C H A P T E R 7 Applications of Integration

93
CHAPTER 7 Applications 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

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

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

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

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

� 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

−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

, 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

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.

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