, 3)−y x= 5−
52y x= + 16−
92y x= 12−y
x2 6 8 10
−4
−2
2
4
6
A � �4
2 � 9
2x � 12 � �x � 5�� dx � �6
4 ��5
2x � 16 � �x � 5�� dx
12 Chapter 7 Applications of Integration
62.
(0, 0) ( , 0)a
( , )b c
x
y = xcb y = ( )x a−c
b a−
y
�12
�base��height� � �ac2
� ac �ac2
� �� a2c
y2 � ay�c
0
� �c
0 �a
cy � a dy
A � �c
0 �b � a
cy � a �
bc
y� dy 63.
Left boundary line:
Right boundary line:
� �2
�24 dy � 4y�
2
�2� 8 � ��8� � 16
A � �2
�2 �� y � 2� � � y � 2�� dy
y � x � 2 ⇔ x � y � 2
y � x � 2 ⇔ x � y � 2
x
y
−1−2−4 2 3 4
−3
−4
1
2
3
4
(4, 2)(0, 2)
(0, −2)(−4, −2)
64.
�152
�52
� ��454
�452
�54
�152 �
�5x2
2 �1
0� ��
5x2
4�
152
x�3
1
� �1
05x dx � �3
1 �
52
x �152 dx
(1, 2)
(0, 0)
(1, −3)
(3, −2)
x
y
2
−1−2 2 3 4
1
−1
−2
−3
−4
A � �1
0�2x � ��3x�� dx � �3
1 ���2x � 4� � 1
2x �
72� dx
65. Answers will vary. If you let and
(a)
(b)
� 2�502� � 1004 sq ft
Area �60
3�10� �0 � 4�14� � 2�14� � 4�12� � 2�12� � 4�15� � 2�20� � 4�23� � 2�25� � 4�26� � 0�
� 3�322� � 966 sq ft
Area �60
2�10� �0 � 2�14� � 2�14� � 2�12� � 2�12� � 2�15� � 2�20� � 2�23� � 2�25� � 2�26� � 0�
n � 10, b � a � 10�6� � 60.�x � 6
66.
(a)
(b)
�43
�296.6� � 395.5 sq mi
Area �32
3�8� �0 � 4�11� � 2�13.5� � 4�14.2� � 2�14� � 4�14.2� � 2�15� � 4�13.5� � 0�
� 2�190.8� � 381.6 sq mi
Area �32
2�8� �0 � 2�11� � 2�13.5� � 2�14.2� � 2�14� � 2�14.2� � 2�15� � 2�13.5� � 0�
b � a � �8��4� � 32�x � 4, n � 8,
Section 7.1 Area of a Region Between Two Curves 13
67.
At
Tangent line: or
The tangent line intersects at
A � �1
�2 �x3 � �3x � 2�� dx � �x4
4�
3x2
2� 2x�
1
�2�
274
x � �2.f �x� � x3
y � 3x � 2y � 1 � 3�x � 1�
f � �1� � 3.�1, 1�,
f��x� � 3x2
(1, 1)
( 2, 8)− −
x1 2 3 4−2−3−4
−6
−8
2
4
6
8
y x= 3 2−
f x x( ) = 3
y f �x� � x3
68.
Tangent line:
Intersection points: and
� ��x 4
4�
3x2
2� 2x�
2
�1� ���4 � 6 � 4� � �
14
�32
� 2� �274
A � �2
�1��x � 2� � �x3 � 2x�� dx � �2
�1��x3 � 3x � 2� dx
�2, 4���1, 1�
y � 1 � 1�x � 1� ⇒ y � x � 2
y� ��1� � 3 � 2 � 1
y� � 3x2 � 2
x
y
(2, 4)
(−1, 1)
−4 2 3 4−1
−2
−3
2
3
4
5
y = x3 − 2x
y = x + 2
y � x3 � 2x, ��1, 1�
69.
At
Tangent line: or
The tangent line intersects at
A � �1
0 � 1
x2 � 1� �1
2 x � 1� dx � �arctan x �
x2
4� x�
1
0�
� � 34
� 0.0354
x � 0.f �x� �1
x2 � 1
y � �12
x � 1y �12
� �12
�x � 1�
f � �1� � �12
.1, 12,
f � �x� � �2x
�x2 � 1�2
x1 2
(0, 1)
1,
12
12
14
32
34
( )
f x( ) =1
2x + 1
12
12
y x= + 1−
y f �x� �1
x2 � 1
70.
Tangent line:
Intersection points:
A � �1�2
0 � 2
1 � 4x2 � ��2x � 2�� dx � �arctan�2x� � x2 � 2x�1�2
0� arctan�1� �
14
� 1 ��
4�
34
� 0.0354
12
, 1, �0, 2�
y � �2x � 2
y � 1 � �2x �12
y�12 �
�822 � �2
y� ��16x
�1 � 4x2�2
x
y
y =
−1 1
1
y = −2x + 22
1 + 4x2(0, 2)
, 112( (
y �2
1 � 4x2, 12
, 1
14 Chapter 7 Applications of Integration
71. on
You can use a single integralbecause
on ��1, 1�.1 � x2
x 4 � 2x2 � 1 ≤
� �x3
3�
x5
5 �1
�1�
415
)x
2
10, )(
,(10)1( , 0
y � �1
�1 �x2 � x 4� dx
A � �1
�1 ��1 � x2� � �x 4 � 2x2 � 1�� dx
��1, 1�x 4 � 2x2 � 1 ≤ 1 � x2 72. on on
Both functions symmetric to origin.
Thus,
� 2 �x2
2�
x4
4 �1
0�
12
A � 2 �1
0 �x � x3� dx
x
( 1, 1)− −
(0, 0)
(1, 1)1
1
−1
−1
y�1
�1 �x3 � x� dx � 0.
�0
�1 �x3 � x� dx � ��1
0 �x3 � x� dx
�0, 1�x3 ≤ x��1, 0�,x3 ≥ x
73. Offer 2 is better because the accumulated salary (areaunder the curve) is larger.
74. Proposal 2 is better since the cumulative deficit (the areaunder the curve) is less.
75.
b � 9 �9
3�4� 3.330
9 � b �9
3�4
�9 � b�3�2 �272
23
�9 � b�3�2 � 9
��9 � b�x �x3
3 ��9�b
0� 9
��9�b
0 ��9 � b� � x2� dx � 9
��9�b
��9�b
��9 � x2� � b� dx � 18
10
6
2
4
226
6
x
,, b9 b ),,( 9 bb ) (
y A � �3
�3 �9 � x2� dx � 36
76.
b � 9 �9�2
� 2.636
9 � b �9�2
�9 � b��9 � b� �812
2 ��9 � b�x �x2
2 �9�b
0�
812
2�9�b
0 ��9 � b� � x� dx �
812
2 �9�b
0 ��9 � x� � b� dx �
812
x
( (9 ), )− −b b (9 , )−b b
−6 −3 3 6
−3
−6
6
9
12
yA � 2 �9
0 �9 � x� dx � 2 �9x �
x2
2 �9
0� 81
Section 7.1 Area of a Region Between Two Curves 15
77. Area of triangle is
Since select a � 4 � 2�2 � 1.172.0 < a < 4,
a � 4 ± 2�2
a2 � 8a � 8 � 0
4 � �a
0�4 � x� dx � �4x �
x2
2 �a
0� 4a �
a2
2
y
x
a
21 3 4
1
2
3
B
O
A
12�4��4� � 8.OAB
78.
a � 4 � 42�3 � 1.48
42�3 � 4 � a
4 � �4 � a�3�2
163
� 2�4
a
�4 � x dx � �43
�4 � x�3�2�4
a�
43
�4 � a�3�2
� 2�4y �y3
3 �2
0� 2�8 �
83� �
323
x
a
−1 1 2 3 4 5−1
−3
1
3
y Total area � �2
�2�4 � y2� dy � 2�2
0�4 � y2� dy
79.
where and is the same as
x( 01, )
2x xf )x(
.0 8.0 6...
0)
02
(0
0
,
4 1.0
0.
0.
.0
6
4
2
y
�1
0 �x � x2� dx � �x2
2�
x3
3 �1
0�
16
.
�x �1n
xi �in
lim���→0
�n
i�1 �xi � xi
2� �x 80.
where and is the same as
x
5
3
2
1
−1
−3 −1 1 3
( 2, 0)− (2, 0)
f x x( ) = 4 − 2
y
�2
�2 �4 � x2� dx � �4x �
x3
3 �2
�2�
323
.
�x �4n
xi � �2 �4in
lim���→0
�n
i�1 �4 � xi
2� �x
81. �5
0 ��7.21 � 0.58t� � �7.21 � 0.45t�� dt � �5
0 0.13t dt � �0.13t2
2 �5
0� $1.625 billion
82.
�2912
billion � $2.417 billion
� �0.01t3
3�
0.16t2
2 �5
0
�5
0 ��7.21 � 0.26t � 0.02t 2� � �7.21 � 0.1t � 0.01t 2�� dt � �5
0 �0.01t 2 � 0.16t� dt
16 Chapter 7 Applications of Integration
83. (a)
(c) billion dollars
(Answers will vary.)
Surplus � �17
12� y1 � y2� dt � 926.4
t
R
Rec
eipt
s (i
n bi
llion
s)
Time (in years)2 4 6 8 10 12
100
200
300
400
500
600
y1 � �270.3151��1.0586�t � 270.3151e0.05695t (b)
(d) No, forever because No, these models are not accurate for the future.According to news, eventually.E > R
1.0586 > 1.0416.y1 > y2
t
E
Exp
endi
ture
s (i
n bi
llion
s)
Time (in years)2 4 6 8 10 12
100
200
300
400
500
600
y2 � �239.9704��1.0416�t � 239.9704e0.04074t
84. (a)
(b)
Percents of families
Perc
ents
of
tota
l inc
ome
x
y
20 40 60 80 100
20
40
60
80
100
y1 � 0.0124x2 � 0.385x � 7.85 (c)
(d) Income inequality � �100
0 �x � y1� dx � 2006.7
Percents of families
Perc
ents
of
tota
l inc
ome
x
y
20 40 60 80 100
20
40
60
80
100
85. 5%:
Difference in profits over 5 years:
Note: Using a graphing utility, you obtain $193,183.
� 893,000�0.2163� � $193,156
� 893,000��25.6805 � 34.0356� � �20 � 28.5714��
�5
0 �893,000e0.05t � 893,000e0.035t� dt � 893,000�e0.05t
0.05�
e0.035t
0.035�5
0
P2 � 893,000e�0.035�t312%:
P1 � 893,000e�0.05�t
86. The total area is 8 times the area of the shaded region to the right. A point is on the upper boundary of the region if
We now determine where this curve intersects the line
Total area � 8 ��2�2�2
0 1 �
x2
4� x dx � 8�x �
x3
12�
x2
2 ��2�2�2
0�
163
�4�2 � 5� � 8�0.4379� � 3.503
x ��4 ± �16 � 16
2� �2 ± 2�2 ⇒ x � �2 � 2�2
x2 � 4x � 4 � 0
x � 1 �x2
4
y � x.
y � 1 �x2
4.
4y � 4 � x2
x2 � 4 � 4y
x2 � y2 � 4 � 4y � y2
�x2 � y2 � 2 � y
�x, y�
x
( , )x y
1 2
1
2y x=
y
Section 7.1 Area of a Region Between Two Curves 17
87. The curves intersect at the point where the slope of equals that of
y2 � 0.08x2 � k ⇒ y�2 � 0.16x � 1 ⇒ x �1
0.16� 6.25
y1, 1.y2
(a) The value of k is given by
k � 3.125.
6.25 � �0.08��6.25�2 � k
y1 � y2
(b)
� 2�6.510417� � 13.02083
� 2�0.08x3
3� 3.125x �
x2
2 �6.25
0
� 2 �6.25
0 �0.08x2 � 3.125 � x� dx
Area � 2 �6.25
0 �y2 � y1� dx
88. (a)
(b)
(c) 5000 V � 5000�12.062� � 60,310 pounds
V � 2A � 2�6.031� � 12.062 m3
� 2 5 �10�5
9� 5.5 � 5 � 6.031 m2
� 2�x �29
�5 � x�3�2�5
0� �x�
5.5
5 A � 2 ��5
0 1 �
13
�5 � x dx � �5.5
5 �1 � 0� dx� 89. (a)
(b)
(c) 5000V � 5000�11.816� � 59,082 pounds
V � 2A � 2�5.908� � 11.816 m3
A � 6.031 � 2�� 116
2
� � 2�� 18
2
� � 5.908
90. True 91. True
92. False. Let and f and g intersectat the midpoint of But
�b
a
� f �x� � g�x�� dx � �2
0�x � �2x � x2�� dx �
23
0.
�0, 2�.�1, 1�,g�x� � 2x � x2.f �x� � x 93. Line:
� 2.7823
��32
�7�
24� 1
� ��cos x �3x2
14��7��6
0
A � �7��6
0 �sin x �
3x7�� dx
x
1
−1
(0, 0)
12
y
4 3
7 6
12
π
π 6π
, −( (
y ��37�
x
94.
is the area of of a circle
Hence, A �4ba �a2
4 � �ab.
��a2
4.
14�a
0
�a2 � x2 dx
y
x
y = b
a
b
x2
a21 −
A � 4�a
0b�1 �
x2
a2 dx �4ba
�a
0
�a2 � x2 dx
18 Chapter 7 Applications of Integration
95. We want to find such that:
But, because is on the graph.
c �49
b �23
9b2 � 4
4 � 3b2 � 8 � 12b2 � 0
b2 �34b4 � �2b � 3b3�b � 0
�b, c�c � 2b � 3b3
b2 �34b4 � cb � 0
�x2 �34 x 4 � cx�
b
0� 0
�b
0��2x � 3x3� � c� dx � 0
x
(b, c)c
y = 2x − 3x3
yc
Section 7.2 Volume: The Disk Method
1. V � � �1
0 ��x � 1�2 dx � � �1
0 �x2 � 2x � 1� dx � ��x3
3� x2 � x�
1
0�
�
3
2. V � � �2
0 �4 � x2�2 dx � � �2
0 �x4 � 8x2 � 16� dx � ��x5
5�
8x3
3� 16x�
2
0�
256�
15
3. V � � �4
1 ��x �2 dx � � �4
1 x dx � ��x2
2 �4
1�
15�
24.
� ��9x �x3
3 �3
0� 18�
V � ��3
0��9 � x2�2 dx � ��3
0�9 � x2� dx
5. V � � �1
0 ��x2�2 � �x3�2� dx � � �1
0 �x4 � x6� dx � ��x5
5�
x7
7 �1
0�
2�
35
6.
x � ±2�2
x2 � 8
8 � 16 � x2
2 � 4 �x2
4
�448�2
15� 132.69
� 2��128�280
�32�2
3� 24�2�
� 2�� x5
80�
2x3
3� 12x�
2�2
0
� 2��2�2
0 � x4
16� 2x2 � 12� dx
V � ��2�2
�2�2 �4 �
x2
4 �2
� �2�2� dx 7.
� ��y2
2 �4
0� 8�
V � � �4
0 ��y �2 dy � ��4
0 y dy
y � x2 ⇒ x � �y
8.
� ��16y �y3
3 �4
0�
128�
3
V � � �4
0 ��16 � y2 �2 dy � ��4
0 �16 � y2� dy
y � �16 � x2 ⇒ x � �16 � y2 9.
V � � �1
0 �y3�2�2 dy � ��1
0 y3 dy � ��y4
4 �1
0�
�
4
y � x2�3 ⇒ x � y3�2
Section 7.2 Volume: The Disk Method 19
10.
� ��y5
5� 2y4 �
16y3
3 �4
1�
459�
15�
153�
5
V � ��4
1 ��y2 � 4y�2 dy � ��4
1 �y4 � 8y3 � 16y2� dy
11.
(a)
(c)
x1 2 3
−1
1
2
3
y
� ��16y �83
y3 �15
y5�2
0�
256�
15
� ��2
0 �16 � 8y2 � y4� dy
V � ��2
0 �4 � y2�2 dy
r� y� � 0R�y� � 4 � y2,
x1 2 3 4
−1
1
2
3
y
� ��4
0 x dx � ��
2x2�
4
0� 8�
V � � �4
0 ��x �2 dx
r�x� � 0R�x� � �x,
x � 4y � 0,y � �x,
(b)
(d)
x1 2 3 4 5
−2
−1
1
2
3
4
y
� ��32y � 4y3 �15
y5�2
0�
192�
5
� ��2
0 �32 � 12y2 � y4� dy
V � ��2
0 ��6 � y2�2 � 4� dy
r�y� � 2R�y� � 6 � y2,
x1 2 3 4
−1
1
2
3
y
� ��16y �15
y5�2
0�
128�
5
V � � �2
0 �16 � y4� dy
r�y� � y2R�y� � 4,
12.
(a)
—CONTINUED—
x−4 −2 2 4
2
4
6
8
y
V � ��8
0 4 �
y2� dy � ��4y �
y2
4 �8
0� 16�
r�y� � �y�2R�y� � 2,
x � 2y � 0,y � 2x2,
(b)
x−4 −2 2 4
2
4
6
8
y
V � ��2
0 4x4 dx � � �4x5
5 �2
0�
128�
5
r�x� � 0R�x� � 2x2,
20 Chapter 7 Applications of Integration
12. —CONTINUED—
(c)
�896�
15
x−4 −2 2 4
2
4
6
y � 4��83
x3 �15
x5�2
0
� ��2
0 �32x2 � 4x4� dx � 4��2
0 �8x2 � x4� dx
V � ��2
0 �64 � �64 � 32x2 � 4x4�� dx
r�x� � 8 � 2x2R�x� � 8, (d)
�16�
3
x−4 −2 4
2
4
6
8
y � ��4y �4�2
3y3�2 �
y2
4 �8
0
� ��8
0 4 � 4�y
2�
y2� dy
V � ��8
0 2 ��y
2 �2
dy
r�y� � 0R�y� � 2 � �y�2,
13. intersect at and
(a)
x
1
2
3
4
−1 1 2 3
y
� ��163
x3 � 2x 4�2
0�
32�
3
� ��2
0�16x2 � 8x3� dx
V � ��2
0 ��4x � x2�2 � x 4� dx
r�x� � x2R�x� � 4x � x2,
�2, 4�.�0, 0�y � 4x � x2y � x2,
(b)
x−2 −1 1 2 3 4
1
2
3
4
5
y
� 8��x4
4�
53
x3 � 3x2�2
0�
64�
3
� 8��2
0�x3 � 5x2 � 6x� dx
V � ��2
0��6 � x2�2 � �6 � 4x � x2�2� dx
r�x� � 6 � �4x � x2�R�x� � 6 � x2,
14. intersect at and
(a)
x−6 −4 −2 2
2
4
8
y
� ��15
x5 � x 4 � 3x3 � 18x2�0
�3�
243�
5
� ��0
�3�x 4 � 4x3 � 9x2 � 36x� dx
V � ��0
�3��6 � 2x � x2�2 � �x � 6�2� dx
r�x� � x � 6R�x� � 6 � 2x � x2,
�0, 6�.��3, 3�y � x � 6y � 6 � 2x � x2,
(b)
x−6 −4 −2 2
2
4
8
y
� �� 15
x5 � x4 � x3 � 9x2�0
�3�
108�
5
� ��0
�3�x4 � 4x3 � 3x2 � 18x� dx
V � ��0
�3��3 � 2x � x2�2 � �x � 3�2� dx
r�x� � �x � 6� � 3R�x� � �6 � 2x � x2� � 3,
Section 7.2 Volume: The Disk Method 21
15.
� 18�
� ��x3
3� 4x2 � 15x�
3
0
x−1 1 2 3 4
1
2
3
5
y � ��3
0 �x2 � 8x � 15� dx
V � ��3
0 ��4 � x�2 � �1�2� dx
r�x� � 1R�x� � 4 � x, 16.
�1447
�
� ��32 � 16 �12828 �
1
1−1−1
2 3 4
2
3
(2, 4)
y
x
� ��16x � x 4 �x7
28�2
0
� ��2
0 �16 � 4x3 �
x6
4 � dx
V � ��1
0 4 �
x3
2 �2
dx
R�x� � 4 �x3
2, r�x� � 0
17.
32.485
� 8 ln 4 �34��
1
1−1−1
2 3 4
2
3
y
x
� ��8 ln 4 �14
� 1�
� ��8 ln�1 � x� �1
1 � x�3
0
� ��3
0 � 8
1 � x�
1�1 � x�2� dx
V � ��3
0 �42 � 4 �
11 � x�
2
� dx
R�x� � 4, r�x� � 4 �1
1 � x18.
x
1
2
3
5
2π π 4π 5ππ9 9 3 9 9
y
� ��8 ln�2 � �3 � � �3 � 27.66
� � ��8 ln 2 � �3 � �3 � � �8 ln 1 � 0 � 0��
� ��8 ln sec x � tan x � tan x���3
0
� ����3
0�8 sec x � sec2 x� dx
V � ����3
0��4�2 � �4 � sec x�2� dx
r�x� � 4 � sec xR�x� � 4,
19.
�208�
3
x1 2 3 4 5
1
−1
2
3
4
5
y � ��y 3
3� 6y 2 � 36y�
4
0
� ��4
0�y2 � 12y � 36� dy
V � ��4
0�6 � y�2 dy
r�y� � 0R�y� � 6 � y, 20.
x1 2 3 4 5
1
−1
2
3
4
5
y
� ��36y �y3
3 �4
0�
368�
3
V � ��4
0��6�2 � �y�2� dy
r�y� � 6 � �6 � y� � yR�y� � 6,
22 Chapter 7 Applications of Integration
21.
�384�
5
−1 5321
−2
−3
1
2
3
4
x
y � 2��y5
5� 4y3 � 32y�
2
0
� 2��2
0�y 4 � 12y2 � 32� dy
V � ��2
�2��6 � y2�2 � �2�2� dy
r�y� � 2R�y� � 6 � y2, 22.
241.59
� 12��13 � 6 ln 3�
x
1
2
3
4
5
6
1 2 3 4 5
y � 36� 133
� 2 ln 13�
� 36��356
� 2 ln 6� � 32
� 2 ln 2��
� 36��y � 2 ln y �1y �
6
2
� 36��6
2 1 �
2y
�1y2� dy
V � ��6
2 6 �
6y�
2
dy
r�y� � 0R�y� � 6 �6y
,
23.
� � ln 4 4.355
� �� ln x � 1 �3
0
� ��3
0
1x � 1
dx
x1 2 3
−1
1
2
y V � ��3
0 1�x � 1�
2
dx
r�x� � 0R�x� �1
�x � 1, 24.
�128�
15
� 2��4x3
3�
x5
5 �2
0
� 2��2
0�4x2 � x4� dx
x
−3
−2
1
2
3
−3 −1 1 2 3
y V � 2��2
0 x�4 � x2�2
dx
r�x� � 0R�x� � x�4 � x2,
25.
�3�
4
� ���1x�
4
1
V � ��4
1 1
x�2
dx
x1 2 3 4
−1
−2
1
2
yr�x� � 0R�x� �1x
, 26.
� 9��� 1x � 1�
8
0� 8�
� 9��8
0�x � 1��2 dx
V � ��8
0 3
x � 1�2
dx
x2 4 6 8
1
2
3
4
yr�x� � 0R�x� �3
x � 1,
27.
��
2�1 � e�2� 1.358
� ���
2e�2x�
1
0
� ��1
0 e�2x dx
V � ��1
0�e�x�2 dx
x1 2
1
2
yr�x� � 0R�x� � e�x, 28.
� ��e4 � 1� 168.38
� ��ex�4
0
� ��4
0 ex dx
V � ��4
0�ex�2�2 dx
x−2 2 4 6
2
4
6
8
yr�x� � 0R�x� � e x�2,
Section 7.2 Volume: The Disk Method 23
29.
The curves intersect at and
� �1523
� �125
3�
277�
3
� ���x 4 �83
x3 � 10x2 � 24x�2
0� ��x 4 �
83
x3 � 10x2 � 24x�3
2
� ��2
0��4x3 � 8x2 � 20x � 24� dx � ��3
2�4x3 � 8x2 � 20x � 24� dx
2
2 3 41−1−2
6
8
10
y
x
(2, 5)
V � ��2
0��5 � 2x � x2�2 � �x2 � 1�2�� dx � ��3
2��x2 � 1�2 � �5 � 2x � x2�2� dx
�2, 5�.��1, 2�
�x � 2��x � 1� � 0
x2 � x � 2 � 0
2x2 � 2x � 4 � 0
x2 � 1 � �x2 � 2x � 5
30.
�883
� �563
� � 48�
� �� x3
12�
5x2
2� 16x�
4
0� ���
x3
12�
5x2
2� 16x�
8
4
� ��4
0 x2
4� 5x � 16� dx � ��8
4 �
x2
4� 5x � 16� dx
2−2−1
1
2
3
4
4 6 8 10
(4, 2)
y
x
V � ��4
0 �4 �
12
x�2
� ��x�2� dx � ��8
4 ���x�2
� 4 �12
x�2
� dx
31.
� 8� �13
�r2h, Volume of cone
��
9�216 � 216 �2163 �
��
9�36y � 6y2 �y3
3 �6
0
��
9�6
0�36 � 12y � y2� dy
V � ��6
0 �1
3 �6 � y��
2
dy
x1 3 4 5 6
1
2
3
4
5
6
yy � 6 � 3x ⇒ x �13
�6 � y� 32.
� �25 �252 � �
25�
2
� ��5y �y2
2 �5
0 (2, 5)
1 2 3 4 5 6 7 8 9
1
2
3
4
5
6
7
8
9
x
y � ��5
0�5 � y� dy
V � ��5
0 ��9 � y �2
� 2 �2 dy
x � �9 � y
x � 3x � 2,y � 0,y � 9 � x2,
33.
Numerical approximation: 4.9348
��
2��� �
�2
2
��
2 �x �12
sin 2x��
0
� ���
0 1 � cos 2x
2 dx
x1 2 3
1
2
3
y V � ���
0�sin x�2 dx 34.
Numerical approximation: 2.4674
��
2 ��
2� ��2
4
��
2 �x �12
sin 2x���2
0
� ����2
0 1 � cos 2x
2 dx
x1 2
1
2
y V � ����2
0�cos x�2 dx
24 Chapter 7 Applications of Integration
35.
Numerical approximation: 10.0359
��
2�e2 � 1�
��
2e2x�2�
2
1
� ��2
1e2x�2 dx
V � ��2
1�ex�1�2 dx 36.
Numerical approximation: 49.0218
� � �e2 � e � 6 � e�2 � e�1�
� � ��e2 � e�2 � 4� � �e�1 � e � 2��
� ��e x � e�x � 2x�2
�1
� ��2
�1�e x � e�x � 2� dx
V � ��2
�1�e x�2 � e�x�2�2 dx
37. V � � �2
0 �e�x2 �2 dx 1.9686 38. V � � �3
1 �ln x�2 dx 3.2332 39.
15.4115
V � � �5
0 �2 arctan�0.2x��2 dx
40.
�3 � 22�3�
5 2.9922
� ��21�3
0�2x � x 4� dx
V � ��21�3
0 ���2x �2 � �x2�2� dx
x � 21�3 1.2599
x3 � 2
x 4 � 2x
1 2
1
2
x
y = x2
y = 2x
y x2 � �2x
41. represents the volume of the solid
generated by revolving the region bounded by about the axis.
y
x
2π
4π
1
x-x � ��2x � 0,y � 0,y � sin x,
����2
0sin2 x dx 42. represents the volume of the solid
generated by revolving the region bounded by about the axis.
x4 8 12 16
1
2
3
4
y
y-y � 4y � 2,x � 0,x � y2,
��4
2y 4 dy
43.
Matches (a)
x
1
2
1 2
yA 3 44.
Matches (b)
x
1
134
34
12
12
14
14
yA 34
Section 7.2 Volume: The Disk Method 25
45.
The volumes are the same because the solid has been translated horizontally. �4x � x2 � 4 � �x � 2�2�
1
1
2
3
4
2 3 4
y
x−2 −1 1
1
2
3
2
y
x
46. (a)
x
y842
4
−4
2
z (b)
x
y88
4
2
z (c)
a < c < b
x
y168
4
4
z
47.
Note:
� 18�
�13
��32�6
V �13
�r2h
� � �
12x3�
6
0� 18�
V � ��6
0 14
x2 dx
x
−2
−1
1
2
3
4
1 2 3 4 5 6
yr�x� � 0R�x� �12
x, 48.
�r2�
3h2 h3 �13
�r2h
� � r2�
3h2 x3�h
0
V � ��h
0 r2
h2 x2 dx
xh
r ( , )h ry = r
hx
yr�x� � 0R�x� �rh
x,
49.
� 2�r3 �13
r3� �43
�r3
� 2��r2x �13
x3�r
0
� 2��r
0�r2 � x2� dx
( , 0)−r ( , 0)rx
y = r x−2 2
y V � ��r
�r
�r2 � x2� dx
r�x� � 0R�x� � �r2 � x2, 50.
��
3�2r3 � 3r2h � h3�
x
r
h
y � � 2r3
3� r2h �
h3
3 �
� ��r3 �r3
3 � � r2h �h3
3 ��
� ��r2y �y3
3 �r
h
� ��r
h
�r2 � y2� dy
V � ��r
h��r2 � y2 �2 dy
r�y� � 0R�y� � �r2 � y2,x � �r2 � y2,
26 Chapter 7 Applications of Integration
51.
� �r2h1 �hH
�h2
3H 2�
� �r2h �h2
H�
h3
3H 2�
� �r2�y �1H
y2 �1
3H 2 y3�h
0
V � ��h
0 �r1 �
yH��
2
dy � �r2 �h
0 1 �
2H
y �1
H 2 y2� dy
−r rx
h
H
yr�y� � 0R�y� � r1 �yH�,x � r �
rH
y � r 1 �yH�,
52. (a)
Let and set
Thus, when the solid is divided into twoparts of equal volume.
x � 2�2,
c � �8 � 2�2
c2 � 8
��c
0 x dx � ��x2
2 �c
0�
�c2
2� 4�.
0 < c < 4
V � ��4
0��x �2 dx � ��4
0 x dx � ��x2
2 �4
0� 8� (b) Set (one third of the volume). Then
To find the other value, set
(two thirds of the volume).
Then
The x-values that divide the solid into three parts of equalvolume are and x � �4�6 ��3.x � �4�3 ��3
d ��32�3
�4�6
3.d 2 �
323
,� d 2
2�
16�
3,
��d
0 x dx �
16�
3
c �4�3
�4�3
3.c2 �
163
,�c2
2�
8�
3,
� �c
0 x dx �
8�
3
53. V � ��2
0 1
8 x2�2 � x�2
dx ��
64 �2
0 x4�2 � x� dx �
�
64�2x5
5�
x6
6 �2
0�
�
30
54.
1031.9016 cubic centimeters
� ��0.1x4
4�
2.2x3
3�
10.9x2
2� 22.2x�
11.5
0� ��2.952x�
15
11.5
V � ��11.5
0��0.1x3 � 2.2x2 � 10.9x � 22.2 �2 dx � ��15
11.5 2.952 dx
x
2
4
6
8
4 8 12 16
yy � ��0.1x3 � 2.2x2 � 10.9x � 22.2,2.95,
0 ≤ x ≤ 11.5 11.5 < x ≤ 15
55. (a)
� 60�
�18�
25 �25x �x3
3 �5
0
�18�
25 �5
0�25 � x2� dx
x−6 −4 −2 2 4 6
−4
−2
2
4
6
8
y V �9�
25 �5
�5�25 � x2� dx
r�x� � 0R�x� �35�25 � x2, (b)
� 50�
�25�
9 �9y �y3
3 �3
0
x
2
2
4
4
6
6
y V �25�
9 �3
0�9 � y2� dy
x ≥ 0r�y� � 0,R�y� �53�9 � y2,
Section 7.2 Volume: The Disk Method 27
56. (a) First find where intersects the parabola:
� ��4b2 �643
b �51215 � � ��64
5�
1283
�323
b � 4b2 � 32b � 64�
� �� x5
80�
2x3
3�
bx3
6� b2x � 8bx � 16x�
4
0
� ��4
0 � x4
16� 2x2 �
bx 2
2� b 2 � 8b � 16� dx
� �4
0 ��4 �
x2
4� b�
2
dx
V � �2�4�b
0 ��4 �
x2
4� b�
2
dx � �4
2�4�b
��b � 4 �x2
4 �2
dx
x � 2�4 � b
x2 � 16 � 4b � 4�4 � b�
x
y
5
42 2
z b � 4 �x2
4
y � b
(b) Graph of
Minimum volume is 17.87 for b � 2.67.
00
4
120
V�b� � ��4b 2 �643
b �51215 � (c)
V � �b� � 8� > 0 ⇒ b �83
is a relative minimum.
V� �b� � ��8b �643 � � 0 ⇒ b �
64�38
�83
� 223
57. Total volume:
Volume of water in the tank:
When the tank is one-fourth of its capacity:
Depth:
When the tank is three-fourths of its capacity the depth is 100 � 32.64 � 67.36 feet.
�17.36 � ��50� � 32.64 feet
y0 �17.36
y03 � 7500y0 � 125,000 � 0
125,000 � 7500y0 � y03 � 250,000
14
500,000�
3 � � �2500y0 �y0
3
3�
250,0003 �
� �2500y0 �y0
3
3�
250,0003 �
� ��2500y �y3
3 �y0
�50
��y0
�50��2500 � y2 �2 dy � ��y0
�50�2500 � y2� dy x
−60
−60
−40
−20
20
20
40
40
60
60
yV �4� �50�3
3�
500,000�
3 ft3
28 Chapter 7 Applications of Integration
58. (a)
Simpson’s Rule:
(b) (c)
00
10
6
V �10
0 � f �x�2 dx 186.35 cm3f �x� � 0.00249x4 � 0.0529x3 � 0.3314x2 � 0.4999x � 2.112
�
3�178.405� 186.83 cm3
� 2�2.9�2 � 4�2.7�2 � 2�2.45�2 � 4�2.2�2 � �2.3�2�V �
3��2.1�2 � 4�1.9�2 � 2�2.1�2 � 4�2.35�2 � 2�2.6�2 � 4�2.85�2
n � 10b � a � 10 � 0 � 10,
V � �10
0� � f �x��2 dx
59. (a) (ii)
is the volume of a right circularcylinder with radius r and height h.
x
( , )h r
y r=y
��h
0 r2 dx (b) (iv)
is the volume of an ellipsoid withaxes and
(0, )a
( , 0)−b ( , 0)bx
y a= 1 −b2x2
y
2b.2a
��b
�b
a�1 �x2
b2 �2
dx (c) (iii)
is the volume of a sphere withradius r.
x
( , 0)−r ( , 0)r
y = r x−2 2
y
��r
�r��r2 � x2 �2 dx
(d) (i)
is the volume of a right circular cone with the radiusof the base as r and height h.
x
( , )h r
y = xrh
y
��h
0 rx
h �2
dx (e) (v)
is the volume of a torus with the radius of its circularcross section as r and the distance from the axis of thetorus to the center of its cross section as R.
x
R
R + r x−2 2
R r x− −2 2
−r r
y
��r
�r��R � �r2 � x2 �2
� �R � �r2 � x2 �2� dx
60. Let and equal the areas of the cross sections of the two solids for Since we have
Thus, the volumes are the same.
V1 � �b
a
A1�x� dx � �b
a
A2�x� dx � V2.
A1�x� � A2�x�,a ≤ x ≤ b.A2�x�A1�x�
Section 7.2 Volume: The Disk Method 29
61.
Base of cross section
(a)
2 + x x− 2
2 + x x− 2
� �4x � 2x2 � x3 �12
x4 �15
x5�2
�1�
8110
V � �2
�1�4 � 4x � 3x2 � 2x3 � x4� dx
� 4 � 4x � 3x2 � 2x3 � x4
A�x� � b2 � �2 � x � x2�2
� �x � 1� � �x2 � 1� � 2 � x � x2
x2 3 4
2
3
4
y
(b)
2 + x x− 2
1
V � �2
�1 �2 � x � x2� dx � �2x �
x2
2�
x3
3 �2
�1�
92
A�x� � bh � �2 � x � x2�1
62.
Base of cross section
(a)
(b)
�32�3
3
� �3�4x �x3
3 �2
�2
2 4 − x22 4 − x2
2 4 − x2
V � �3�2
�2�4 � x2� dx
� �3�4 � x2�
A�x� �12
bh �12
�2�4 � x2 ���3�4 � x2 �
� 4�4x �x3
3 �2
�2�
1283
V � �2
�24�4 � x2� dx
2 4 − x2
2 4 − x2
A�x� � b2 � �2�4 � x2 �2
� 2�4 � x2
x
3
1
1
−1
−3
3−3
y
(c)
(d)
4 − x2
4 − x2
2
V � �2
�2�4 � x2� dx � �4x �
x3
3 �2
�2�
323
�12
�2�4 � x2 ���4 � x2 � � 4 � x2A�x� �12
bh
4 − x22
V ��
2�2
�2�4 � x2� dx �
�
2 �4x �x3
3 �2
�2�
16�
3
��
2��4 � x2 �2
��
2�4 � x2�A�x� �
12
�r2
30 Chapter 7 Applications of Integration
63.
Base of cross section
(a)
(c)
1 − y31 − y3
1 − y3
V ��34
�1
0�1 � 3�y �2 dy �
�34 1
10� ��340
��34
�1 � 3�y �2
A�y� �12
bh �12
�1 � 3�y ��32 � �1 � 3�y �
� �y �32
y4�3 �35
y5�3�1
0�
110
� �1
0�1 � 2y1�3 � y 2�3� dy
V � �1
0�1 � 3�y �2 dy
1 − y3
1 − y3
A�y� � b2 � �1 � 3�y �2
� 1 � 3�y
x1
1
14
14
12
12
34
34
y
(b)
(d)
a
b
1 − y3
V ��
2 �1
0�1 � 3�y �2 dy �
�
2 110� �
�
20
��
2�1 � 3�y �2 A�y� �
12
�ab ��
2�2��1 � 3�y �1 � 3�y
2
1 − y3
V �18
� �1
0 �1 � 3�y �2 dy �
�
8 1
10� ��
80
A�y� �12
�r2 �12
� 1 � 3�y2 �2
�18
� �1 � 3�y �2
64. The cross sections are squares. By symmetry, we can setup an integral for an eighth of the volume and multiply by 8.
�163
r3
� 8�r2y �13
y3�r
0
V � 8�r
0�r2 � y2� dy
y
x
A�y� � b2 � ��r2 � y2 �2
65.
r
R
RR2 2− r
�43
� �R2 � r2�3�2
� 2� ��R2 � r2�3�2 ��R2 � r2�3�2
3 �
� 2� ��R2 � r2�x �x3
3 ��R 2 �r2
0
� 2� ��R2 �r2
0 �R2 � r2 � x2� dx
V � � ��R2 �r2
��R2 �r2
���R2 � x2 �2� r2 � dx
Section 7.2 Volume: The Disk Method 31
66.
r � 5�1 � 2�2�3 3.0415
25�1 � 2�2�3� � r2
25 �25
�22�3� � r2
25 � r2 � 1252 �2�3
�25 � r2�3�2 �1252
43
��25 � r2�3�2 �12
43�� �125� 67. V � ��1
0y2 dy � �
y3
3 �1
0�
�
3
68.
� ��1 �13� �
23
�
� ��y2 �y3
3 �1
0
� ��1
0�2y � y2� dy
V � ��1
0�12 � �1 � y�2� dy 69.
�2�
15
� ��13
�15�
� ��x3
3�
x5
5 �1
0
V � ��1
0�x2 � x 4� dx
70.
� ��15� �
�
5
� ��x2 � x3 �x5
5 �1
0
� ��1
0�2x � 3x2 � x 4� dx
� ��1
0�1 � 2x2 � x 4 � 1 � 2x � x2� dx
V � ��1
0��1 � x2�2 � �1 � x�2� dx 71.
��
2
� ��1 �12�
� ��y �y 2
2 �1
0
V � ��1
0�1 � y� dy
72.
��
6
� ��1 �43
�12�
� ��y �43
y3�2 �y2
2 �1
0
� ��1
0�1 � 2�y � y� dy
V � ��1
0�1 � �y �2 dy 73.
��
6
� ��12
�13�
� ��y2
2�
y3
3 �1
0
V � ��1
0� y � y2� dy
32 Chapter 7 Applications of Integration
74.
��
6
� ��43
�32
�13�
� ��43
y3�2 �3y2
2�
y3
3 �1
0
� ��1
0 �2�y � 3y � y2� dy
� ��1
0 �1 � 2y � y2 � 1 � 2�y � y� dy
V � ��1
0 ��1 � y�2 � �1 � �y �2� dy 75. (a) When represents a square.
When represents a circle.
(b)
To approximate the volume of the solid, form slices,each of whose area is approximated by the integralabove. Then sum the volumes of these slices.n
n
A � 2 �1
�1 �1 � x a�1�a dx � 4 �1
0 �1 � xa�1�a dx
y � �1 � x a�1�a
x
a = 2
a = 1
−1
−1
1
1
y
a � 2: x 2 � y 2 � 1
a � 1: x � y � 1
76. (a) Since the cross sections are isosceles right triangles:
(b)
As V →�. → 90,
V �tan
2 �r
�r
�r2 � y2� dy � tan �r
0 �r2 � y2� dy � tan �r2y �
y3
3 �r
0�
23
r3 tan
A�x� �12
bh �12�r2 � y2��r2 � y2 tan � �
tan 2
�r2 � y2�
V �12
�r
�r
�r2 � y2� dy � �r
0 �r2 � y2� dy � �r2y �
y3
3 �r
0�
23
r3
A�x� �12
bh �12
��r2 � y2 ���r2 � y2 � �12
�r2 � y2�x
y
77. (a)
R
x
y
� 8�R�r
0
�r2 � y2 dy
� 2��r
04R�r2 � y2 dy
V � 2��r
0 � �R � �r2 � y2�2
� �R � �r2 � y2� 2� dy
x � R ± �r2 � y2
�x � R�2 � y2 � r2 (b) is one-quarter of the area of a circle of
radius
V � 8�R�14�r2� � 2�2r2R
r, 14�r2.
�r
0
�r2 � y2 dy
Section 7.3 Volume: The Shell Method
Section 7.3 Volume: The Shell Method 33
1.
� �2� x3
3 �2
0�
16�
3
V � 2��2
0 x�x� dx
h�x� � xp�x� � x, 2.
� 2��x2
2�
x3
3 �1
0�
�
3
� 2��1
0 �x � x2� dx
V � 2��1
0 x�1 � x� dx
h�x� � 1 � xp�x� � x, 3.
� �4�
5x5�2�
4
0�
128�
5
� 2��4
0 x3�2 dx
V � 2��4
0 x�x dx
h�x� � �xp�x� � x,
4.
� 2��2x2 �x 4
4 �2
0� 8�
� 2��2
0 �4x � x3� dx
V � 2��2
0 x�4 � x2� dx
h�x� � 8 � �x2 � 4� � 4 � x2p�x� � x, 5.
� ��
2x4�
2
0� 8�
V � 2��2
0 x3 dx
x−1 1 2 3
1
2
3
4
yh�x� � x2p�x� � x,
6.
� ��x4
4 �6
0� 324�
V � 2��6
0 12
x3 dx
1−1−3
3
6
9
12
15
18
2 3 4 5 6
y
x
h�x� �12
x2p�x� � x, 7.
� 4��23
x3 �14
x4�2
0�
16�
3
� 4��2
0 �2x2 � x3� dx
x
1
2
3
4
−1 1 2 3
y V � 2��2
0 x�4x � 2x2� dx
h�x� � �4x � x2� � x2 � 4x � 2x2p�x� � x,
8.
� 2��2x2 �14
x4�2
0� 8�
V � 2��2
0 �4x � x3� dx
x21−1−2
1
2
3
yh�x� � 4 � x2p�x� � x, 9.
�8�
3
� 2�� x4
4�
43
x3 � 2x2�2
0
V � 2��2
0 �x3 � 4x2 � 4x� dx
� x2 � 4x � 4
h�x� � 4 � �4x � x2�
x
1
2
3
4
−1 1 2 3
yp�x� � x
10.
� 2��2x2 �23
x3�2
0�
16�
3
� 2��2
0 �4x � 2x2� dx
V � 2��2
0 x�4 � 2x� dx
x
1
2
3
4
−1 1 2 3
yh�x� � 4 � 2xp�x� � x,
34 Chapter 7 Applications of Integration
11.
� 0.986
� �2� 1 �1�e
� ���2�e�x2�2�1
0
� �2� �1
0 e�x2�2 x dx
x1
1
14
14
12
34
34
12
y V � 2��1
0 x 1
�2�e�x2�2 dx
h�x� �1
�2�e�x2�2p�x� � x, 12.
� ��2� cos x��
0� 4�
� 2���
0 sin x dx
V � 2���
0 x �sin x
x � dx
1
−1
2
3
x
4π π
2π
43π
yh�x� �sin x
xp�x� � x,
13.
� 2��y2 �y3
3 �2
0�
8�
3
� 2��2
0 �2y � y2� dy
V � 2��2
0 y�2 � y� dy
h�y� � 2 � yp�y� � y, 14. on
� 2���y2 �y3
3 �0
�2�
8�
3
� 2��0
�2 ��2y � y2� dy
V � 2��0
�2 ��y��2 � y� dy
h�y� � 4 � �2 � y� � 2 � y
��2, 0���p�y� ≥ 0p�y� � �y,
15. and if
and if
� 2��y2
2 �1�2
0� 2��y �
y2
2 �1
1�2�
�
4�
�
4�
�
2
V � 2� �1�2
0 y dy � 2� �1
1�2 �1 � y� dy
12
≤ y ≤ 1.h�y� �1y
� 1p�y� � y
x21
1
12
14
12
32
34
y0 ≤ y < 12
.h�y� � 1p�y� � y
16.
� 2��128 � 64� � 128�
� 2��8y2 �y4
4 �4
0
� 2��4
0�16y � y3� dy
1
−1
−2
−3
−4
2
3
4
4 8 12
y
x
V � 2��4
0y�16 � y2� dy
h�y� � 16 � y2 p�y� � y, 17.
�6�
7�27� �
768�
7
� �2�37y7�3�
8
0
� 2��8
0y 4�3 dy
V � 2��8
0y 3�y dy
x−2 2 4 6
2
4
6
8
yp�y� � y, h�y� � 3�y
18.
�4�
5�9�5�2 �
4�
5�243� �
9725
�
� 2�25y 5�2�
9
0
V � 2��9
0y�y dy � 2��9
0y3�2 dy
x
y
3 6 9
3
6
9
p�y� � y, h�y� � �y
Section 7.3 Volume: The Shell Method 35
19.
� 2��8 �163 � �
16�
3
� 2��2y2 �23
y3�2
0
� 2��2
0�4y � 2y2� dy
x
y
−1 1 2 3 4
−1
1
2
3
4
(2, 2)
V � 2��2
0y�4 � 2y� dy
p�y� � y, h�y� � �4 � y� � �y� � 4 � 2y 20.
� 2��4 �83
� 4� �16�
3
� 2�� y2 �y3
3�
y 4
4 �2
0
� 2��2
0�2y � y2 � y3� dy
x
y
(2, 2)
−1−2 1 2
−1
1
2
3
V � 2��2
0y�2 � y � y2� dy
p�y� � y, h�y� � y � �y2 � 2� � 2 � y � y2
21.
x
1
2
3
4
1 2 3
y
� 4��x 4
4� 2x3 � 4x2�
2
0� 16�
� 2� �2��2
0�x3 � 6x2 � 8x� dx
V � 2��2
0�4 � x��4x � 2x2� dx
h�x� � 4x � x2 � x2 � 4x � 2x2p�x� � 4 � x, 22.
x
1
2
3
4
−1 1 3
y
� 2��4x2 �83
x3 �12
x4�2
0�
16�
3
� 2��2
0�8x � 8x2 � 2x3� dx
V � 2��2
0�2 � x��4x � 2x2� dx
h�x� � 4x � x2 � x2 � 4x � 2x2p�x� � 2 � x,
23.
x
−1
1
2
3
4
1 2 3 4
y
� 2��x4
4� 3x3 � 10x2�
4
0� 64�
� 2��4
0�x3 � 9x2 � 20x� dx
V � 2��4
0�5 � x��4x � x2� dx
h�x� � 4x � x2p�x� � 5 � x, 24.
x1 2 3 4
4
5
−2
−1
1
2
3
y
� 2��4x3�2 �25
x5�2�4
0�
192�
5
� 2� �4
0�6x1�2 � x3�2� dx
V � 2� �4
0�6 � x��x dx
h�x� � �xp�x� � 6 � x,
25. The shell method would be easier: shells
Using the disk method: �Note: V �128�
3 �V � ��4
0 ��2 � �4 � x �2
� �2 � �4 � x �2 dx
V � 2��4
0�4 � � y � 2�2� y dy
36 Chapter 7 Applications of Integration
26. The shell method is easier:
Using the disk method, and �Note: V � ��8�ln 2�2 � 8 ln 2 � 3� �V � ��3
0�ln�4 � y��2 dy.x � ln�4 � y�
V � 2��ln 4
0x�4 � e x� dx
27. (a) Disk
x
2
2
6
4
−1 1 3
8
y
V � ��2
0x6 dx � ��x7
7 �2
0�
128�
7
r�x� � 0 R�x� � x3,
(b) Shell
x
2
2
6
4
−1 1 3
8
y
V � 2��2
0x 4 dx � 2��x5
5 �2
0�
64�
5
h�x� � x3p�x� � x,
(c) Shell
� 2��x 4 �15
x5�2
0�
96�
5
� 2��2
0�4x3 � x4� dx
V � 2��2
0�4 � x�x3 dx
x
2
2
6
4
1 3 4
8
yh�x� � x3p�x� � 4 � x,
28. (a) Disk
� �100�
3 � 1125
� 1� �49615
�
� 100��x�3
�3�5
1
� 100��5
1x�4 dx
V � ��5
110
x2 2
dx
R�x� �10x2 , r�x� � 0
1−1−2
2
4
6
8
10
2 3 4 5
y
x
(b) Shell
� 20��ln x �5
1� 20� ln 5
� 20��5
1 1x dx
V � 2��5
1x10
x2 dx
R�x� � x, r�x� � 0
(c) Disk
� ��1003x3 �
200x �
5
1�
190415
�
V � ��5
1�102 � 10 �
10x2
2
� dx
R�x� � 10, r�x� � 10 �10x2
Section 7.3 Volume: The Shell Method 37
29. (a) Shell
(b) Same as part (a) by symmetry
(0, )a
( , 0)ax
y
� 2� �a3
2�
4a3
5�
a3
3 � ��a3
15
� 2� � a2
y2 �4a1�2
5y5�2 �
y3
3 �a
0
� 2� �a
0 �ay � 2a1�2y3�2 � y2� dy
V � 2� �a
0 y�a � 2a1�2 y1�2 � y� dy
h�y� � �a1�2 � y1�2�2 p�y� � y,
(c) Shell
(0, )a
( , 0)ax
y
� 2� �a2x �43
a3�2x3�2 �45
a1�2x5�2 �13
x3�a
0�
4� a3
15
� 2��a
0�a2 � 2a3�2x1�2 � 2a1�2x3�2 � x2� dx
V � 2��a
0�a � x��a1�2 � x1�2�2 dx
h�x� � �a1�2 � x1�2�2 p�x� � a � x,
30. (a) Disk
� 2� a3 �95
a3 �97
a3 �13
a3 �32� a3
105
� 2� �a2x �95
a4�3x5�3 �97
a2�3x7�3 �13
x3�a
0
� 2��a
0�a2 � 3a4�3x2�3 � 3a2�3x4�3 � x2� dx
V � ��a
�a
�a2�3 � x2�3�3 dx
r�x� � 0 R�x� � �a2�3 � x2�3�3�2,
x
( , 0)a
(0, )a
( , 0)−a
y (b) Same as part (a) by symmetry
x
( , 0)a
(0, )a
(0, )−a
y
31. Answers will vary.
(a) The rectangles would be vertical.
(b) The rectangles would be horizontal.
32. (a)
x
y
2
2 5
−2
z (b)
xy
2
55
z (c)
a < c < b
xy
2
5 10
z
38 Chapter 7 Applications of Integration
33.
This integral represents the volume of the solid generatedby revolving the region bounded by and about the x-axis by using the disk method.
represents this same volume by using the shell method.
Disk method
x
4
3
2
1
−153 421
y
2��2
0y �5 � �y2 � 1�� dy
x � 5y � 0,y � �x � 1,
��5
1�x � 1� dx � ��5
1 ��x � 1 �2 dx 34.
represents the volume of the solid generated by revolvingthe region bounded by and aboutthe y-axis by using the shell method.
represents this same volume by using the disk method.
Disk method
x
4
3
2
1
−153 421
y
��2
0�16 � �2y�2� dy � ��2
0��4�2 � �2y�2� dy
x � 4y � 0,y � x�2,
2��4
0xx
2 dx
35. (a)
(b)
V � 2� �1
0 x�1 � x4�3�3�4 dx � 1.5056
y � �1 � x 4�3�3�4
y � 0x � 0,x 4�3 � y 4�3 � 1,
y x )= (1 − 4/3 3/4
−0.25
−0.25 1.5
1.5 36. (a)
(b) V � 2� �1
0 x�1 � x3 dx � 2.3222
x1
1
1
1
1
1
3
3
4
4
2
2
4
4
y
37. (a)
(b) V � 2� �6
2 x 3��x � 2�2�x � 6�2 dx � 187.249
−1
−1
7
y x x= ( 2) ( 6)− −3 2 2
7 38. (a)
(b) V � 2� �3
1
2x1 � e1�x dx � 19.0162
x1 2 3 4
4
3
2
1
y
39.
Matches (d)
x1 2
1
2
yVolume � 7.5
x � 2x � 0,y � 0,y � 2e�x, 40.
Matches (e)
x
2
1
4π
2π
yVolume � 1
x ��
4x � 0,y � 0,y � tan x,
Section 7.3 Volume: The Shell Method 39
41.
Now find such that:
(Quadratic Formula)
Take since the other root is too large.
Diameter: 2�4 � 2�3 � 1.464
x0 � �4 � 2�3 � 0.73205,
x02 � 4 ± 2�3
x04 � 8x0
2 � 4 � 0
1 � 2x02 �
14
x04
1 � 2 �x2 �18
x 4�x0
0
� � 2� �x0
0 2x �
12
x3 dx
x2
2
1
1
y
x0
V � 2��2
0x2 �
12
x2 dx � 2��2
0 2x �
12
x3 dx � 2��x2 �18
x 4�2
0� 4� �total volume�
h�x� � 2 �12
x2p�x� � x,
42. Total volume of the hemisphere is By the Shell Method,Find such that:
Diameter: 2�9 � 182�3 � 2.920
x0 � �9 � 182�3 � 1.460
�9 � x02� 3�2 � 18
� ��23
�9 � x2�3�2�x0
0� 18 �
23
�9 � x02� 3�2
6 � ��x0
0 �9 � x2�1�2��2x� dx
x
2
1
321−1−1
−3
−2
−3 −2
y 6� � 2� �x0
0 x�9 � x2 dx
x0h�x� � �9 � x2.p�x� � x,
12�4
3��r3 �23� �3�3 � 18�.
43.
� 4� 2 � �2� 23�1 � x2�3�2�
1
�1� 4� 2
� 8� �
2 � 2��1
�1x�1 � x2�1�2��2� dx
� 8��1
�1
�1 � x2 dx � 4��1
�1x�1 � x2 dx
V � 4��1
�1�2 � x��1 � x2 dx 44.
� 2� 2r2R
� 4�R� r2
2 � �2�23�r2 � x2�3�2�
r
�r
� 4�R�r
�r
�r2 � x2 dx � 4��r
�r x�r2 � x2 dx
V � 4� �r
�r
�R � x��r2 � x2 dx
45. (a)
Hence,
—CONTINUED—
�x sin x dx � sin x � x cos x � C.
ddx
�sin x � x cos x � C� � cos x � x sin x � cos x � x sin x
40 Chapter 7 Applications of Integration
45. —CONTINUED—
(b) (i)
x
y
−1.0
−
0.5
1.0
3 4π
2π π
4π
4π
� 2� ��1 � 0� � 0� � 2�
� 2��sin x � x cos x���2
0
V � 2����2
0x sin x dx
p�x� � x, h�x� � sin x (ii)
� 6� ��� � 6� 2
� 6��sin x � x cos x��
0
� 6���
0x sin x dx
x
y
− 5 4π
2π π
4π
−1
−2
1
2
V � 2���
0x�3 sin x� dx
p�x� � x, h�x� � 2 sin x � ��sin x� � 3 sin x
46. (a)
Hence, �x cos x dx � cos x � x sin x � C.
ddx
�cos x � x sin x � C� � �sin x � sin x � x cos x � x cos x
(b) (i)
x
y
−1 1
−1
2
� 2.1205
� 4��cos x � x sin x �x 4
4 �0.8241
0
V � 2�2���0.8241
0x�cos x � x2� dx
x2 � cos x ⇒ x � ±0.8241 (ii)
x
y
−1−2 1 2
1
2
3
4
� 6.2993
� 2��1.511
0�4 cos x � 4x sin x �
�x � 2�3
3 �1.511
0
V � 2��1.511
0x�4 cos x � �x � 2�2� dx
4 cos x � �x � 2�2 ⇒ x � 0, 1.5110
47.
(a) Plane region bounded by
(b) Revolved about the axis
Other answers possible
−1 21 3 4
1
2
3
4
x
y
y-
y � x2, y � 0, x � 0, x � 2
2��2
0x3 dx � 2��2
0x�x2� dx 48.
(a) Plane region bounded by
(b) Revolved about the axis
Other answers possible
1
1(1, 1)
x
y
x = y
x-
x � �y, x � 1, y � 0
2��1
0� y � y3�2� dy � 2��1
0y�1 � �y � dy
Section 7.3 Volume: The Shell Method 41
49.
(a) Plane region bounded by
(b) Revolved around line
Other answers possible
x
y
−1−2−3 1 2 3 4 5
1
2
3
4
5
6
−2
x = 6 − y
y � �2
x � �6 � y, x � 0, y � 0
2��6
0� y � 2��6 � y dy 50.
(a) Plane region bounded by
(b) Revolved about the line
x
y
1 2 3
1
2
3y = ex
4
x � 4
y � e x, y � 0, x � 0, x � 1
2��1
0�4 � x�e x dx
51. Disk Method
xr−r
r
y
� ��r2y �y3
3 �r
r�h�
13
�h2�3r � h�
V � � �r
r�h
�r2 � y2� dy
r�y� � 0
R�y� � �r2 � y2
52.
Note: If then volume is that of a sphere.a � b,
�4�b3a
a3 �43
�a2b
�4�b
a ��a2 � x2�3�2
3 �a
0
�4�b
a �a
0
�a2 � x2 x dx
V � 2�2���a
0xb�1 �
x2
a2 dx
p�x� � x, h�x� � b�1 �x2
a2
y � ±b�1 �x2
a2
y2
b2 � 1 �x2
a2
a
b
x
y x2
a2 �y2
b2 � 1
53. (a)
(b)
limn→�
�abn�b � �
limn→�
R1�n� � limn→�
n
n � 1� 1
R1�n� �abn�1�n��n � 1��
�abn�b �n
n � 1
� abn�11 �1
n � 1 � abn�1 nn � 1
� abn�1 � abn�1
n � 1
� �abnx � axn�1
n � 1�b
0
Area region � �b
0�abn � axn� dx (c) Disk Method:
(d)
(e) As the graph approaches the line x � 1.n →�,
limn→�
��b2��abn� � �
limn→�
R2�n� � limn→�
nn � 2 � 1
R2�n� ��abn�2�n��n � 2��
��b2��abn� � nn � 2
� 2�a�bn�2
2�
bn�2
n � 2� � �abn�2 nn � 2
� 2�a�bn
2x2 �
xn�2
n � 2�b
0
� 2�a�b
0�xbn � xn�1� dx
V � 2��b
0x�abn � axn� dx
42 Chapter 7 Applications of Integration
54. (a) (ii)
is the volume of a right circular cone with the radiusof the base as r and height h.
x
(0, )h
( , 0)r
y h= 1 − xr( (
y
2� �r
0 hx1 �
xr dx (b) (v)
is the volume of a torus with the radius of its circularcross section as r and the distance from the axis of thetorus to the center of its cross section as R.
x
x R=
y r x= 2 2−
y r x= − −2 2
( , 0)r
( , 0)−r
y
2� �r
�r
�R � x��2�r2 � x2 � dx
(c) (iii)
is the volume of a sphere withradius
x
y r x= 2 2−
y r x= − −2 2
( , 0)r
y
r.
2� �r
0 2x�r2 � x2 dx (d) (i)
is the volume of a right circularcylinder with a radius of r and a height of h.
x
( , )r h
y
2� �r
0 hx dx (e) (iv)
is the volume of an ellipsoid withaxes and
x
y = 1 −
1 −
( , 0)b
(0, )a
(0, )−a
x2
x2
b2
b2
a
y a= −
y
2b.2a
2� �b
0 2ax�1 � �x2�b2� dx
55. (a)
(b) Top line:
Bottom line:
(Note that Simpson’s Rule is exact for this problem.)
� 121,475 cubic feet
� 2� �26,0003 � � 2� �32,000
3 �
� 2���x3
6� 25x2�
20
0� 2���2x3
3� 40x2�
40
20
� 2� �20
0 �1
2x2 � 50x dx � 2� �40
20 ��2x2 � 80x� dx
V � 2� �20
0 x�1
2x � 50 dx � 2� �40
20 x��2x � 80� dx
y � 40 �0 � 40
40 � 20�x � 20� � �2�x � 20� ⇒ y � �2x � 80
y � 50 �40 � 5020 � 0
�x � 0� � �12
x ⇒ y � �12
x � 50
�20�
3�5800� � 121,475 cubic feet
�2� �40�
3�4� �0 � 4�10��45� � 2�20��40� � 4�30��20� � 0�
V � 2� �4
0 x f �x� dx
Section 7.3 Volume: The Shell Method 43
56. (a)
� 1,366,593 cubic feet
�2� �200�
3�8� �0 � 4�25��19� � 2�50��19� � 4�75��17� � 2�100�15 � 4�125��14� � 2�150��10� � 4�175��6� � 0�
V � 2��200
0x f �x� dx
57.
(a)
� �� x 4
4�
8x3
3� 8x2�
4
0�
64�
3
� ��4
0�x3 � 8x2 � 16x� dx
V � ��4
0x�4 � x�2 dx
y2 � ��x�4 � x�2 � ��4 � x��x
y1 � �x�4 � x�2 � �4 � x��x
1 2 3 4 5 6−1
−2
−3
−4
1
2
3
4
x
y y2 � x�4 � x�2, 0 ≤ x ≤ 4
(b)
� 4��85
x5�2 �27
x7�2�4
0�
2048�
35
� 4��4
0�4x3�2 � x5�2� dx
V � 4��4
0x�4 � x��x dx (c)
� 4��323
x3�2 �165
x5�2 �27
x7�2�4
0�
8192�
105
� 4��4
0�16�x � 8x3�2 � x5�2� dx
V � 4��4
0�4 � x��4 � x��x dx
(b)
−20 225
−6
24
d � �0.000561x2 � 0.0189x � 19.39(c)
(d) Number gallons � V�7.48� � 10,048,221 gallons
� 1,343,345 cubic feet
V � 2��200
0xd�x� dx � 2� �213,800�
58.
(a)
� ��x 4
4�
5x3
3 �0
�5�
625�
12
V � ��0
�5 x
2�x � 5� dx
y2 � ��x2�x � 5� � �x�x � 5
y1 � �x2�x � 5� � x�x � 5
−2−4−6 2
−4
4
x
y y2 � x2�x � 5�, �5 ≤ x ≤ 0
(b)
Let
� 4��27
u7�2 � 4u5�2 �503
u3�2�5
0�
1600�5�
21
� 4��5
0�u5�2 � 10u3�2 � 25u1�2� du
V � 4��5
0�u � 5�2�u du
u � x � 5, du � dx.
V � 4��0
�5 x�x�x � 5 � dx (c)
Let
� 4���27
u7�2 � 2u5�2�5
0�
400�5�
7
� 4��5
0��u5�2 � 5u3�2� du
V � 4��5
0��u��u � 5��u du
u � x � 5, du � dx.
V � 4��0
�5��5 � x�x�x � 5 dx
44 Chapter 7 Applications of Integration
59.
c � 2 �c �14
yields no volume.� �4c � 1��c � 2� � 0
4c2 � 9c � 2 � 0
4c � 1 � 2c�c �14�
V1 � V2 ⇒ 4c � 1c
� � 2��c �14�
V2 � 2��c
1�4 x�1
x� dx � 2�x�c
1�4� 2��c �
14�
c14
x
y V1 � ��c
1�4
1x2 dx � ���
1x�
c
1�4� ���
1c
� 4� �4c � 1
c�
Section 7.4 Arc Length and Surfaces of Revolution
1.
(a)
(b)
� �135
x�5
0� 13
s � �5
01 � � 12
5 �2
dx
y� �125
y �125
x
� 13
d � �5 � 0�2 � �12 � 0�2
�0, 0�, �5, 12� 2.
(a)
(b)
� �53
x�7
1� 10
s � �7
11 � � 4
3�2
dx
y� �43
y �43
x �23
� 10
d � �7 � 1�2 � �10 � 2�2
�1, 2�, �7, 10� 3.
�23
�8 � 1� 1.219
� �23
�1 � x�3�2�1
0
s � �1
0 1 � x dx
y� � x1�2, 0 ≤ x ≤ 1
y �23
x3�2 � 1
4.
�2
27�823�2 � 1� 54.929
� � 227
�1 � 9x�3�2�9
0
s � �9
0
1 � 9x dx
y� � 3x1�2, 0 ≤ x ≤ 9
y � 2x3�2 � 3 5.
� 55 � 22 8.352
�32 �
23
�x2�3 � 1�3�2�8
1
�32
�8
1 x2�3 � 1� 2
3x1�3� dx
� �8
1 x2�3 � 1
x2�3 dx
s � �8
1 1 � � 1
x1�3�2
dx
y� �1
x1�3, 1 ≤ x ≤ 8
y �32
x2�3 6.
�3316
2.063
� �18
x 4 �1
4x2�2
1
� �2
1 � 1
2x3 �
12x3� dx
s � �b
a
1 � �y� �2 dx
1 � �y� �2 � � 12
x3 �1
2x3�2
, �1, 2�
y� �12
x3 �1
2x3, 1 ≤ x ≤ 2
y �x4
8�
14x2
Section 7.4 Arc Length and Surfaces of Revolution 45
7.
� � 110
x5 �1
6x3�2
1�
779240
3.2458
� �2
1 � 1
2x4 �
12x4� dx
� �2
1 �1
2x4 �
12x4�
2
dx
s � �b
a
1 � �y��2 dx
1 � �y��2 � �12
x4 �1
2x4�2
, 1 ≤ x ≤ 2
y� �12
x4 �1
2x4
y �x5
10�
16x3 8.
� 103�2 � 23�2 28.794
� �32
�23
�x2�3 � 1�3�2�27
1
�32�
27
1
x2�3 � 1� 23x1�3� dx
� �27
1x2�3 � 1
x2�3 dx
s � �27
11 � � 1
x1�3�2
dx
y� � x�1�3, 1 ≤ x ≤ 27
y �32
x2�3 � 4
9.
� ln�2 � 1� � ln�2 � 1� 1.763
� �ln csc x � cot x �3��4
��4
s � �3��4
��4 csc x dx
1 � �y� �2 � 1 � cot2 x � csc2 x
y� �1
sin x cos x � cot x
y � ln�sin x�, ��
4,
3�
4 � 10.
� ln�2 � 3 � 1.3170
� ln sec x � tan x ���3
0
� ���3
0 sec x dx
s � ���3
0
sec2 x dx
1 � �y� �2 � 1 � tan2 x � sec2 x
y� ��sin xcos x
� �tan x
y � ln�cos x�, 0 ≤ x ≤�
3
11.
�12 �ex � e�x�
2
0�
12 �e2 �
1e2� 3.627
�12
�2
0 �ex � e�x� dx
s � �2
0�1
2�ex � e�x��
2
dx
1 � �y� �2 � �12
�ex � e�x��2
, �0, 2�
y� �12
�ex � e�x�, �0, 2�
y �12
�ex � e�x� 12.
� ln�4�33�4� � ln
169
� 2 ln�43� 0.57536
� ln�sinh�x���ln 3
ln 2� ln�4
3� � ln�34�
� �ln 3
ln 2 ex � e�x
ex � e�x dx � �ln 3
ln 2 coth x dx
s � �b
a1 � �dy
dx�2
dx � �ln 3
ln 2 1 � e2x
e2x � 1 dx
�1 � 2e2x � e4x
�1 � e2x�2 � �1 � e2x
1 � e2x�2
1 � �dydx�
2
� 1 �4e2x
1 � 2e2x � e4x
dydx
�e x
e x � 1�
e x
e x � 1�
�2e x
e2x � 1�
2e x
1 � e2x
y � ln�ex � 1ex � 1� � ln�e x � 1� � ln�e x � 1�
46 Chapter 7 Applications of Integration
13.
� �y3
3� y�
4
0�
643
� 4 �763
� �4
0�y2 � 1� dy
� �4
0
y 4 � 2y2 � 1 dy
s � �4
0
1 � y2�y2 � 2� dy
dxdy
� y�y2 � 2�1�2
x �13
�y2 � 2�3�2, 0 ≤ y ≤ 4 14.
�12�
163
� 4� �12�
23
� 2� �103
� �12�
23
y3�2 � 2y1�2��4
1
s � �4
1 12�y �
1y� dy
�14
�y � 2 � y�1� �14�y �
1y�
2
1 � �dxdy�
2
� 1 �14
y �14
y�1 �12
dxdy
�12
y1�2 �12
y�1�2
x �13
�y3�2 � 3y1�2�
x �13y�y � 3�, 1 ≤ y ≤ 4
15. (a)
−1
−1
3
5
y � 4 � x2, 0 ≤ x ≤ 2 (b)
L � �2
0
1 � 4x2 dx
1 � �y� �2 � 1 � 4x2
y� � �2x (c) L 4.647
16. (a)
−3 2
−5
5
y � x2 � x � 2, �2 ≤ x ≤ 1 (b)
L � �1
�2 2 � 4x � 4x2 dx
1 � �y� �2 � 1 � 4x2 � 4x � 1
y� � 2x � 1 (c) L 5.653
17. (a)
−1
−1
4
2
y �1x
, 1 ≤ x ≤ 3 (b)
L � �3
11 �
1x4 dx
1 � �y� �2 � 1 �1x4
y� � �1x2 (c) L 2.147
18. (a)
−1 2
−2
y = 1x + 1
5
y �1
1 � x, 0 ≤ x ≤ 1 (b)
L � �1
01 �
1�1 � x�4 dx
1 � �y� �2 � 1 �1
�1 � x�4
y� � �1
�1 � x�2(c) L 1.132
Section 7.4 Arc Length and Surfaces of Revolution 47
19. (a)
−0.5
2
3 2�
2�−
y � sin x, 0 ≤ x ≤ � (b)
L � ��
0
1 � cos2 x dx
1 � �y� �2 � 1 � cos2 x
y� � cos x (c) L 3.820
20. (a)
−2
−2
2
y x= cos
2
y � cos x, ��
2 ≤ x ≤
�
2(b)
L � ���2
���2 1 � sin2x dx
1 � �y� �2 � 1 � sin2 x
y� � �sin x (c) 3.820
21. (a)
−1
−1
3
3
1 ≥ x ≥ e�2 0.135
y � �ln x
0 ≤ y ≤ 2x � e�y, (b)
L � �1
e�21 �
1x2 dx
1 � �y� �2 � 1 �1x2
y� � �1x
(c) L 2.221
Alternatively, you can do all the computations with respect to y.
(a) 0 ≤ y ≤ 2x � e�y, (b)
L � �2
0
1 � e�2y dy
1 � �dxdy�
2
� 1 � e�2y
dxdy
� �e�y (c) L 2.221
22. (a)
−1 6
−6
2
y � ln x, 1 ≤ x ≤ 5 (b)
L � �5
11 �
1x2 dx
1 � �y� �2 � 1 �1x2
y� �1x
(c) L 4.367
23. (a)
−3
1.5−0.5
3
y � 2 arctan x, 0 ≤ x ≤ 1 (b)
L � �1
01 �
4�1 � x2�2 dx
y� �2
1 � x2 (c) L 1.871
48 Chapter 7 Applications of Integration
24. (a)
−10
−5
10
10
y � 36 � x 2, 33 ≤ x ≤ 6
x � 36 � y2 , 0 ≤ y ≤ 3 (b)
� �3
0
636 � y2
dy
L � �3
01 �
y2
36 � y2 dy
��y
36 � y 2
dxdy
�12
�36 � y2��1�2��2y� (c) L 3.142 ��!�
Alternatively, you can convert to a function of
Although this integral is undefined at a graphing utility still gives L 3.142.x � 0,
L � �6
33 1 �
x2
36 � x2 dx � �6
33
636 � x2
dx
y� �dydx
� �x
36 � x2
y � 36 � x2
x.
25.
Matches (b)
s 5
x
(0, 5)
(2, 1)
−1 1 2 3 4
5
4
3
2
1
5x2 + 1
y =
y�2
01 � � d
dx �5
x2 � 1��2
dx 26.
Matches (e)
s 1
x
84
4
2
1
3π
, 1( (
(0, 0)
y x= tan
π
π
y���4
0 1 � � d
dx�tan x��
2
dx
27.
(a)
(b)
(c) (Simpson’s Rule, )
(d) 64.672
n � 10s � �4
0
1 � �3x2�2 dx � �4
0
1 � 9x4 dx 64.666
64.525
d � �1 � 0�2 � �1 � 0�2 � �2 � 1�2 � �8 � 1�2 � �3 � 2�2 � �27 � 8�2 � �4 � 3�2 � �64 � 27�2
d � �4 � 0�2 � �64 � 0�2 64.125
�0, 4�y � x3,
28.
(a)
(b)
(c)
(d) 160.287
s � �4
0
1 � �4x�x2 � 4��2 dx 159.087
160.151
d � �1 � 0�2 � �9 � 16�2 � �2 � 1�2 � �0 � 9�2 � �3 � 2�2 � �25 � 0�2 � �4 � 3�2 � �144 � 25�2
d � �4 � 0�2 � �144 � 16�2 128.062
f �x� � �x2 � 4�2, �0, 4�
Section 7.4 Arc Length and Surfaces of Revolution 49
29. (a)
(c)
Divide into two intervals.
�127
�403�2 � 133�2 � 16� � 10.5131
s1 � s2 �1
27�403�2 � 8 � 8 � 133�2�
� �1
27�8 � 133�2� � 1.4397
� �1
27�43�2 � 133�2�
� �1
27�9x2�3 � 4�3�2�
0
�1
� �1
18�0
�1�9x2�3 � 4�1�2 6
x1�3 dx
��13 �0
�1
�9x2�3 � 41
x1�3 dx, �x ≤ 0�
s1 � �0
�1�9x2�3 � 4
9x2�3 dx��1, 0�:
��1, 8�
1 � f ��x�2 � 1 �4
9x2�3 �9x2�3 � 4
9x2�3
f ��x� �23
x�1�3
10
−2
−2
10
f �x� � x2�3 (b) No, is not defined.
�127
�403�2 � 8� � 9.0734
�1
27�403�2 � 43�2�
�1
27�9x2�3 � 4�3�2�
8
0
�13�
8
0
�9x2�3 � 41
x1�3 dx, �x ≥ 0�
s2 � �8
0�9x2�3 � 4
9x2�3 dx�0, 8�:
f ��0�
30.
Total length: s � 4�12� � 48
� 2�8
0x�2�3 dx � 6x1�3�
8
0� 12
14
s � �8
0� 4
x2�3 dx
1 � �y� �2 � 1 �4 � x2�3
x2�3 �4
x2�3
y� �32
�4 � x2�3�1�2�23
x�1�3 ���4 � x2�3�1�2
x1�3
y � �4 � x2�3�3�2, 0 ≤ x ≤ 8
y2�3 � 4 � x2�3
x2�3 � y2�3 � 4 31. (a)
(b)
(c)
L4 � �4
0�1 �
25256
x3 dx � 6.063y4� �5
16x 3�2,
L3 � �4
0�1 �
x2
4 dx � 5.916y3� �
12
x,
L2 � �4
0�1 �
9x16
dx � 5.759y2� �34
x1�2,
L1 � �4
0
�2 dx � 5.657y1� � 1,
y1, y2, y3, y4
x−1 1 2 3 4 5
−1
1
2
3
4
5
y
y1
y2
y3
y4
50 Chapter 7 Applications of Integration
32. Let and
Equivalently, and
Numerically, both integrals yield L � 2.0035.
L2 � �1
0
1 � e2y dy � �1
0
1 � e2x dx.dxdy
� ey,0 ≤ y ≤ 1,x � ey,
L1 � �e
11 �
1x2 dx.y� �
1x
1 ≤ x ≤ e,y � ln x,
33.
When Thus, the fleeing object has traveledunits when it is caught.
The pursuer has traveled twice the distance that the fleeing object has traveled when it is caught.
� 12 �
23
x3�2 � 2x1�2�1
0�
43
� 2�23�
s � �1
0 x � 12x1�2 dx �
12
�1
0 �x1�2 � x�1�2� dx
1 � �y� �2 � 1 ��x � 1�2
4x�
�x � 1�2
4x
y� �13 �
32
x1�2 �32
x�1�2� � �12�
x � 1x1�2
23
y �23.x � 0,
y �13
�x3�2 � 3x1�2 � 2� 34.
Thus, there are square feet of roofing on the barn.
100�47� � 4700
� 20�e �1e� 47 ft
� �10�ex�20 � e�x�20��20
�20
�12
�20
�20 �ex�20 � e�x�20� dx
s � �20
�20� 1
2�ex�20 � e�x�20��
2
dx
� � 12
�ex�20 � e�x�20��2
1 � �y� �2 � 1 �14
�ex�10 � 2 � e�x�10�
y� � �12
�ex�20 � e�x�20�
y � 31 � 10�ex�20 � e�x�20�
35.
� 2�20� sinh x
20�20
0� 40 sinh�1� 47.008 m
L � �20
�20 cosh
x20
dx � 2�20
0 cosh
x20
dx
1 � �y� �2 � 1 � sinh2 x
20� cosh2
x20
y� � sinh x
20
y � 20 cosh x
20, �20 ≤ x ≤ 20 36.
(Use Simpson’s Rule with or a graphing utility.)n � 100
1480
s � �299.2239
�299.2239
1 � ��0.6899619478 sinh 0.0100333x�2 dx
y� � �0.6899619478 sinh 0.0100333x
y � 693.8597 � 68.7672 cosh 0.0100333x
37.
� 3 arcsin 23
2.1892
� �3 arcsin x3�
2
0� 3�arcsin
23
� arcsin 0�
s � �2
0 9
9 � x2 dx � �2
0
39 � x2
dx
1 � �y� �2 �9
9 � x2
y� ��x
9 � x2
y � 9 � x2 38.
14
�2��5�� 7.8540 � s
7.8540
� �5 arcsin x5�
4
�3� 5�arcsin
45
� arcsin��35��
s � �4
�3 25
25 � x2 dx � �4
�3
525 � x2
dx
1 � �y� �2 �25
25 � x2
y� ��x
25 � x2
y � 25 � x2
Section 7.4 Arc Length and Surfaces of Revolution 51
39.
��
9�8282 � 1� 258.85
� ��
9�1 � x4�3�2�
3
0
��
6 �3
0 �1 � x4�1�2�4x3� dx
S � 2� �3
0 x3
31 � x4 dx
y� � x2, �0, 3�
y �x3
340.
�8�
3�103�2 � 53�2� 171.258
�83
��x � 1�3�2�9
4
� 4��9
4
x � 1 dx
S � 2��9
42x1 �
1x dx
y� �1x
, �4, 9�
y � 2x
41.
� 2� � x6
72�
x2
6�
18x2�
2
1�
47�
16
� 2� �2
1 �x5
12�
x3
�1
4x3� dx
S � 2� �2
1 �x3
6�
12x��
x2
2�
12x2� dx
1 � �y� �2 � �x2
2�
12x2�
2
, �1, 2�
y� �x2
2�
12x2
y �x3
6�
12x
42.
� �2�58
x2�6
0� 95�
S � 2� �6
0 x25
4 dx
1 � �y� �2 �54
, �0, 6�
y� �12
y �x2
43.
��
27�145145 � 1010 � 199.48
� � �
27�9x4�3 � 1�3�2�
8
1
��
18 �8
1 �9x4�3 � 1�1�2�12x1�3� dx
�2�
3 �8
1 x1�39x4�3 � 1 dx
S � 2� �8
1 x1 �
19x4�3 dx
y� �1
3x2�3, �1, 8�
y � 3x � 2 44.
��
6�373�2 � 1� 117.319
� ��
6�1 � 4x2�3�2�
3
0
��
4�3
0�1 � 4x2�1�2 �8x� dx
S � 2��3
0x1 � 4x2 dx
y� � �2x
y � 9 � x2, �0, 3�
45.
14.4236
S � 2� ��
0 sin x1 � cos2 x dx
y� � cos x, �0, ��
y � sin x 46.
22.943
S � 2� �e
1 xx2 � 1
x2 dx � 2��e
1
x2 � 1 dx
1 � �y� �2 �x2 � 1
x2 , �1, e�
y� �1x
y � ln x
52 Chapter 7 Applications of Integration
47. A rectifiable curve is one that has a finite arc length. 48. The precalculus formula is the distance formula betweentwo points. The representative element is
��xi �2 � ��yi�2 �1 � ��yi
�xi�2
�xi.
49. The precalculus formula is the surface area formula forthe lateral surface of the frustum of a right circular cone.The representative element is
2� f �di ��xi2 � �yi
2 � 2� f �di �1 � ��yi
�xi�
2
�xi.
50. The surface of revolution given by will be larger. is larger for f1.r�x�
f1
51.
� �2�r2 � h2
r �x2
2 ��r
0� �rr2 � h2
S � 2� �r
0 xr2 � h2
r2 dx
1 � �y� �2 �r2 � h2
r2
y� �hr
y �hxr
52.
� 2� �r
�r
r dx � �2� rx�r
�r� 4� r2
S � 2� �r
�r
r2 � x2 r2
r2 � x2 dx
1 � �y� �2 �r2
r2 � x2
y� ��x
r2 � x2
y � r2 � x2
53.
See figure in Exercise 54.
� 6� �3 � 5 � 14.40
� ��6�9 � x2�2
0
� �3� �2
0
�2x9 � x2
dx
S � 2� �2
0
3x9 � x2
dx
1 � �y� �2 �3
9 � x2
y� ��x
9 � x2
y � 9 � x2 54. From Exercise 53 we have:
� 2�rh �where h is the height of the zone�
� 2r� �r � r2 � a2 � � 2r2� � 2r�r2 � a2
� ��2r�r2 � x2�a
0
� �r� �a
0
�2x dxr2 � x2
S � 2� �a
0
rxr2 � x2
dx
x
x y r2 2 2+ =
r
r h−
h
y
55.
Amount of glass needed: V ��
27 �0.015
12 � 0.00015 ft3 0.25 in.3
��
3 �1�3
0 � 1
3� 2x � 9x2� dx �
�
3 � 13
x � x2 � 3x3�1�3
0�
�
27 ft 2 0.1164 ft2 16.8 in.2
S � 2� �1�3
0 � 1
3x1�2 � x3�2� 1
36�x�1�2 � 9x1�2�2 dx �
2�
6 �1�3
0 �1
3x1�2 � x3�2��x�1�2 � 9x1�2� dx
1 � �y� �2 � 1 �1
36�x�1 � 18 � 81x� �
136
�x�1�2 � 9x1�2�2
y� �16
x�1�2 �32
x1�2 �16
�x�1�2 � 9x1�2�
y �13
x1�2 � x3�2
Section 7.4 Arc Length and Surfaces of Revolution 53
56. (a)
Ellipse:
y2 � �21 �x2
9
y1 � 21 �x2
9−5 5
−4 9x2 2y
4+ = 1
4x2
9�
y2
4� 1 (b)
L � �3
01 �
4x2
81 � 9x2 dx
��2x
91 � �x2�9��
�2x
39 � x2
y� � 2�12��1 �
x2
9 ��1�2
��2x9 �
y � 21 �x2
9, 0 ≤ x ≤ 3
(c) You cannot evaluate this definite integral, since the integrand is not defined at Simpson’s Rule will not work for the same reason. Also, the integrand does not have an elementary antiderivative.
x � 3.
57. (a) We approximate the volume by summing six disks of thickness 3 and circumference equal to the average of the given circumferences:
(b) The lateral surface area of a frustum of a right circular cone is For the first frustum:
Adding the six frustums together:
� 1168.64
224.30 � 208.96 � 208.54 � 202.06 � 174.41 � 150.37
�58 � 512 ��9 � � 7
2��2
�1�2
� �51 � 482 ��9 � � 3
2��2
�1�2
�70 � 662 ��9 � � 4
2��2
�1�2
� �66 � 582 ��9 � � 8
2��2
�1�2
�
S �50 � 65.52 ��9 � �15.5
2� �2
�1�2
� �65.5 � 702 ��9 � �4.5
2��2
�1�2
�
� �50 � 65.52 ��9 � �65.5 � 50
2� �2
�1�2
.
r
R
sh
S1 ��32 � �65.5 � 502� �
2
�1�2
� 502�
�65.52� �
�s�R � r�.
�3
4��21813.625� � 5207.62 cubic inches
�3
4��57.752 � 67.752 � 682 � 622 � 54.52 � 49.52�
�3
4� ��50 � 65.52 �
2
� �65.5 � 702 �
2
� �70 � 662 �
2
� �66 � 582 �
2
� �58 � 512 �
2
� �51 � 482 �
2
�
V �6
i�1� ri
2�3� � �6
i�1�� Ci
2��2
�3� �3
4� �
6
i�1Ci
2
Ci
(c)
−1
−1 19
20
r � 0.00401y3 � 0.1416y2 � 1.232y � 7.943 (d)
1179.5 square inches
S � �18
0 2�r�y�1 � r��y�2 dy
V � �18
0�r2 dy 5275.9 cubic inches
54 Chapter 7 Applications of Integration
58. (a)
(b)
(Answers will vary.)
(c)
(Answers will vary.)
L � �400
0
1 � f ��x�2 dx 794.9 feet
Area � �400
0f �x� dx 131,734.5 square feet 3.0 acres �1 acre � 43,560 square feet�
y � f �x� � 0.0000001953x4 � 0.0001804x3 � 0.0496x2 � 4.8323x � 536.9270
59. (a)
(b)
� 2� �b
1 x4 � 1
x3 dx
� 2� �b
1 1x1 �
1x4 dx
S � 2� �b
1 1x1 � �� 1
x2�2
dx
x1 b
2
1
y = x1
y
V � � �b
1 1x2 dx � ���
x�b
1� ��1 �
1b� (c)
(d) Since
we have
and Thus,
limb→� 2��b
1 x4 � 1
x3 dx � �.
limb→� ln b → �.
�b
1 x4 � 1
x3 dx > �b
1 1x dx � �ln x�
b
1� ln b
x4 � 1x3 >
x4
x3 �1x > 0 on �1, b�
limb→� V � lim
b→� � �1 �1b� � �
60. (a) Area of circle with radius
Area of sector with central angle (in radians):
S �
2�A �
2���L2� �
12
L2
L: A � �L2 (b) Let s be the arc length of the sector, which is the circumference of the base of the cone. Here,
and you have
S �12
L2 �12
L2� sL� �
12
Ls �12
L �2�r� � �rL.
s � L � 2�r,
(c) The lateral surface area of the frustum is the difference of the large cone and the small one.
By similar triangles, Hence,
� �L�r1 � r2 �.
S � �r2L � �L1�r2 � r1� � �r2L � �Lr1
L � L1
r2�
L1
r1 ⇒ Lr1 � L1�r2 � r1�.
� �r2L � �L1�r2 � r1�
S � �r2�L � L1� � �r1L1L
L
r
r
1
12
61. Individual project 62. Essay
Section 7.4 Arc Length and Surfaces of Revolution 55
63.
[Surface area of portion above the -axis]x
� ��12�
5�4 � x2�3�5�2�
8
0�
192�
5
� 4��8
0 �4 � x2�3�3�2
x1�3 dx
S � 2��8
0�4 � x2�3�3�2 4
x2�3 dx
1 � �y� �2 � 1 �4 � x2�3
x2�3 �4
x2�3
y� �32
�4 � x2�3�1�2��23
x�1�3� ���4 � x2�3�1�2
x1�3
y � �4 � x2�3�3�2, 0 ≤ x ≤ 8
y2�3 � 4 � x2�3
x2�3 � y2�3 � 4 64.
��
12�64 � 64 � 64� �
16�
3
��
12�16x � 4x2 � x3�4
0
��
12 �4
0�16 � 8x � 3x2� dx
� 2��4
0 �4 � x��4 � 3x�
24 dx
S � 2��4
0 �4 � x�x
12�
�4 � 3x�48x
dx
�48x � 16 � 24x � 9x2
48x�
�4 � 3x�2
48x, x 0
1 � �y� �2 � 1 ��4 � 3x�2
48x
y� ��4 � 3x�3
12x
y ��4 � x�x
12
y2 �1
12x�4 � x�2, 0 ≤ x ≤ 4
65.
By symmetry,
w−w
h(w, h)
x
y
C � 2�w
0 1 �
4h2
w 4 x2 dx.
h � kw2 ⇒ k �h
w2 ⇒ 1 � �y� � � 1 �4h2
w 4 x2
1 � �y� �2 � 1 � 4k2x2
y � kx2, y� � 2kx 66.
1444.5 meters
� 2�700
01 �
4�155�2x2
7004 dx
C � 2�w
01 �
4h2
w 4 x2 dx
67. Let be the point on the graph of where the tangent linemakes an angle of with the axis.
L � �4�9
01 �
94 x dx �
827�22 � 1�
x0 �49
y� �32 x1�2 � 1
y � x3�2
x-45�
45�
(x0 , y0)
(0, 0)
y2 = x3
x
yy2 � x3�x0, y0�
Section 7.5 Work
56 Chapter 7 Applications of Integration
1.
� 1000 ft � lb
W � Fd � �100��10� 2.
� 11,200 ft � lb
W � Fd � �2800��4� 3.
� 448 joules (newton-meters)
W � Fd � �112��4�
6. is the work done by a force F moving an
object along a straight line from to x � b.x � a
W � �b
a
F�x� dx 7. Since the work equals the area under the force function,you have �c� < �d� < �a� < �b�.
4. W � Fd � �9�2000��� 12�5280�� � 47,520,000 ft � lb 5. Work equals force times distance, W � FD.
8. (a)
(b)
(c)
(d) W � �9
0 �x dx �
23
x3�2�9
0�
23
�27� � 18 ft � lbs
W � �9
0
127
x2 dx �x3
81�9
0� 9 ft � lbs
� 160 ft � lbs
W � �7
0 20 dx � �9
7 ��10x � 90� dx � 140 � 20
W � �9
0 6 dx � 54 ft � lbs 9.
� 30.625 in. � lb 2.55 ft � lb
�245
8 in. � lb
W � �7
0 54
x dx � 58
x2�7
0
k �54
5 � k�4�
F�x� � kx
10.
� 35 in. � lb
W � �9
5 54
x dx � 58
x2�9
511.
� 87.5 joules or Nm
� 8750 n � cm
� �50
20 253
x dx �25x2
6 �50
20
W � �50
20 F�x� dx
250 � k�30� ⇒ k �253
F�x� � kx 12.
� 28,000 n � cm � 280 Nm
� �70
0 807
x dx �40x2
7 �70
0
W � �70
0 F�x� dx
800 � k�70� ⇒ k �807
F�x� � kx
13.
W � �12
0 209
x dx � 109
x2�12
0� 160 in. � lb �
403
ft � lb
k �209
20 � k�9�
F�x� � kx 14.
W � 2 �4
0 15x dx � 15x2�
4
0� 240 ft � lb
15 � k�1� � k
F�x� � kx
15.
Note: 4 inches foot��13�
W � �7�12
1�3 324x dx � 162x2�
7�12
1�3� 37.125 ft � lbs
W � 18 � �1�3
0 kx dx �
kx2
2 �1�3
0�
k18
⇒ k � 324 16.
W � �5�24
1�6 540x dx � 270x2�
5�24
1�6� 4.21875 ft � lbs
W � 7.5 � �1�6
0 kx dx �
kx2
2 �1�6
0�
k72
⇒ k � 540
Section 7.5 Work 57
17. Assume that Earth has a radius of 4000 miles.
F�x� �80,000,000
x 2
k � 80,000,000
5 �k
�4000�2
F�x� �kx2 (a)
(b)
1395.3 mi � ton 1.47 � 1010 ft � ton
W � �4300
4000 80,000,000
x2 dx
487.8 mi � tons 5.15 � 109 ft � lb
W � �4100
4000 80,000,000
x2 dx � �80,000,000x �
4100
4000
18.
limh→�
W � 20,000 mi�ton 2.1 � 1011 ft � lb
W � �h
4000
80,000,000x2 dx � �
80,000,000x �
h
4000�
�80,000,000h
� 20,000
19. Assume that Earth has a radius of 4000 miles.
F�x� �160,000,000
x2
k � 160,000,000
10 �k
�4000�2
F�x� �kx2 (a)
(b)
3.57 � 1011 ft � lb
3.38 � 104 mi � ton
� 33,846.154 mi � ton
W � �26,000
4000 160,000,000
x2 dx � �160,000,000x �
26,000
4000 �6,153.846 � 40,000
3.10 � 1011 ft � lb
2.93 � 104 mi � ton
� 29,333.333 mi � ton
W � �15,000
4000 160,000,000
x2 dx � �160,000,000x �
15,000
4000 �10,666.667 � 40,000
20. Weight on surface of moon:
Weight varies inversely as the square of distance from the center of the moon. Therefore:
95.652 mi � ton 1.01 � 109 ft � lb
W � �1150
1100 2.42 � 106
x2 dx � �2.42 � 106
x �1150
1100� 2.42 � 106� 1
1100�
11150�
k � 2.42 � 106
2 �k
�1100�2
F�x� �kx2
16�12� � 2 tons
21. Weight of each layer:
Distance:
(a)
(b) W � �4
0 62.4�20��4 � y� dy � 4992y � 624y2�
4
0� 9984 ft � lb
W � �4
2 62.4�20��4 � y� dy � 4992y � 624y2�
4
2� 2496 ft � lb
4 � y
x1 2 3 4 5 6
6
5
4
3
2
1
4 − y
y62.4�20� �y
58 Chapter 7 Applications of Integration
22. The bottom half had to be pumped a greater distance than the top half.
23. Volume of disk:
Weight of disk of water:
Distance the disk of water is moved:
� 39,200��12� � 470,400� newton–meters
� 39,200�5y �y2
2 �4
0
W � �4
0�5 � y��9800�4� dy � 39,200��4
0�5 � y� dy
5 � y
9800�4�� �y
� �2�2 �y � 4� �y 24. Volume of disk:
Weight of disk:
Distance the disk of water is moved: y
� 862,400� newton–meters
� 39,200� �22�
W � �12
10y�9800��4�� dy � 39,200�y2
2 �12
10
9800�4�� �y
4� �y
26. Volume of disk:
Weight of disk:
Distance:
(a)
(b)
� 49
�62.4���14
y4��6
4 7210.7� ft � lb
W �49
�62.4�� �6
4 y3 dy
� 49
�62.4���14
y4��2
0 110.9� ft � lb
W �49
�62.4�� �2
0 y3 dy
y
62.4��23
y�2
�y
x
y
4321−1−2−3−4
7
5
4
3
2
y�� 23
y�2
�y
27. Volume of disk:
Weight of disk:
Distance:
x
4
8
10
642−2−2
−6 −4
y
y
� 20,217.6� ft � lb
� 62.4� �6
0 �36y � y3� dy � 62.4� 18y2 �
14
y4�6
0
W � 62.4� �6
0 y�36 � y2� dy
y
62.4��36 � y2� �y
� ��36 � y2 �2 �y
25. Volume of disk:
Weight of disk:
Distance:
� 2995.2� ft � lb
�49
�62.4��2y3 �14
y4�6
0
W �4�62.4��
9 �6
0 �6 � y�y2 dy
6 � y
62.4��23
y�2
�y
x
6 − y
4321−1−2−3−4
7
5
4
3
2
y� �23
y�2
�y
28. Volume of each layer:
Weight of each layer:
Distance:
x1 2 3 4
4
3
2
1
(2, 3)
y x= 3 3−
y � 3106.35 ft � lb
� 53.11172 �
� 53.118y �3y2
2�
y3
3 �3
0
� 53.1�3
0�18 � 3y � y2� dy
W � �3
053.1�6 � y�� y � 3� dy
6 � y
53.1� y � 3� �y
y � 33
�3� �y � � y � 3� �y
Section 7.5 Work 59
29. Volume of layer:
Weight of layer:
Distance:
The second integral is zero since the integrand is odd and the limits of integration are symmetric to the origin. The first integral represents the area of a semicircle of radius Thus, the work is
W � 336�132 ���3
2�2
�12� � 2457� ft � lb.
32.
� 336132
�1.5
�1.5 �9
4� y2 dy � �1.5
�1.5 �9
4� y2 y dy� W � �1.5
�1.5 42�8��9
4� y2 �13
2� y� dy
132
� y
W � 42�8���9�4� � y2 �y
x
4
2
6
8
642−2−2
−4
−6 −4
132
− y
Tractor
yV � lwh � 4�2���9�4� � y2 �y
30. Volume of layer:
Weight of layer:
Distance:
The second integral is zero since the integrand is odd and the limits of integration are symmetric to the origin. The first integral represents the area of a semicircle of radius Thus, the work is
W � 1008�192 ���5
2�2
�12� � 29,925� ft � lb 94,012.16 ft � lb.
52.
� 1008192
�2.5
�2.5 �25
4� y2 dy � �2.5
�2.5 �25
4� y2 ��y� dy�
W � �2.5
�2.5 42�24��25
4� y2 �19
2� y� dy
192
� y
W � 42�24���25�4� � y2 �y
x
6
3
12
963−3−3
−6
−9 −6
192
− y
Ground
level
yV � 12�2���25�4� � y2 �y
31. Weight of section of chain:
Distance:
� 337.5 ft � lb
� �32
�15 � y�2�15
0
W � 3 �15
0 �15 � y� dy
15 � y
3 �y 32. The lower 10 feet of chain are raised 5 feet with a constant force.
The top 5 feet will be raised with variable force.
Weight of section:
Distance:
W � W1 � W2 � 150 �752
�3752
ft � lb
W2 � 3 �5
0 �5 � y� dy � �3
2�5 � y�2�
5
0�
752
ft � lb
5 � y
3 �y
W1 � 3�10�5 � 150 ft � lb
33. The lower 5 feet of chain are raised 10 feet with a constant force.
The top 10 feet of chain are raised with a variable force.
Weight per section:
Distance:
W � W1 � W2 � 300 ft � lb
� �32
�10 � y�2�10
0� 150 ft � lb W2 � 3 �10
0 �10 � y� dy
10 � y
3 �y
W1 � 3�5��10� � 150 ft � lb
34. The work required to lift the chain is (fromExercise 31). The work required to lift the 500-poundload is The work required to liftthe chain with a 100-pound load attached is
W � 337.5 � 7500 � 7837.5 ft � lbs.
W � �500��15� � 7500.
337.5 ft � lb
60 Chapter 7 Applications of Integration
35. Weight of section of chain:
Distance:
�34
�15�2 � 168.75 ft � lb
W � 3 �7.5
0 �15 � 2y� dy � �3
4�15 � 2y�2�
7.5
0
15 � 2y
3 �y 36.
�34
�12�2 � 108 ft � lb
W � 3 �6
0 �12 � 2y� dy � �3
4�12 � 2y�2�
6
0
37. Work to pull up the ball:
Work to wind up the top 15 feet of cable: force is variable
Weight per section:
Distance:
Work to lift the lower 25 feet of cable with a constantforce:
� 7987.5 ft � lb
W � W1 � W2 � W3 � 7500 � 112.5 � 375
W3 � �1��25��15� � 375 ft � lb
� 112.5 ft � lb
W2 � �15
0 �15 � x� dx � �1
2�15 � x�2�
15
0
15 � x
1 �y
W1 � 500�15� � 7500 ft � lb 38. Work to pull up the ball:
Work to pull up the cable: force is variable
Weight per section:
Distance:
W � W1 � W2 � 20,000 � 800 � 20,800 ft � lb
� 800 ft � lb
W2 � �40
0 �40 � x� dx � �1
2�40 � x�2�
40
0
40 � x
1 �y
W1 � 500�40� � 20,000 ft � lb
39.
� 2000 ln�32� 810.93 ft � lb
� 2000 ln V �3
2
W � �3
2 2000
V dV
k � 2000
1000 �k2
p �kV
40.
2746.53 ft � lb
� 2500 ln 3
� 2500 ln V�3
1
W � �3
1
2500V
dV
2500 �k1
⇒ k � 2500
p �kV
41.
�3k4
�units of work�
� k2 � x�
1
�2� k�1 �
14�
W � �1
�2
k�2 � x�2 dx
F�x� �k
�2 � x�2
42. (a)
(b)
(c)
(d) when is a maximum when feet.
(e) W � �2
0 F�x� dx 25,180.5 ft � lbs
x 0.524F�x�x 0.524 feet.F�x� � 0
0 20
25,000
F�x� � �16,261.36x4 � 85,295.45x3 � 157,738.64x2 � 104,386.36x � 32.4675
24,88.889 ft � lbW 2 � 03�6� �0 � 4�20,000� � 2�22,000� � 4�15,000� � 2�10,000� � 4�5000� � 0�
W � FD � �8000���2� � 16,000� ft � lbs
Section 7.6 Moments, Centers of Mass, and Centroids 61
43. W � �5
0 1000�1.8 � ln�x � 1�� dx � 3249.44 ft � lb 44. W � �4
0 �ex2
� 1100 � dx � 11,494 ft � lb
45. W � �5
0 100x125 � x3 dx � 10,330.3 ft � lb 46. W � �2
0 1000 sinh x dx � 2762.2 ft � lb
Section 7.6 Moments, Centers of Mass, and Centroids
1. x �6��5� � 3�1� � 5�3�
6 � 3 � 5� �
67
2. x �7��3� � 4��2� � 3�5� � 8�6�
7 � 4 � 3 � 8�
1711
3. x �1�7� � 1�8� � 1�12� � 1�15� � 1�18�
1 � 1 � 1 � 1 � 1� 12 4. x �
12��6� � 1��4� � 6��2� � 3�0� � 11�8�12 � 1 � 6 � 3 � 11
� 0
5. (a)
(b) x �12��6 � 3� � 1��4 � 3� � 6��2 � 3� � 3�0 � 3� � 11�8 � 3�
12 � 1 � 6 � 3 � 11�
�9933
� �3
x ��7 � 5� � �8 � 5� � �12 � 5� � �15 � 5� � �18 � 5�
5� 17 � 12 � 5
6. The center of mass is translated k units as well.
7.
x � 6 feet
125x � 750
50x � 750 � 75x
50x � 75�L � x� � 75�10 � x� 8. (Person on left)
x � 323 feet
750x � 2750
200x � 2750 � 550x
200x � 550�5 � x�
9.
x
2
1
321−1−1
−2
−3
−4
−3 −2
m1
m2
m3
(2, 2)
( 3, 1)−
(1, 4)−
y �x, y � � �109
, �19�
y �5�2� � 1�1� � 3��4�
5 � 1 � 3� �
19
x �5�2� � 1��3� � 3�1�
5 � 1 � 3�
109
10.
xm1
m2
m3
(1, 1)−
(5, 5)
( 4, 0)−
−4 −2−2
4 6
8
6
4
2
y �x, y � � �0, 0�
y �10��1� � 2�5� � 5�0�
10 � 2 � 5� 0
x �10�1� � 2�5� � 5��4�
10 � 2 � 5� 0
11.
�x, y � � �58
, 1316�
y �3��3� � 4�5� � 2�1� � 1�0� � 6�0�
3 � 4 � 2 � 1 � 6�
1316 x
642
6
8−2−2
−4
m3m4
m1
m2
m5 (7, 1)(0, 0)
( 2, 3)− −
(5, 5)
( 3, 0)−
y
x �3��2� � 4�5� � 2�7� � 1�0� � 6��3�
3 � 4 � 2 � 1 � 6�
58
62 Chapter 7 Applications of Integration
12.
�x, y� � �6227
, 6427�
y �12�3� � 6�5� � �152��8� � 15��2�
12 � 6 � �152� � 15�
9640.5
�6427
x
m1
m4
m3
m2
(2, 3)
(2, 2)−
(6, 8)
( 1, 5)−
−2−2
2 6 8
8
6
2
y x �12�2� � 6��1� � �152��6� � 15�2�
12 � 6 � �152� � 15�
9340.5
�6227
13.
�x, y � � �125
, 34�
x �My
m�
64�
5 � 316�� �
125
My � � �4
0 xx dx � ��
25
x52�4
0�
64�
5
y �Mx
m� 4�� 3
16�� �34
Mx � � �4
0 x2
�x � dx � ��x2
4 �4
0� 4�
x1 2 3 4
4
3
2
1( , )x y
y m � � �4
0 x dx � �2�
3x32�
4
0�
16�
3
14.
�x, y � � �32
, 35�
x �My
m�
2�
�43�� �3
2
My � ��2
0x�1
2x2� dx �
12
��2
0x3 dx � ��
8x 4�
2
0� 2�
y �Mx
m�
�45���43�� �
3
5
Mx � ��2
0 12 �
12
x2��12
x2� dx ��
8�2
0x 4 dx � � �
40x5�
2
0�
3240
� �45
�
y
x1
1
2
2
32
35
,( )
m � ��2
0 12
x2 dx � ��x3
6 �2
0�
43
�
15.
�x, y � � �35
, 1235�
x �My
m�
�
20 �12� � �
35
My � � �1
0 x�x2 � x3� dx � � �1
0 �x3 � x4� dx � ��x4
4�
x5
5 �1
0�
�
20
y �Mx
m�
�
35 �12� � �
1235
Mx � � �1
0 �x2 � x3�
2�x2 � x3� dx �
�
2 �1
0 �x4 � x6� dx �
�
2�x5
5�
x7
7�1
0�
�
35
x1
1
14
14
12
34
34
12
( , )x y
y
(1, 1)
m � � �1
0 �x2 � x3� dx � ��x3
3�
x4
4 �1
0�
�
12
Section 7.6 Moments, Centers of Mass, and Centroids 63
16.
�x, y � � �25
, 12�
x �My
m�
�
15 �6�� �
25
My � � �1
0 x�x � x� dx � � �1
0 �x32 � x2� dx � ��2
5x52 �
x3
3 �1
0�
�
15
y �Mx
m�
�
12 �6�� �
12
Mx � � �1
0 �x � x�
2�x � x� dx �
�
2 �1
0 �x � x2� dx �
�
2 �x2
2�
x3
3 �1
0�
�
12
x1
1
14
14
12
34
34
12
( , )x y
y m � � �1
0 �x � x� dx � ��2
3x32 �
x2
2�1
0�
�
6
17.
�x, y � � �32
, 225 �
x �My
m�
27�
4 � 29�� �
32
���x4
4� x3�
3
0�
27�
4 My � � �3
0 x ���x2 � 4x � 2� � �x � 2�� dx � � �3
0 ��x3 � 3x2� dx �
y �Mx
m�
99�
5 � 29�� �
225
��
2 �x5
5� 2x4 �
11x3
3� 6x2�
3
0�
99�
5
��
2 �3
0 ��x2 � 5x � 4���x2 � 3x� dx �
�
2 �3
0 �x4 � 8x3 � 11x2 � 12x� dx
Mx � � �3
0 ���x2 � 4x � 2� � �x � 2�
2 ����x2 � 4x � 2� � �x � 2�� dx
x−1 1 2 3 4
1
2
3
4
5
6
5
( , )x y
y
(3, 5)
m � � �3
0 ���x2 � 4x � 2� � �x � 2�� dx � ���x3
3�
3x2
2 �3
0�
9�
2
18.
�x, y� � �185
, 52�
2−2−1
1
2
3
4
5
4 6 8 10
y
x
185
52
,( )
(9, 4)
(0, 1)
x �My
m�
�815���92�� �
18
5; y �
Mx
m�
�454���92�� �
5
2
My � ��9
0x�x � 1 �
13
x � 1� dx � ��9
0 �x32 �
13
x2� dx � ��25
x52 �19
x3�9
0 � ��486
5� 81� �
815
�
��
2�x2
6�
x3
27�
43
x32�9
0�
�
2 �272
� 27 � 36� �454
�
��
2�9
0 �x �
13
x32 �13
x32 �19
x2 � 2x �23
x� dx ��
2�9
0 �1
3x �
19
x2 � 2x� dx
Mx � ��9
0 x � 1 � �13�x � 1
2 �x � 1 �13
x � 1� dx ��
2�9
0 �x �
13
x � 2��x �13
x� dx
� ��23
x32 �x2
6 �9
0� ��18 �
272 � �
92
� m � ��9
0 ��x � 1� � �1
3x � 1�� dx � ��9
0 �x �
13
x� dx
64 Chapter 7 Applications of Integration
19.
�x, y � � �5, 107 �
x �My
m� 96�� 5
96�� � 5
My � � �8
0 x�x23� dx � ��3
8x83�
8
0� 96�
y �Mx
m�
192�
7 � 596�� �
107
Mx � � �8
0 x23
2�x23� dx �
�
2�37
x73�8
0�
192�
7
x
2
−2
4
6
4 6 82
( , )x y
y m � � �8
0 x 23 dx � ��3
5x53�
8
0�
96�
5
20.
By symmetry, and
�x, y � � �0, 207 �
y �512�
7 � 5128�� �
207
Mx � 2� �8
0 �4 � x23
2 ��4 � x23� dx � ��16x �37
x73�8
0�
512�
7
x � 0.My
x
−4
−4−8
12
4
8
8
( , )x y
ym � 2� �8
0 �4 � x23� dx � 2��4x �
35
x53�8
0�
128�
5
21.
By symmetry, and
�x, y � � �85
, 0�y � 0.Mx
x �My
m�
256�
15 � 332�� �
85
My � 2� �2
0 �4 � y2
2 ��4 � y2� dy � ��16y �83
y3 �y5
5 �2
0�
256�
15x
−2
−1
1
1
32
2
( , )x y
y m � 2� �2
0 �4 � y2� dy � 2��4y �
y3
3�2
0�
32�
3
22.
�x, y � � �25
, 1�
y �Mx
m�
4�
3 � 34�� � 1
Mx � � �2
0 y�2y � y2� dy � ��2y3
3�
y4
4 �2
0�
4�
3
x �My
m�
8�
15 �34�� �
25
My � � �2
0 �2y � y2
2 ��2y � y2� dy ��
2�4y3
3� y4 �
y5
5 �2
0�
8�
15
x1 2
1
2
( , )x y
y m � � �2
0 �2y � y2� dy � ��y2 �
y3
3�2
0�
4�
3
Section 7.6 Moments, Centers of Mass, and Centroids 65
23.
�x, y � � ��35
, 32�
y �Mx
m�
27�
4 � 29�� �
32
Mx � � �3
0 y ��2y � y2� � ��y�� dy � � �3
0 �3y2 � y3� dy � ��y3 �
y4
4 �3
0�
27�
4
x �My
m� �
27�
10 � 29�� � �
35
��
2 �3
0 �y4 � 4y3 � 3y2� dy �
�
2�y5
5� y4 � y3�
3
0� �
27�
10
My � � �3
0 ��2y � y2� � ��y��
2��2y � y2� � ��y�� dy �
�
2 �3
0 �y � y2��3y � y2� dy
x
−1
−1−2−3
3
1
1
( , )x y
y
(−3,3) m � � �3
0 ��2y � y2� � ��y�� dy � ��3y2
2�
y3
3�3
0�
9�
2
24.
�x, y � � �85
, 12�
y �Mx
m�
9�
4 � 29�� �
12
� � �2
�1 �2y � y2 � y3� dy � ��y2 �
y3
3�
y4
4 �2
�1�
9�
4
Mx � � �2
�1 y ��y � 2� � y2� dy
x �My
m�
36�
5 � 29�� �
85
��
2 �2
�1 ��y � 2�2 � y4� dy �
�
2��y � 2�3
3�
y5
5 �2
�1�
36�
5
My � � �2
�1 ��y � 2� � y2�
2��y � 2� � y2� dy
1
1
−1
2
2
3
3 4x
( , )x y
y m � � �2
�1 ��y � 2� � y2� dy � ��y2
2� 2y �
y3
3 �2
�1�
9�
2
25.
My � �1
0 �x2 � x3� dx � �x3
3�
x4
4 �1
0� �1
3�
14� �
112
Mx �12
�1
0 �x2 � x4� dx �
12�
x3
3�
x5
5 �1
0�
12 �
13
�15� �
115
A � �1
0 �x � x2� dx � �1
2x2 �
x3
3 �1
0�
16
26.
My � �4
1 x�1
x� dx � �x�4
1� 3
Mx �12
�4
1
1x2 dx � �1
2 ��1x��
4
1� ��1
8�
12� �
38
A � �4
1 1x dx � �ln x �
4
1� ln 4
27.
My � �3
0 �2x2 � 4x� dx � �2x3
3� 2x2�
3
0� 18 � 18 � 36
� �2x3
3� 4x2 � 8x�
3
0� 18 � 36 � 24 � 78
Mx �12
�3
0 �2x � 4�2 dx � �3
0 �2x2 � 8x � 8� dx
A � �3
0 �2x � 4� dx � �x2 � 4x�
3
0� 9 � 12 � 21
66 Chapter 7 Applications of Integration
28.
by symmetry.My � 0
� �12�
x5
5�
8x3
3� 16x�
2
�2� ��32
5�
643
� 32� � �25615
Mx �12
�2
�2 �x2 � 4��4 � x2� dx � �
12
�2
�2 �x4 � 8x2 � 16� dx
A � �2
�2 ��x2 � 4� dx � 2 �2
0 �4 � x2� dx � �8x �
2x3
3 �2
0� 16 �
163
�323
29.
Therefore, the centroid is �3.0, 126.0�.
y �Mx
m� 126.0
x �My
m� 3.0
My � � �5
0 10x2125 � x3 dx � �
10�
3 �5
0
125 � x3 ��3x2� dx �12,5005�
9� 3105.6�
� 50� �5
0x2�125 � x3� dx �
3,124,375�
24� 130,208�
Mx � � �5
0 �10x125 � x3
2 ��10x125 � x3 � dx
−50
−1 6
400 m � � �5
0 10x125 � x3 dx � 1033.0�
30.
Therefore, the centroidis �2.2, 0.3�.
y �Mx
m� 0.3
−1
−1
5
3 x �My
m� 2.2
My � � �4
0 x2e�x2 dx � 5.1732�
��
2 �4
0 x2e�x dx � 0.7619�
Mx � � �4
0 �xe�x2
2 ��xe�x2� dx
m � � �4
0 xe�x2 dx � 2.3760� 31.
by symmetry. Therefore,the centroid is �0, 16.2�.x � 0
−25
−5
25
50 y �Mx
m� 16.18
�25�
2 �20
�20 �400 � x2�23 dx � 20064.27
Mx � � �20
�20 5 3400 � x2
2�5 3400 � x2 � dx
m � � �20
�20 5 3400 � x2 dx � 1239.76�
32.
by symmetry. Therefore, the centroid is �0, 0.8�.x � 0
y �Mx
m� 0.8
Mx � � �2
�2 12 �
8x2 � 4��
8x2 � 4� dx � 32� �2
�2
1�x2 � 4�2 dx � 5.14149�
−3 3
−1
3 m � � �2
�2
8x2 � 4
dx � 6.2832�
Section 7.6 Moments, Centers of Mass, and Centroids 67
33.
From elementary geo-metry, is thepoint of intersection ofthe medians.
�b3, c3�
x
( , 0)a( , 0)−a
( , )b c
( , )x y
y = ( + )x acb a+
y = cb a− ( )x a−
y �x, y � � �b3
, c3�
�2c �
y2
2�
y3
3c�c
0�
c3
�1ac
�c
0 y��2a
cy � 2a� dy �
2c �c
0 �y �
y2
c � dy
y �1ac
�c
0 y��b � a
cy � a� � �b � a
cy � a�� dy
�1
2ac �2ab
cy2 �
4ab3c2 y3�
c
0�
12ac �
23
abc� �b3
�1
2ac �c
0 �4ab
cy �
4abc2 y2� dy
x � � 1ac�
12
�c
0 ��b � a
cy � a�
2
� �b � ac
y � a�2
� dy
1A
�1ac
A �12
�2a�c � ac 34.
This is the point of intersection of the diagonals.
x
( , )x y
y = cb
x
cb ( )x a−y =
(0, 0)
( , )b c
( , 0)a
( + , )a b c
y
�x, y � � �b � a2
, c2�
y �1ac
�c
0 y��b
cy � a� � �b
cy�� dy � �1
c y2
2 �c
0�
c2
�1
2ac�abc � a2c� �
12
�b � a�
�1
2ac�abc
y2 � a2y�c
0
�1
2ac �c
0 �2ab
cy � a2� dy
x �1ac
12
�c
0 ��b
cy � a�
2
� �bc
y�2
� dy
1A
�1ac
A � bh � ac
35.
Thus,
The one line passes through and It’s equation is The other line
passes through and It’s equation is is the point of
intersection of these two lines.
�x, y �y �a � 2b
cx � b.�c, a � b�.�0, �b�
y �b � a
2cx �
a2
.�c, b2�.�0, a2�
� x, y � � ��a � 2b�c3�a � b� ,
a2 � ab � b2
3�a � b� �.
�1
3�a � b� �b2 � 2ab � a2 � 3ab � 3a2 � 3a2� �a2 � ab � b2
3�a � b�
x
( , )x y
cy = b a− x a+
(0, 0)
( , 0)c
( , )c b
(0, )a
b
a
y
�1
3c�a � b� ��b2 � 2ab � a2�c � 3ac�b � a� � 3a2c�
�1
c�a � b� ��b � a
c �2
x3
3�
2a�b � a�c
x2
2� a2x�
c
0�
1c�a � b� �
�b � a�2c3
� ac�b � a� � a2c�
y �2
c�a � b� 12
�c
0 �b � a
cx � a�
2
dx �1
c�a � b��c
0 ��b � a
c �2
x2 �2a�b � a�
cx � a2� dx
�2
c�a � b���b � a�c2
3�
ac2
2 � �2
c�a � b��2bc2 � 2ac2 � 3ac2
6 � �c�2b � a�3�a � b� �
�a � 2b�c3�a � b�
2c�a � b� �
b � ac
x3
3�
ax2
2 �c
0 x �
2c�a � b��
c
0 x�b � a
cx � a� dx �
2c�a � b� �
c
0 �b � a
cx2 � ax� dx �
1A
�2
c�a � b�
A �c2
�a � b�
68 Chapter 7 Applications of Integration
36. by symmetry.
r
r
x
y �x, y � � �0, 4r3��
�1
�r2�r2x �x3
3�r
�r�
1�r2 �4r3
3 � �4r3�
y �2
�r2 12
�r
�r
�r2 � x2 �2 dx
1A
�2
�r2
A �12
�r2
x � 0 37. by symmetry.
b
−a ax
y �x, y � � �0, 4b3��
�1
�ab �b2
a2��a2x �x3
3 �a
�a�
b�a3�4a3
3 � �4b3�
y �2
�ab 12
�a
�a
� ba
a2 � x2�2
dx
1A
�2
�ab
A �12
�ab
x � 0
38.
�x, y � � �14
, 710�
�32
�1
0 �1 � 4x2 � 4x3 � x4� dx �
32�x �
43
x3 � x4 �x5
5 �1
0�
710
�32
�1
0 �1 � �2x � x2�2� dx y � 3 �1
0 �1 � �2x � x2��
2�1 � �2x � x2�� dx
x � 3 �1
0 x �1 � �2x � x2�� dx � 3 �1
0 �x � 2x2 � x3� dx � 3�x2
2�
23
x3 �x4
4 �1
0�
14
1A
� 3
A � �1
0 �1 � �2x � x2�� dx �
13
39. (a)
(b) by symmetry.
(c) because is odd.
(d) since there is more area above than below.y �b2
y > b2
bx � x3My � �b
�b
x�b � x2� dx � 0
x � 0
−1−2−3−4−5 1 2 3 4 5x
y
y b=
(e)
y �Mx
A�
4b2b5
4bb3�
35
b
� �bb �bb
3 �2 � 4bb
3
A � �b
�b
�b � x2� dx � �bx �x3
3 �b
�b
� b2b �b2b
5�
4b2b5
� �b
�b
b2 � x4
2 dx �
12�b2x �
x5
5 �b
�b
Mx � �b
�b
�b � x2��b � x2�
2 dx
Section 7.6 Moments, Centers of Mass, and Centroids 69
40. (a) by symmetry.
because is an odd function.
(c)
y �Mx
A�
4n b�4n�1�2n�4n � 1�4n b�24n�1�2n�2n � 1� �
2n � 14n � 1
b
� 2�b � b12n �b�2n�1�2n
2n � 1 � �4n
2n � 1b�2n�1�2n
A � � 2nb
�2nb
�b � x2n� dx � 2 �bx �x2n�1
2n � 1�2nb
0
� b2b12n �b�4n�1�2n
4n � 1�
4n4n � 1
b�4n�1�2n
�12 �b2x �
x4n�1
4n � 1��2nb
�2nb
Mx � �2nb
�2nb
�b � x2n��b � x2n�
2 dx � �2nb
�2nb
12
�b2 � x4n� dx
bx � x2n�1
My � �2nb
�2nb
x�b � x2n� dx � 0
My � 0 (b) because there is more area above
than below.
(d)
(e)
(f) As the figure gets narrower.
x
y b=
y x= 2n
b−2nb
2n
y
n →�,
limn→�
y � limn→�
2n � 14n � 1
b �12
b
y �b2
y > b2
n 1 2 3 4
917 b7
13 b59 b3
5 by
41. (a) by symmetry.
(b) (Use nine data points.)
(c)
�x, y � � �0, 12.85�
y �Mx
A�
23,697.681843.54
� 12.85
y � ��1.02 � 10�5�x 4 � 0.0019x2 � 29.28
�x, y � � �0, 12.98�
y �Mx
A�
72,160355603
�72,1605560
�12.98
Mx � �40
�40 f �x�2
2 dx �
403�4� �302 � 4�29�2 � 2�26�2 � 4�20�2 � 0� �
103
�7216� �72,160
3
A � 2 �40
0 f �x�dx �
2�40�3�4� �30 � 4�29� � 2�26� � 4�20� � 0� �
203
�278� �5560
3
x � 0
42. Let be the top curve, given by The bottom curve is
—CONTINUED—
d�x�.l � d.f �x�
x 0 0.5 1.5 2.0
f 2.0 1.93 1.73 1.32 0
d 0.50 0.48 0.43 0.33 0
1.0
70 Chapter 7 Applications of Integration
42. —CONTINUED—
(a)
�x, y � � �0, 1.078�
y �Mx
A�
4.97974.62
� 1.078
�16
�29.878� � 4.9797
�2
3�4� �3.75 � 4�3.4945� � 2�2.808� � 4�1.6335� � 0�
� �2
0 � f �x�2 � d�x�2� dx
Mx � �2
�2 f �x� � d�x�
2� f �x� � d�x�� dx
�13
�13.86� � 4.62
� 22
3�4� �1.50 � 4�1.45� � 2�1.30� � 4�.99� � 0�
Area � 2 �2
0 � f �x� � d�x�� dx (b)
(c)
f
d
−2 20
3
�x, y � � �0, 1.068�
y �Mx
A�
4.91334.59998
� 1.068
d�x� � �0.02648x4 � 0.01497x2 � .4862
f �x� � �0.1061x4 � 0.06126x2 � 1.9527
43. Centroids of the given regions: and
�x, y � � �4 � 3�
4 � �, 0� � �1.88, 0�
y �4�0� � ��0�
4 � �� 0
x �4�1� � ��3�
4 � ��
4 � 3�
4 � �
Area: A � 4 � �
x1 3
2
1
−1
−2
y�3, 0��1, 0�
44. Centroids of the given regions: and
�x, y � � �2514
, 1514�
y �3�32� � 2�12� � 2�1�
7�
1527
�1514
x �3�12� � 2�2� � 2�72�
7�
2527
�2514
Area: A � 3 � 2 � 2 � 7
x
1
2
3
4
41 2 3
m1m3m2
y�72
, 1��12
, 32�, �2,
12�,
45. Centroids of the given regions: and
�x, y � � �0, 13534 � � �0, 3.97�
y �15�32� � 12�5� � 7�152�
34�
13534
x �15�0� � 12�0� � 7�0�
34� 0
Area: A � 15 � 12 � 7 � 34
x−4 −3 −2 −1 1 2 3 4
7
6
5
4
3
2
1
y�0, 152 ��0,
32�, �0, 5�,
Section 7.6 Moments, Centers of Mass, and Centroids 71
46.
By symmetry,
�x, y � � �0, 789304� � �0, 2.595�
y ��74��716� � �28764��5516�
�74� � �28764� �16,5696384
�789304
x � 0.
m2 �78 �6 �
78� �
28764
, P2 � �0, 5516�
x1
2
3
4
5
−1 12
32
32
− 12
−
m1
m2
ym1 �78
�2� �74
, P1 � �0, 716�
47. Centroids of the given regions: and
Mass:
�x, y � � �2 � 3�
2 � �, 0� � �2.22, 0�
y � 0
x �4�1� � 2��3�
4 � 2��
2 � 3�
2 � �
4 � 2�
�3, 0��1, 0� 48. Centroids of the given regions: and
Mass:
�x, y � � �8 � 3�
8 � �, 0� � �1.56, 0�
x �8�1� � ��3�
8 � ��
8 � 3�
8 � �
y � 0
8 � �
�1, 0��3, 0�
49. is distance between center of circle and axis.is area of circle. Hence,
V � 2�rA � 2��5��16�� � 160�2 � 1579.14.
A � ��4�2 � 16�y-r � 5 50. V � 2�rA � 2��3��4�� � 24�2
51.
V � 2�rA � 2��83��8� �
128�
3� 134.04
r � y �83
�116�16x �
x3
3�4
0�
83
y � �18�
12
�4
0 �4 � x��4 � x� dx
x
1
2
3
4
41 2 3
( , )x y
yA �12
�4��4� � 8
52.
V � 2�rA � 2��225 ��32
3 � �1408�
15� 294.89
r � x �225
x �My
A�
70415
323�
22
5
� 2�645
�323 � �
70415
My � 2�4
0�u � 2�u du � 2�4
0�u32 � 2u12� du � 2�2
5u52 �
43
u32�4
0
Let u � x � 2, x � u � 2, du � dx:
My � �6
2�x�2x � 2 dx � 2�6
2xx � 2 dx
2
2
4
6
4 6
y
x
(6, 4)
( , )x y
A � �6
22x � 2 dx �
43
�x � 2�32�6
2�
323
72 Chapter 7 Applications of Integration
53.
x �My
m, y �
Mx
m
Mx � m1 y1 � . . . � mn yn
My � m1x1 � . . . � mnxn
m � m1 � . . . � mn 54. A planar lamina is a thin flat plate of constant density.The center of mass is the balancing point on thelamina.
�x, y�
55. (a) Yes.
(b) Yes.
(c) Yes.
(d) No
�x, y� � �56, � 5
18��x, y� � �5
6 � 2, 518� � �17
6 , 518�
�x, y� � �56, 5
18 � 2� � �56, 41
18� 56. Let be a region in a plane and let be a line such that does not intersect the interior of If is the distancebetween the centroid of and then the volume of thesolid of revolution formed by revolving about is
where is the area of R.AV � 2�rALR
VL,RrR.
LLR
57. The surface area of the sphere is The arclength of C is The distance traveled by the centroid is
This distance is also the circumference of the circle of radius
Thus, and we have Therefore, thecentroid of the semicircle is �0, 2r��.y � r2 � x2
y � 2r�.2�y � 4r
d � 2�y
y.
d �Ss
�4�r2
�r� 4r.
x
(0, )y
−r
r
r
y
s � �r.S � 4�r2. 58. The centroid of the circle is The distance traveled
by the centroid is The arc length of the circle is alsoTherefore,
x
−2
−1
−1 1
1
3
2
d
C
y
S � �2���2�� � 4�2.2�.2�.
�1, 0�.
59.
Centroid:
As The graph approaches the x-axis and the line as
x
1
1
y x= n (1, 1)
y
n →�.x � 1�x, y � → �1, 14�.n →�,
�n � 1n � 2
, n � 1
4n � 2�
y �Mx
m�
n � 12�2n � 1� �
n � 14n � 2
x �My
m�
n � 1n � 2
My � � �1
0 x�xn� dx � �� �
xn�2
n � 2�1
0�
�
n � 2
Mx ��
2 �1
0 �xn�2 dx � ��
2�
x2n�1
2n � 1�1
0�
�
2�2n � 1�
m � �A ��
n � 1
A � �1
0 xn dx � � xn�1
n � 1�1
0�
1n � 1
60. Let be the shaded triangle with vertices and Let be the large triangle with vertices
and consists of the region minus the region
Centroid of
Centroid of
Area:
by symmetry.
−4 −3 −2 −1
2
6
7
43(0, 0)
(0, 3)
(1, 4) (4, 4)
(−4, 4)
(−1, 4) T
V
x
y
�x, y� � �0, 135 �
y �135
15y �117
3
15y � 1�113 � � 16�8
3�x � 0
V � 16 � 1 � 15
U: �0, 83�; Area � 16
T: �0, 113 �; Area � 1
T.UV�0, 0�.�4, 4�,��4, 4�,
U�0, 3�.�1, 4�,��1, 4�,T
Section 7.7 Fluid Pressure and Fluid Force
Section 7.7 Fluid Pressure and Fluid Force 73
1. F � PA � �62.4�5���3� � 936 lb 2. F � PA � �62.4�5���16� � 4992 lb
3.
� 62.4�2��6� � 748.8 lb
F � 62.4�h � 2��6� � �62.4��h��6� 4.
� 62.4�4��48� � 11,980.8 lb
F � 62.4�h � 4��48� � �62.4��h��48�
5.
� 1123.2 lb
� 249.6�3y �y2
2 �3
0
� 249.6 �3
0 �3 � y� dy
F � 62.4 �3
0 �3 � y��4� dy
L�y� � 4
x1 2 3 4
4
3
2
1
yh�y� � 3 � y 6.
Force is one-third that of Exercise 5.
�43
�62.4��3y2
2�
y3
3 �3
0� 374.4 lb
�43
�62.4� �3
0 �3y � y2� dy
F � 62.4 �3
0 �3 � y�� 4
3y dy
L�y� �43
y
x1−1−2 2
4
2
1
yh�y� � 3 � y
7.
� 748.8 lb
� 124.8�3y �y3
9 �3
0
x1−1−2 2
4
2
1
y � 124.8 �3
0 �3 �
y2
3 dy
F � 2�62.4� �3
0 �3 � y�� y
3� 1 dy
L�y� � 2� y3
� 1h�y� � 3 � y 8.
x1−1
−3
1
y
� �62.4�23�4 � y2�32�
0
�2� 332.8 lb
F � 62.4 �0
�2��y��2��4 � y2 dy
L�y� � 2�4 � y2
h�y� � �y
9.
� 1064.96 lb
� 124.8�8y32
3�
2y52
5 �4
0
x1−1−2 2
3
1
y � 124.8 �4
0 �4y12 � y32� dy
F � 2�62.4� �4
0 �4 � y��y dy
L�y� � 2�y
h�y� � 4 � y 10.
� 748.8 lbx
1−1
−1
−2
−4
y � �62.4�49�9 � y2�32�
0
�3
� 62.4�23 �0
�3 �9 � y2�12��2y� dy
F � 62.4 �0
�3 ��y� 4
3�9 � y2 dy
L�y� �43�9 � y2
h�y� � �y
74 Chapter 7 Applications of Integration
11.
� 9800�8y � y2�2
0� 117,600 newtons
F � 9800 �2
0 2�4 � y� dy
L�y� � 2
x1−1−2 2
3
yh�y� � 4 � y
12.
� 44,100�3�2 � 2� newtons
� 19,600�9�2�2 � 1�4
�9��2 � 1�
4 � � 19,600��y2
2� 3�2y �
y3
3 �3�22
0� �3�2y � 18y �
y3
3�
6�2 � 12
y�3�2
3�22�
F � 2�9800���3�22
0�1 � 3�2 � y�y dy � �3�2
3�22�1 � 3�2 � y��3�2 � y� dy�
L2�y� � 2�3�2 � y� �upper part�
L1�y� � 2y �lower part�
2323
2 1+ 3
2
3
3
y
x1−1−2−3 2 3
h�y� � �1 � 3�2� � y
13.
x3−3 6 9
9
6
3
y � 2,381,400 newtons
� 9800�72y � 7y2 �2y3
9 �9
0
F � 9800 �9
0 �12 � y��6 �
2y3 dy
L�y� � 6 �2y3
h�y� � 12 � y 14.
� 171,500 newtons
� 9800�6y �y2
2 �5
0
F � 9800 �5
0 1�6 � y� dy
L�y� � 1
x−3 −2 −1 1 2 3
6
yh�y� � 6 � y
15.
� 1407�2y �y2
2 �2
0� 2814 lb
� 1407 �2
0 �2 � y� dy
F � 140.7 �2
0 �2 � y��10� dy
L�y� � 10
x
3
4
642−2−1
−2
−6 −4
yh�y� � 2 � y
Section 7.7 Fluid Pressure and Fluid Force 75
17.
� 844.2�4y �y2
2 �4
0� 6753.6 lb
� 844.2 �4
0 �4 � y� dy
F � 140.7 �4
0 �4 � y��6� dy
L�y� � 6
x
3
1
5
321−1−1
−3 −2
yh�y� � 4 � y 18.
� 1055.25 lb
� 140.7 �452
� 15�
� 140.7 ��52
y2 �59
y3�0
�3
x2 3 4 6
−1
−2
−3
−4
−5
1
y
(5, −3)
� 140.7�0
�3��5y �
53
y2 dy
F � 140.7 �0
�3 ��y��5 �
53
y dy
L�y� � 5 �53
y
h�y� � �y
19.
� ��214 �2
3�9 � 4y2�32�0
�32� 94.5 lb
�428
�0
�32 �9 � 4y2�12��8y� dy
F � 42 �0
�32 ��y��9 � 4y2 dy
L�y� � 2�12�9 � 4y2
x
2
1
21−2 −1
−2
−1
yh�y� � �y
20.
The second integral is zero since it is an odd function and the limits of integration are symmetric to the origin. The first integral is twice the area of a semicircle of radius
Thus, the force is 63�94�� � 141.75� � 445.32 lb.
��9 � 4y2 � 2��94� � y2 �
32.
F � 42 �32
�32 �3
2� y�9 � 4y2 dy � 63 �32
�32 �9 � 4y2 dy �
214
�32
�32 �9 � 4y2 ��8y� dy
L�y� � 2�12�9 � 4y2
h�y� �32
� y
16.
� 3376.8 lb
� ��140.7��4�3 �2
3�9 � y2�32�0
�3
��140.7��4�
3 �0
�3 �9 � y2 ��2y� dy
F � 140.7 �0
�3 ��y��2��4
3�9 � y2 dy
L�y� � 2�43�9 � y2 x
2−2
−6
−4
2
yh�y� � �y
76 Chapter 7 Applications of Integration
22. (a)
(b) F � wk�r2 � �62.4��5���32� � 2808� lbs
F � wk�r2 � �62.4��7���22� � 1747.2� lbs
23.
� wb�ky �y2
2 �h2
�h2� wb�hk� � wkhb
F � w �h2
�h2 �k � y�b dy
L�y� � b
x
2
2
2 2
h
h
b b
water level
−
−
y
k
h�y� � k � y
24. (a)
(b)
� �62.4��172 ��5��10� � 26,520 lbs
F � wkhb
� �62.4��112 ��3��5� � 5148 lbs
F � wkhb 25. From Exercise 23:
F � 64�15��1��1� � 960 lb
26. From Exercise 21:
F � 64�15�� �12�2
� 753.98 lb
27.
Using Simpson’s Rule with we have:
� 3010.8 lb
F � 62.4�4 � 03�8� �0 � 4�3.5��3� � 2�3��5� � 4�2.5��8� � 2�2��9� � 4�1.5��10� � 2�1��10.25� � 4�0.5��10.5� � 0�
n � 8
F � 62.4 �4
0 �4 � y�L�y� dy
h�y� � 4 � y
28.
Solving for x, you obtain
� 2�124.8� �3
0 �3 � y�� y
5 � y dy � 546.265 lb
F � 62.4�2� �3
0 �3 � y�� 4y
5 � y dy
L�y� � 2� 4y5 � y
x � �4y�5 � y�.
y � 5x2�x2 � 4�
x
2
4
1
5
321−1−1
−3 −2
yh�y� � 3 � y
21.
The second integral is zero since its integrand is odd and the limits of integration are symmetric to the origin. The first integral is the area of a semicircle with radius r.
F � w��2k� �r2
2� 0� � wk�r2
� w�2k �r
�r
�r2 � y2 dy � �r
�r
�r2 � y2 ��2y� dy� F � w �r
�r
�k � y��r2 � y2 �2� dy
L�y� � 2�r2 � y2
x
water level
r
r−r
−r
yh�y� � k � y
Review Exercises for Chapter 7 77
29.
� 6448.73 lb
F � 62.4 �4
0 2�12 � y��42�3 � y2�3�3�2 dy
L�y� � 2�42�3 � y2�3�3�2
x−6 −4 −2 2 4 6
10
8
6
4
yh�y� � 12 � y
30.
� 62.4�7 �4
0 �12 � y��16 � y2 dy � 21373.7 lb
F � 62.4 �4
0 �12 � y��7�16 � y2� dy
L�y� � 2�7�16 � y2�
2� �7�16 � y2�
x−6 −4 −2 2 4 6
10
8
6
yh�y� � 12 � y
31. (a) If the fluid force is one-half of 1123.2 lb, and the (b) The pressure increases with increasing depth.height of the water is then
b2 � 4.5 ⇒ b � 2.12 ft.
b2 �b2
2� 2.25
�by �y2
2 �b
0� 2.25
�b
0 �b � y� dy � 2.25
F � 62.4 �b
0 �b � y��4� dy �
12
�1123.2�
L�y� � 4
h�y� � b � y
b,
32. Fluid pressure is the force per unit of area exerted by afluid over the surface of a body.
33. see page 508.F � Fw � w�d
c
h�y�L�y� dy,
34. The left window experiences the greater fluid force because its centroid is lower.
1.
3 42
1
5,,125x
y
(1, 0)(5, 0)
(1, 1)
A � �5
1 1x2 dx � ��1
x�5
1�
45
Review Exercises for Chapter 7
2.
� �4x �1x�
5
1�2�
815
x1 2 3 4
1
2
3
5
6
6
(5, 4)
5, 125( )
12
, 4 )(
yA � �5
1�2 4 �
1x2 dx
78 Chapter 7 Applications of Integration
3.
��
4� ��
4 ��
2
� �arctan x�1
�12
1
1 1
1
x
,,2
111,,
2
y
(−1, 0) (1, 0)
A � �1
�1
1x2 � 1
dx 4.
� ��y � 1�3
3 �1
0�
13
� �1
0 �y � 1�2 dy
( 1, 1)−
x
1
−2
12
32
32
− 12
−
y � �1
0 �y2 � 2y � 1� dy
A � �1
0 ��y2 � 2y� � ��1�� dy
5.
�12
� 2�12
x2 �14
x4�1
01
1
1
1x
,( )11
0,0(
)1,
)
1(
yA � 2 �1
0 �x � x3� dx 6.
x
−1
−2
1
2
3
2 3 4 5
(2, 1)−
(5, 2)
y�
92
� �2y �12
y2 �13
y3�2
�1
� �2
�1 �2 � y � y2� dy
A � �2
�1 ��y � 3� � �y2 � 1�� dy
7.
4
6
3211x
)2,, e(2
)2,, e(0
1),(0
y
� e2 � 1
� �xe2 � ex�2
0
A � �2
0 �e2 � ex� dx 8.
3
1
xπ2
π3
π6
2π3
5π6
π6 , 2( ) 5π
6, 2( )
y
� 2�2�
3� ln�2 � �3�� � 1.555
� 2�� � 0� � ��
3� ln�2 � �3��
� 2�2x � ln csc x � cot x ���2
��6
A � 2 ���2
��6 �2 � csc x� dx
Review Exercises for Chapter 7 79
10.
x−2 −1 2
2
3
12
12
12 3
5π
37π
π3
,
,
,
(
(
(
)
)
)
y
��
3� 2�3
� �y2
� sin y�5��3
��3� �sin y �
y2�
7��3
5��3
A � �5��3
��3 1
2� cos y dy � �7��3
5��3 cos y �
12 dy
12. Point of intersection is given by:
(.7832, .4804)
−1 3
−2
4
� 1.189
� �3x � 2x2 �13
x3 �14
x4�0.783
0
A � �0.783
0 �3 � 4x � x2 � x3� dx
x3 � x2 � 4x � 3 � 0 ⇒ x � 0.783.
9.
4π
2
x
y
21
4π )) ,
215
4π )) , −
�4�2
� 2�2
� 1�2
�1�2 � � 1
�2�
1�2
� ��cos x � sin x�5��4
��4
A � �5��4
��4 �sin x � cos x� dx
11.
−4 10
−16
(0, 3)
(8, 3)20
� �8x2 �23
x3�8
0�
5123
� 170.667
� �8
0 �16x � 2x2� dx
A � �8
0 ��3 � 8x � x2� � �x2 � 8x � 3�� dx
13.
−1
(0, 1)
(1, 0)2
2
−1
� �x �43
x3�2 �12
x2�1
0�
16
� 0.1667
� �1
0 �1 � 2x1�2 � x� dx
A � �1
0 �1 � �x�2 dx
y � �1 � �x�214.
( 2, 8)−
(0, 0)
(2, 8)
−4 4
−2
10
� 2�43
x3 �15
x5�2
0�
12815
� 8.5333
� 2 �2
0 �4x2 � x4� dx
A � 2 �2
0 �2x2 � �x4 � 2x2�� dx
80 Chapter 7 Applications of Integration
16.
x2
2
3 4 5
1
−1
−2
3
(1, 0)
(5, 2)
y
� �23
�x � 1�3�2 �14
�x � 1�2�5
1�
43
� �5
1 ��x � 1 �
x � 12 � dx
A � �2
0 ��2y � 1� � �y2 � 1�� dy
y �x � 1
2 ⇒ x � 2y � 1
y � �x � 1 ⇒ x � y2 � 115.
�43
� �y2 �13
y3�2
0
� �2
0�2y � y2� dy
x
3
1
2−2 −1
−1
(0, 2)
(0, 0)
y A � �2
0 �0 � �y2 � 2y�� dy
� �0
�12�x � 1 dx
A � �0
�1 ��1 � �x � 1 � � �1 � �x � 1 �� dx
x � y2 � 2y ⇒ x � 1 � �y � 1�2 ⇒ y � 1 ± �x � 1
17.
� �1
0 3y dy � �3
2y2�
1
0�
32
A � �1
0 ��y � 2� � �2 � 2y�� dy
y � x � 2 ⇒ x � y � 2, y � 1
y � 1 �x2
⇒ x � 2 � 2y
� �2
0 x2
dx � �3
2 �3 � x� dx
A � �2
0 �1 � 1 �
x2� dx � �3
2 �1 � �x � 2�� dx
(0, 1) (3, 1)
(2, 0)x
3
3
1
2
y
20. (a)
1250
80
y � �6.8335��1.2235�t � �6.8335�e0.2017t (b)
Difference: billion dollars�20
15�R2 � y� dt � 12.06
R2 � 5 � 6.83e0.2t
18.
� �13
y3 � y�2
0�
143
A � �2
0 �y2 � 1� dy
x � y2 � 1
A � �1
0 2 dx � �5
1 �2 � �x � 1� dx
x2 3 4 5
1
3
5
4
(1, 0)
(5, 2)
y
19. Job 1 is better. The salary for Job 1 is greater than the salary for Job 2 for all the years except the first and 10th years.
Review Exercises for Chapter 7 81
21. (a) Disk
(c) Shell
x2
2
3
4
1
1
3
y
� 2� �2x2 �x3
3 �4
0�
64�
3
� 2� �4
0 �4x � x2� dx
V � 2� �4
0 �4 � x�x dx
x2
2
3 4
4
1
1
3
y
V � � �4
0 x2 dx � ��x3
3 �4
0�
64�
3
(b) Shell
(d) Shell
x2
2
3
4
4 5
5
1
−1
1
3
y
� 2� �3x2 �13
x3�4
0�
160�
3
� 2� �4
0 �6x � x2� dx
V � 2� �4
0 �6 � x�x dx
x2
2
3 4
4
1
1
3
y
V � 2� �4
0 x2 dx � �2�
3x3�
4
0�
128�
3
22. (a) Shell
(b) Shell
�8�
3
� 2��23
y3 �14
y4�2
0
� 2� �2
0 �2y2 � y3� dy
x2 3
4
4
1
1
3
y V � 2� �2
0 �2 � y�y2 dy
x2
2
3
4
4
1
1
3
y
V � 2� �2
0 y3 dy � ��
2y4�
2
0� 8�
(c) Disk
(d) Disk
�176�
15
� ��15
y5 �23
y3�2
0
� � �2
0 �y4 � 2y2� dy
x2
2
3
4
5
4
1
1
3
y V � � �2
0 ��y2 � 1�2 � 12� dy
x2
2
3
4
4
1
1
3
y
V � � �2
0 y4 dy � ��
5y5�
2
0�
32�
5
82 Chapter 7 Applications of Integration
23. (a) Shell
x2−2
−2
2
3
4
−4
1
1−1−3
y
� �3��12
23�16 � x2�3�2�
4
0� 64�
V � 4� �4
0 x3
4�16 � x2 dx
(b) Disk
x2−2
−2
2
3
4
−4
1
1−1−3
y
�9�
8 �16x �x3
3 �4
0� 48�
V � 2� �4
0 �3
4�16 � x2�
2
dx
24. (a) Shell
(0, )b
( , 0)a
x
x2 y2
a2 b2+ = 1
y �43
�a2b
� ��4�b3a
�a2 � x2�3�2�a
0
��2�b
a �a
0 �a2 � x2�1�2��2x� dx
V � 4� �a
0 �x� b
a�a2 � x2 dx
(b) Disk
(0, )b
( , 0)a
x
x2 y2
a2 b2+ = 1
y �43
�ab2
�2�b2
a2 �a2x �13
x3�a
0
V � 2� �a
0 b2
a2 �a2 � x2� dx
25. Shell
� ���
4� 0� �
�2
4
� �� arctan�x2��1
0
� � �1
0
�2x��x2�2 � 1
dx
V � 2� �1
0
xx4 � 1
dx
x1
1
y 26. Disk
��2
2
� 2��
4� 0
� �2� arctan x�1
0
V � 2� �1
0 � 1�1 � x2�
2
dx
x1 2
2
3
−1−2
−1
y
Review Exercises for Chapter 7 83
27. Shell:
� 4� �13
u3 �12
u2 � 3u � 3 ln�1 � u��2
0�
4�
3�20 � 9 ln 3� � 42.359
� 4� �2
0 u3 � 2u1 � u
du � 4� �2
0 u2 � u � 3 �
31 � u du
V � 2� �6
2
x
1 � �x � 2 dx � 4� �2
0 �u2 � 2�u
1 � u du
dx � 2u du
x � u2 � 2
u � �x � 2
V � 2��6
2
x
1 � �x � 2 dx
x1
1
2
2
3
3 654
−1
−2
−3
y
28. Disk
x1
1
y
� ��
2e2 ��
2 ��
2 1 �1e2
� � �1
0 e�2x dx � ���
2e�2x�
1
0
V � � �1
0 �e�x�2 dx
29. Since
x
−1
−1
y
� ��25
u5�2 �23
u3�2�1
0�
415
A � ��1
0 �u � 1��u du � ��1
0 �u3�2 � u1�2� du
dx � du
x � u � 1
u � x � 1
y ≤ 0, A � ��0
�1 x�x � 1 dx.
30. (a) Disk
x
−1
−1
y
� ��x4
4�
x3
3 �0
�1�
�
12
� � �0
�1 �x3 � x2� dx
V � � �0
�1 x2�x � 1� dx
(b) Shell
� 4��17
u7 �25
u5 �13
u3�1
0�
32�
105
� 4� �1
0 �u6 � 2u4 � u2� du
� 4� �1
0 �u2 � 1�2u2du
V � 2� �0
�1 x2�x � 1 dx
dx � 2u du
x � u2 � 1x
−1
−1
y u � �x � 1
84 Chapter 7 Applications of Integration
31. From Exercise 23(a) we have:
Disk:
By Newton’s Method, and the depth of the gasoline is 3 � 1.042 � 1.958 ft.y0 � �1.042
y03 � 27y0 � 27 � 0
9y0 �13
y03 � ��27 � 9� � 9
�9y �13
y3�y0
�3� 9
19
�y0
�3 �9 � y2� dy � 1
� �y0
�3 169
�9 � y2� dy � 16�
14
V � 16�
V � 64� ft3
36. Since has this integral represents the length of the graph of tan x from to This length is a little over 1 unit. Answers (b).
x � ��4.x � 0f��x� � sec2 x,f �x� � tan x
32.
Since we have Thus,
2 a x− 22
2 a x− 22
2 a x− 22
a � 3�5�32
� 1.630 meters.
a3 � �5�3��2.�4�3 a3��3 � 10,
� �34a3
3
V � �3 �a
�a
�a2 � x2� dx � �3�a2x �x3
3 �a
�a
� �3 �a2 � x2�
A�x� �12
bh �12
�2�a2 � x2���3�a2 � x2 �
34.
s � �3
1 1
2x2 �
12x2 dx � �1
6x3 �
12x�
3
1�
143
1 � �y��2 � 12
x2 �1
2x22
y� �12
x2 �1
2x2
y �x3
6�
12x
33.
�815
�1 � 6�3 � � 6.076
� 2�25
u5�2 �23
u3�2�3
1�
415�u3�2�3u � 5��
3
1
� 2 �3
1 �u3�2 � u1�2� du
s � �4
0 �1 � �x dx � 2 �3
1 �u�u � 1� du
dx � 2�u � 1� du
x � �u � 1�2
u � 1 � �x
1 � � f��x��2 � 1 � �x
f��x� � x1�4
f �x� �45
x5�4
35.
by Simpson’s Rule or graphing utility�� � 4018.2 ft
�1
20�2000
�2000�400 � 9 sinh2 x
2000 dx
s � �2000
�2000 �1 � � 3
20 sinh x
2000�2
dx
y� �3
20sinh x
2000
y � 300 cosh x2000 � 280, �2000 ≤ x ≤ 2000
Review Exercises for Chapter 7 85
38.
� 4��23�x � 1�3�2�
3
0�
56�
3
S � 2� �3
0 2�x�x � 1
x dx � 4� �3
0 �x � 1 dx
1 � �y��2 � 1 �1x
�x � 1
x
y� �1�x
y � 2�x37.
x2
2
3 4
4
1
1
3
(4, 3)
y
S � 2� �4
0 3
4x�25
16 dx � �15�
8 x2
2 �4
0� 15�
1 � �y��2 �2516
y� �34
y �34
x
39.
� 50 in. � lb � 4.167 ft � lb
W � �5
0 4x dx � �2x2�
5
0
F � 4x
4 � k�1�
F � kx 40.
� 225 in. � lb � 18.75 ft � lb
W � �9
0 509
x dx � �259
x2�9
0
F �509
x
50 � k�9� ⇒ k �509
F � kx
42. We know that
Depth of water:
Time to drain well: minutes
Volume of water pumped in Exercise 41: 391.7 gallons
Work � 78� ft � ton
x �588�52�391.7
� 78
391.752�
�588x�
�49��12� � 588 gallons pumped
t �150
3.064� 49
�3.064t � 150
dhdt
�9� dV
dt �9� � 8
7.481 � �3.064 ft�min.
dVdt
��
9 dhdt
V � �r2h � �19h
dVdt
�4 gal�min � 12 gal�min
7.481 gal�ft3� �
87.481
ft3�min
41. Volume of disk:
Weight of disk:
Distance:
� 104,000� ft � lb � 163.4 ft � ton
W �62.4�
9 �150
0 �175 � y� dy �
62.4�
9 �175y �y2
2 �150
0
175 � y
62.4�13
2
�y
�13
2
�y
86 Chapter 7 Applications of Integration
43. Weight of section of chain:
Distance moved:
W � 5�10
0�10 � x� dx � ��5
2�10 � x�2�10
0� 250 ft � lb
10 � x
5 �x
44. (a) Weight of section of cable:
Distance:
(b) Work to move 300 pounds 200 feet vertically:
� 40 ft � ton � 30 ft � ton � 70 ft � ton
Total work � work for drawing up the cable � work of lifting the load
200�300� � 60,000 ft � lb � 30 ft � ton
W � 4 �200
0 �200 � x� dx � ��2�200 � x�2�
200
0� 80,000 ft � lb � 40 ft � ton
200 � x
4 �x
45.
a �3�80�
64�
154
� 3.75
80 � �4
0 ax2 dx �
ax3
3 �4
0�
643
a
W � �b
a
F�x� dx 46.
� 51 ft � lbs
� ��9 � 54� � ��96 � 192 � 54 � 144�
� ��19
x2 � 6x�9
0� ��2
3x2 � 16x�
12
9
W � �9
0 �2
9x � 6 dx � �12
9 �4
3x � 16 dx
F�x� � ���2�9�x � 6,��4�3�x � 16,
0 ≤ x ≤ 9 9 ≤ x ≤ 12
W � �b
a
F�x� dx
47.
�x, y � � a5
, a5
�3a2�a2x �
83
a3�2x3�2 � 3ax2 �85
a1�2x5�2 �13
x3�a
0�
a5
�3a2 �a
0 �a2 � 4a3�2x1�2 � 6ax � 4a1�2x3�2 � x2� dx
y � 6a2 1
2 �a
0 ��a � �x �4 dx
xa
a
( , )x y
y x �6a2 �a
0 x��a � �x �2 dx �
6a2 �a
0 �ax � 2�ax3�2 � x2� dx �
a5
1A
�6a2
� �ax �43�ax3�2 �
12
x2�a
0�
a2
6A � �a
0 ��a � �x �2 dx � �a
0 �a � 2�ax1�2 � x� dx
Review Exercises for Chapter 7 87
48.
( , )x y
−3 3 6
3
6
9
x
y
�x, y � � 1, 175
�3
64�9x � 6x2 �43
x3 �15
x5�3
�1�
175
y � 332
12
�3
�1 ��2x � 3�2 � x4� dx �
364
�3
�1�9 � 12x � 4x2 � x4� dx
x �3
32 �3
�1 x�2x � 3 � x2� dx �
332
�3
�1 �3x � 2x2 � x3� dx �
332�
32
x2 �23
x3 �14
x4�3
�1� 1
1A
�3
32
A � �3
�1 ��2x � 3� � x2� dx � �x2 � 3x �
13
x3�3
�1�
323
50.
x
−2
2
2
6
4
4
6 8
( , )x y
y
�x, y � � 103
, 4021
�12
516�
37
x7�3 �1
12x3�
8
0�
4021
y � 516
12
�8
0 x4�3 �
14
x2 dx
�5
16�38
x8�3 �16
x3�8
0�
103
x �5
16 �8
0 xx2�3 �
12
x dx
1A
�5
16
A � �8
0 x2�3 �
12
x dx � �35
x5�3 �14
x2�8
0�
165
49. By symmetry,
x−a a
a
( , )x y
2
y
�x, y � � 0, 2a2
5
�6
8a3 a5 �23
a5 �15
a5 �2a2
5
�6
8a3�a4x �2a2
3x3 �
15
x5�a
0
�6
8a3 �a
0 �a4 � 2a2x2 � x4� dx
y � 34a31
2 �a
�a
�a2 � x2�2 dx
1A
�3
4a3
A � 2 �1
0 �a2 � x2� dx � 2�a2x �
x3
3 �a
0�
4a3
3
x � 0.
88 Chapter 7 Applications of Integration
51. by symmetry.
For the trapezoid:
For the semicircle:
Let then and When When
Thus, we have:
The centroid of the blade is 2�9� � 49�3�� � 9� , 0.
x �180 � 4�4 � 9��
3�
12�9 � �� �
2�9� � 49�3�� � 9�
x�18 � 2�� � 60 �4�4 � 9��
3
� 2��12
23�4 � u2�3�2�
2
0� 12���2�2
4 � �16
3� 12� �
4�4 � 9��3
My � 2 �2
0 �u � 6��4 � u2 du � 2 �2
0 u�4 � u2 du � 12 �2
0 �4 � u2 du
u � 2.x � 8,u � 0.x � 6,dx � du.x � u � 6u � x � 6,
My � �8
6 x ��4 � �x � 6�2 � ���4 � �x � 6�2 �� dx � 2 �8
6 x�4 � �x � 6�2 dx
m � 12����2�2 � 2�
� �6
0 1
3x2 � 2x dx � �x3
9� x2�
6
0� 60
My � �6
0 x�1
6x � 1 � �1
6x � 1� dx
m � ��4��6� � �1��6�� � 18
y � 0
(6, 2)
(6, 2)−
x
−4
−1
−2
−3
1
1
2
2
3
3
4
4
5 7
y = x + 116
y
52. Wall at shallow end:
Wall at deep end:
Side wall:
5−5
5
10
10
15
15
20
20
25 30 35 40 45x
y
F � F1 � F2 � 72,800 lb
F2 � 62.4 �5
0 �10 � y�8y dy � 62.4 �5
0 �80y � 8y2� dy
F1 � 62.4 �5
0 y�40� dy � ��1248�y2�
5
0� 31,200 lb
F � 62.4 �10
0 y�20� dy � ��624�y2�
10
0� 62,400 lb
F � 62.4 �5
0 y�20� dy � ��1248� y2
2 �5
0� 15,600 lb
53. Let
x
( , )x yd
c
D
gf
y
� �Area��depth of centroid�
� �Area��D � y �
� ��d
c
� f �y� � g�y�� dy��D ��d
cy � f �y� � g�y�� dy
�d
c
� f �y� � g�y�� dy � � ��d
c
D � f �y� � g�y�� dy � �d
c
y � f �y� � g�y�� dy� F � �d
c
�D � y�� f �y� � g�y�� dy
� weight per cubic volume.D � surface of liquid;
Problem Solving for Chapter 7 89
Problem Solving for Chapter 7
1.
limc→0�
TR
� limc→0�
12
c3
16
c3
� 3
R � �c
0�cx � x2� dx � �cx2
2�
x3
3 �c
0�
c3
2�
c3
3�
c3
6
T �1
2c�c2� �
1
2c3
54. F � 62.4�16��5 � 4992� lb
2.
Let be the intersection of the line and the parabola.
Then, or
m � 1 � �12�
1�3
0.2063
�12�
1�3
� 1 � m
12
� �1 � m�3
� �1 � m�2�2 � 2m�
� �1 � m�2�6 � 4�1 � m� � 6m�
1 � 6�1 � m�2 � 4�1 � m�3 � 6m�1 � m�2
��1 � m�2
2�
�1 � m�3
3� m
�1 � m�2
2
112
� �x2
2�
x3
3� m
x2
2 �1�m
0
12�
16� � �1�m
0�x � x2 � mx� dx
c � 1 � m.mc � c�1 � c� ⇒ m � 1 � c
�c, mc�
R � �1
0x�1 � x� dx � �x2
2�
x3
3 �1
0�
12
�13
�16
3. (a)
(b)
V � 2�2 r2R
� � �4R�14
�r2 � �2r2R
� ��r
04Rr2 � y2 dy
12
V � �r
0 �� �R � r2 � y2�2
� � �R � r2 � y2�2� dy
�x � R�2 � y2 � r2 ⇒ x � R ± r2 � y2
� 8���
4� � 2�2 ⇒ V � 4�2
� 8��1
0
1 � y2 dy �Integral represents 1�4 �area of circle��
� ��1
0 ��4 � 41 � y2 � �1 � y2�� � �4 � 41 � y2 � �1 � y2��� dy
12
V � �1
0 �� �2 � 1 � y2�2
� � �2 � 1 � y2�2� dy
90 Chapter 7 Applications of Integration
4.
For
�52�
3
S � 2�2���1
0x1 � � 1 � 2x2
221 � x2�2
dx
x > 0, y� �1 � 2x2
221 � x2
−0.5
−0.5 0.5 1.5−1.5
−0.25
0.25
0.5
y
x
y � ± x 1 � x2
22
8y2 � x2�1 � x2� 5.
which does not depend on r!
r h2
x2 + y2 = r2
r2 − h2
4r
��4�
3 ��
h3
8 � ��h3
6
� �2� �23
�r2 � x2�3�2�r
r2��h2�4�
V � 2�2���r
r2��h2�4� xr2 � x2 dx
6. By the Theorem of Pappus,
� 2� �d �12w2 � l2� lw
V � 2�r A
7. (a) Tangent at A:
To find point B:
Tangent at B:
To find point C:
Area of
Area of
Area of �area Sarea R
� 16�S � 16�area of R�
S � �4
�2�12x � 16 � x3� dx � 108
R � �1
�2�x3 � 3x � 2� dx �
274
�x � 2�2�x � 4� � 0 ⇒ C � �4, 64�
x3 � 12x � 16 � 0
x3 � 12x � 16
y � 12x � 16
y � 8 � 12�x � 2�
y � x3, y� � 3x2
�x � 1�2�x � 2� � 0 ⇒ B � ��2, �8�
x3 � 3x � 2 � 0
x3 � 3x � 2
y � 3x � 2
y � 1 � 3�x � 1�
y � x3, y� � 3x2 (b) Tangent at
To find point B:
Tangent at B:
To find point C:
Area of
Area of
Area of S � 16�area of R�
S � �4a
�2a
�12a2x � 16a3 � x3� dx � 108a4
R � �a
�2a
�x3 � 3a2x � 2a3� dx �274
a4
⇒ C � �4a, 64a3�
�x � 2a�2�x � 4a� � 0
x3 � 12a2x � 16a3 � 0
y � 12a2x � 16a3
y � 8a3 � 12a2�x � 2a�
⇒ B � ��2a, �8a3�
�x � a�2�x � 2a� � 0
x3 � 3a2x � 2a3 � 0
y � 3a2x � 2a3
y � a3 � 3a2�x � a�A�a, a3�:
Problem Solving for Chapter 7 91
8.
f �x� � 2ex�2 � 2
f �0� � 0 ⇒ C � �2
f �x� � 2ex�2 � C
f��x� � e x�2
f��x�2 � ex
9.
(a)
(b)
(c)
(d)
This is the length of the curve from to x � 2.x � 1y � x3�2
�222722 �
132713 2.0858 s�2� � �2
11 �
94
t dt � � 827�1 �
94
t�3�2
�2
1
s�x� � �x
11 � �3
2t1�2�
2
dt � �x
11 �
94
t dt
�ds�2 � �1 � f��x�2��dx�2 � �1 � �dydx�
2
��dx�2 � �dx�2 � �dy�2
ds � 1 � f��x�2 dx
s��x� �dsdx
� 1 � f��x�2
s�x� � �x
�
1 � f��t�2 dt
10. Let be the density of the fluid and the density of the iceberg. The buoyant force is
where is a typical cross section and g is the accelerationdue to gravity. The weight of the object is
or 89.3%�0
�f
�submerged volume
total volume�
0.92 � 103
1.03 � 103 � 0.893
�f g�0
�h
A�y� dy � �0 g�L�h
�h
A�y� dy
F � W
W � �0 g�L�h
�h
A�y� dy.
A�y�
F � �f g�0
�h
A�y� dy
�0�f 11. (a) by symmetry
(b)
(c) �x, y� � �2, 0�limb→
x � limb→
2b
b � 1� 2
�x, y� � � 2bb � 1
, 0� x �2�b � 1��b�b2 � 1��b2 �
2bb � 1
My � 2�6
1
1x2 dx �
2�b � 1�b
m � 2�b
1
1x3 dx �
b2 � 1b2
3
2
−1
−2
−3
4 5 632
1
x
y
�x, y� � �127
, 0� x �5�3
35�36�
127
m � 2�6
1
1x3 dx � ��
1x2�
6
1�
3536
My ��6
1x� 1
x3 � ��1x3�� dx ��6
1
2x2 dx � ��2
1x�
6
1�
53
y � 0
92 Chapter 7 Applications of Integration
12. (a) by symmetry
(b)
�x, y� � �32
, 0�limb→
x �32
�x, y� � � 3b�b � 1�2�b2 � b � 1�, 0� x �
�b2 � 1��b2
2�b3 � 1��3b3 �3b�b � 1�
2�b2 � b � 1�
m � 2�b
1
1x4 dx �
2�b3 � 1�3b3
My � 2�b
1
1x3 dx �
b2 � 1b2
�x, y� � �6343
, 0� x �35�36
215�324�
6343
m � 2�6
1
1x4 dx �
215324
My � 2�6
1x
1x4 dx � 2�6
1
1x3 dx �
3536
1
−1−1 2 3 4 5
−2
−3
2
3
y
x
y = x41
y = − x41
y � 0
13. (a)
(b) W � area � 3 � �1 � 1� � 2 �12
� 712
W � area � 2 � 4 � 6 � 12
15. Point of equilibrium:
Consumer surplus
Producer surplus � �80
0�10 � 0.125x� dx � 400
� �80
0��50 � 0.5x� � 10� dx � 1600
�P0, x0� � �10, 80�
x � 80, p � 10
50 � 0.5x � 0.125x
14. (a) Trapezoidal:
(b) Simpson’s: Area 1603�8��0 � 4�50� � 2�54� � 4�82� � 2�82� � 4�73� � 2�75� � 4�80� � 0� � 10,4131
3 sq ft
Area 1602�8��0 � 2�50� � 2�54� � 2�82� � 2�82� � 2�73� � 2�75� � 2�80� � 0� � 9920 sq ft
16. Point of equilibrium:
Consumer surplus
Producer surplus � �20
0�840 � 42x� dx � 8400
� �20
0��1000 � 0.4x2� � 840� dx � 2133.33
�P0, x0� � �840, 20�
x � 20, p � 840
1000 � 0.4x2 � 42x
Problem Solving for Chapter 7 93
17. We use Exercise 23, Section 7.7, which gives for a rectangle plate.
Wall at shallow end
From Exercise 23:
Wall at deep end
From Exercise 23:
Side wall
From Exercise 23:
Total force: F1 � F2 � 46,592 lb
� 26,624 lb
� 624 �4
0 �8y � y2� dy � 624�4y2 �
y3
3 �4
0
F2 � 62.4 �4
0 �8 � y��10y� dy
F1 � 62.4�2��4��40� � 19,968 lb
5
5
10
10
15
15
20
20
25 40 45x
y = 8x = 40
y = x110
yF � 62.4�4��8��20� � 39,936 lb
F � 62.4�2��4��20� � 9984 lb
F � wkhb
18. (a) Answers will vary.
(c) See the article by Professor Larson Riddle at http://ecademy.agnesscott.edu/lriddle/arc/contest.htmOne such function is
�arc length 2.9195� f3�x� �8�x � x2
f2�x� ��
2 sin ��x�
f1�x� � 6�x � x2�
(b) arc length
arc length 3.3655f2
3.2490f1