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C H A P T E R 6Applications of Integration

Section 6.1 Area of a Region Between Two Curves . . . . . . . . . . 264

Section 6.2 Volume: The Disk Method . . . . . . . . . . . . . . . . . 271

Section 6.3 Volume: The Shell Method . . . . . . . . . . . . . . . . 278

Section 6.4 Arc Length and Surfaces of Revolution . . . . . . . . . . 282

Section 6.5 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287

Section 6.6 Moments, Centers of Mass, and Centroids . . . . . . . . . 290

Section 6.7 Fluid Pressure and Fluid Force . . . . . . . . . . . . . . . 297

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305

P A R T I I

Page 2: 1

C H A P T E R 6Applications of Integration

Section 6.1 Area of a Region Between Two CurvesSolutions to Even-Numbered Exercises

264

2. A � �2

�2 ��2x � 5� � �x2 � 2x � 1�� dx � �2

�2 ��x2 � 4� dx

4. A � �1

0 �x2 � x3� dx 6. A � 2�1

0 ��x � 1�3 � �x � 1�� dx

8.

x−2 2

−2

2

y

�1

�1 ��1 � x2� � �x2 � 1�� 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 12.

π4

π4−

π4− 2

2,( )

π4− , 2( ) π

4 , 2( )

π4

22,( )

y

x

2

3

���4

���4�sec2 x � cos x� dx

14.

Matches (a)

x

1

3

4

1 2 3

(0, 2)

(4, 0)

y

A � 1

g�x� � 2 � x

f �x� � 2 �12 x 16.

2

2

4

6

8

4 6 10

y

x

(2, 9)

2,

(8, 6)

(8, 0)

92( )

� �64 � 112 � 80� � �1 � 7 � 20� � 18

� �x3

8�

7x2

4� 10x

8

2

� �8

2�3

8x2 �

72

x � 10 dx

A � �8

2��10 �

12

x � ��38

x�x � 8� dx

Page 3: 1

Section 6.1 Area of a Region Between Two Curves 265

18. The points of intersection are given by

(3, 4)

(0, 1)

1 2

2

3

4

5

6

3 4 5 6

y

x

� �9 �272

�92

� ��x3

3�

3x2

2 3

0

� �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

20. The points of intersection are given by:

x−1 1 2 3 4 5

3

4

5

6

(0, 2)

(3, 5)

y

� �3

0 ��x2 � 3x� dx � ��x3

3�

32

x23

0�

92

� �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

22.

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

24. The points of intersection are given by

� 2��12

� 1 � 0 � ��34 �

12

� 2�x2

2� x �

34

�x � 1�4�31

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

Page 4: 1

266 Chapter 6 Applications of Integration

26. The points of intersection are given by:

� �3

0 �3y � y2� dy � �3

2y2 �

13

y33

0�

92

� �3

0 ��2y � y2� � ��y�� dy

A � �3

0 � f �y� � g�y�� dy

y�y � 3� � 0 when y � 0, 3

x

−1

1

3

−3 −2 1

( 3, 3)−

(0, 0)

y 2y � y2 � �y

28.

� �� 16 � y23

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

30.

� 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

32. The point of intersection is given by:

Numerical Approximation: 2.0

x−2 −1 2

−2

−1

( 1, 2)−

(1, 0)

(1, 2)−

y

� �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

34. The points of intersection are given by:

Numerical Approximation: 8.533

( 2, 8)−

(0, 0)

(2, 8)

−4 4

−2

10

� 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

Page 5: 1

Section 6.1 Area of a Region Between Two Curves 267

36.

The points of intersection are given by:

Numerical Approximation: 8.267 � 0.617 � 0.883 � 9.767

�29330

�24830

�3760

�5360

�2

1 ��x3 � 4x� � �x4 � 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 � 4xx

−1−3−4 1 3 4

−3

−4

3

4

f

g

yf �x� � x4 � 4x2, g�x� � x3 � 4x

38.

Numerical Approximation: 6.908

� 6.908

� 3 ln 10

� �3 ln�x2 � 1�3

0(0, 0)

3, 95( )

−1 5

−1

3 A � �3

0 � 6x

x2 � 1� 0 dx 40.

−1 5

−1

2

A � �4

0 x 4 � x

4 � x dx � 3.434

42.

π2

π6

π2

−1, −1( )

π6 2

1, ( )

y

x

� � 34

� 32 � �0� �

3 34

� 1.299

� �12

sin 2x � cos x��6

���2

A � ���6

���2�cos 2x � sin x� dx 44.

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

46. From the graph we see that f and g intersect twice at and

� 2 �1 �1

ln 3 � 0.180

� �x2 � x �1

ln 3�3x�

1

0

� �1

0 ��2x � 1� � 3x� dx

A � �1

0 �g�x� � f �x�� dx

x � 1.x � 0

x

2

3

−1 1 2

(0, 1)

(1, 3)

y

48.

−2

(0, 1)

(π, 1)2

�4

�45

� ��2 cos x �12

sin 2x�

0� 4

A � ��

0 ��2 sin x � cos 2x� � 0� dx

Page 6: 1

268 Chapter 6 Applications of Integration

50.

(1, 0)

(5, 1.29)

−2

0 6

2

� 5.181� 2�ln 5�2 � �2�ln x�25

1

A � �5

1 �4 ln x

x� 0 dx 52. (a)

(b)

No, it cannot be evaluated by hand.

(c) 1.2556

−1

−1

2

4 y xe= x

A � �1

0 x ex dx.

x � 1x � 0,y � 0,y � x ex,

54. F�x� � �x

0�1

2t2 � 2 dt � �1

6t3 � 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

56. 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

58.

(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

� � 74

x2 � 7x4

2� ��7

4x2 � 21x

6

4� 7 � 7 � 14

� �4

2 � 7

2x � 7 dx � �6

4 ��7

2x � 21 dx

A � �4

2 �� 9

2x � 12 � �x � 5� dx � �6

4 ���5

2x � 16 � �x � 5� dx

Page 7: 1

Section 6.1 Area of a Region Between Two Curves 269

60.

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

f �x� �1

x2 � 1

x1 2

(0, 1)

1,

12

12

14

32

34

( )

f x( ) =1

2x + 1

12

12

y x= + 1−

y

64. on

on

Both functions symmetric to origin

Thus,

A � 2 �1

0 �x � x3� dx � 2 �x2

2�

x4

4 1

0�

12

�1

�1 �x3 � x� dx � 0.

�0

�1 �x3 � x� dx � ��1

0 �x3 � x� dx.

�0, 1�x3 ≤ x

��1, 0�x3 ≥ x

x

( 1, 1)− −

(0, 0)

(1, 1)1

1

−1

−1

y

62. Answers will vary. See page 417.

66. Proposal 2 is better, since the cummulative deficit (the area under the curve) is less.

68.

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

A � 2 �9

0 �9 � x� dx � 2 �9x �

x2

2 9

0� 81

x

( (9 ), )− −b b (9 , )−b b

−6 −3 3 6

−3

−6

6

9

12

y

Page 8: 1

270 Chapter 6 Applications of Integration

70.

where and is the same as

�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

x

5

3

2

1

−1

−3 −1 1 3

( 2, 0)− (2, 0)

f x x( ) = 4 − 2

y

72.

�2912

billion � $2.417 billion

� �0.01t3

3�

0.16t2

2 5

0

�5

0 ��7.21 � 0.26t � 0.02t2� � �7.21 � 0.1t � 0.01t2�� dt � �5

0 �0.01t2 � 0.16t� dt

74. 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

0.0355

0

P2 � 893,000 e�0.035�t312 %:

P1 � 893,000 e�0.05�t

76. The curves intersect at the point where the slope of equals that of

y2 � 0.08x2 � k ⇒ y�2 � 0.16x � 1 ⇒ x �1

.16� 6.25

y1, 1.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(a) The value of k is given by

k � 3.125.

6.25 � �0.08��6.25�2 � k

y1 � y2

78. (a)

(b) (c) 5000V � 5000�11.816� � 59,082 poundsV � 2A � 2�5.908� � 11.816 m3

A � 6.031 � 2 �� � 116

2

� 2 �� �18

2

� 5.908

80. True

Page 9: 1

Section 6.2 Volume: The Disk Method

Section 6.2 Volume: The Disk Method 271

2. V � � �2

0 �4 � x2�2 dx � � �2

0 �x4 � 8x2 � 16� dx � ��x5

5�

8x3

3� 16x�

2

0�

256�

15

4.

� ��9x �x3

3 �3

0� 18�

V � ��3

0��9 � x2�2 dx � ��3

0�9 � x2� dx 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

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 10.

�459�

15�

153�

5

� ��y5

5� 2y4 �

16y3

3 �4

1

V � ��4

1 ��y2 � 4y�2 dy � ��4

1 �y4 � 8y3 � 16y2� dy

12. 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,(a)

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,

(c)

x−4 −2 2 4

2

4

6

y

� 4��83

x3 �15

x5�2

0�

896�

15

� ��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)

x−4 −2 4

2

4

6

8

y

� ��4y �4�2

3y3�2 �

y2

4 �8

0

�16�

3

� ��8

04 � 4�y

2�

y2 dy

V � ��8

0 2 ��y

2 2

dy

r�y� � 0R�y� � 2 � �y�2,

Page 10: 1

272 Chapter 6 Applications of Integration

14. intersect at and �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,(a)

x−6 −4 −2 2

2

4

8

y

� ��15

x5 � x4 � 3x3 � 18x2�0

�3�

243�

5

� ��0

�3�x4 � 4x3 � 9x2 � 36x� dx

V � ��0

�3��6 � 2x � x2�2 � �x � 6�2 dx

r�x� � x � 6R�x� � 6 � 2x � x2,

16.

1

1−1−1

2 3 4

2

3

(2, 4)

y

x

� ��32 � 16 �12828 � �

1447

� ��16x � x4 �x7

28�2

0

� ��2

0�16 � 4x3 �

x6

4 � dx

V � ��2

04 �

x3

2 2

dx

R�x� � 4 �x3

2, r�x� � 0 18.

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,

20.

� ��36y �y3

3 �4

0�

368�

3

V � ��4

0��6�2 � �y�2 dy

r�y� � 6 � �6 � y� � yR�y� � 6,

x1 2 3 4 5

1

−1

2

3

4

5

y

Page 11: 1

Section 6.2 Volume: The Disk Method 273

22.

x

1

2

3

4

5

6

1 2 3 4 5

y

� 12��13 � 6 ln 3� � 241.59 � 36� 133

� 2 ln 13

� 36��356

� 2 ln 6 � 32

� 2 ln 2�

� 36��y � 2 ln�y� �1y �

6

2

� 36��6

21 �

2y

�1y2 dy

V � ��6

26 �

6y

2

dy

r�y� � 0R�y� � 6 �6y

, 24.

x

−3

−2

1

2

3

−3 −1 1 2 3

y

�128�

15

� 2��4x3

3�

x5

5 �2

0

� 2��2

0�4x2 � x4� dx

V � 2��2

0�x�4 � x2 2

dx

r�x� � 0R�x� � x�4 � x2,

26.

x2 4 6 8

1

2

3

4

y

� 9��� 1x � 1�

8

0� 8�

� 9� �8

0�x � 1��2 dx

V � ��8

0 3

x � 12

dx

r�x� � 0R�x� �3

x � 1, 28.

x−2 2 4 6

2

4

6

8

y

� ��e4 � 1� � 168.38

� ��ex�4

0

� ��4

0 ex dx

V � ��4

0�ex�2�2 dx

r�x� � 0R�x� � e x�2,

30.

�883

� �563

� � 48�

� �� x3

12�

5x2

2� 16x�

4

0� ���

x3

12�

5x2

2� 16x�

8

4

� ��4

0x2

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

x2

� ��x�2� dx � ��8

4���x�2

� 4 �12

x2

� dx

Page 12: 1

274 Chapter 6 Applications of Integration

32.

� �25 �252 �

25�

2

� ��5y �y2

2 �5

0

� ��5

0�5 � y� dy

V � ��5

0���9 � y �2

� 22 dy

x � �9 � y

(2, 5)

1 2 3 4 5 6 7 8 9

1

2

3

4

5

6

7

8

9

x

yx � 3x � 2,y � 0,y � 9 � x2,

34.

x1 2

1

2

y

V � � ���2

0 �cos x 2 dx � 2.4674 36. V � � �3

1 �ln x 2 dx � 3.2332

38. V � � �5

0 �2 arctan �0.2x� 2 dx � 15.4115

40.

Matches (b)

x

1

134

34

12

12

14

14

yA � 34 42. or V � �d

c

A�y� dyV � �b

a

A�x� dx

44. (a)

x

y842

4

−4

2

z (b)

x

y88

4

2

z (c)

a < c < b.

x

y168

4

4

z

46.

�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,

Page 13: 1

Section 6.2 Volume: The Disk Method 275

48.

x

r

h

y

��

3�2r3 � 3r2h � h3�

� � 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, 50. (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 thirdsof 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

52.

� 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

y � ��0.1x3 � 2.2x2 � 10.9x � 22.2,2.95,

0 ≤ x ≤ 11.5 11.5 < x ≤ 15

x

2

4

6

8

4 8 12 16

y

x

y

5

42 2

z

54. (a) First find where intersects the parabola:

—CONTINUED—

� ��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�

b � 4 �x2

4

y � b

Page 14: 1

276 Chapter 6 Applications of Integration

(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

56. (a)

Simpson’s Rule:

��

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

(b)

00

10

6

f �x� � 0.00249x4 � 0.0529x3 � 0.3314x2 � 0.4999x � 2.112

(c) V � �10

0 � f �x�2 dx � 186.35 cm3

58. V �12

�10��2��3� � 30 m3

54. —CONTINUED—

—CONTINUED—

60.

Base of Cross Section

(a)

� 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 (b)

2 4 − x22 4 − x2

2 4 − x2

� �3�4x �x3

3 �2

�2�

32�33

V � �3�2

�2�4 � x2� dx

� �3 �4 � x2�

A�x� �12

bh �12

�2�4 � x2 ���3�4 � x2 �

Page 15: 1

Section 6.2 Volume: The Disk Method 277

62. 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

A�y� � b2 � ��r2 � y2�2y

x

(c)

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 (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 � x2

A�x� �12

bh

60. —CONTINUED—

64.

�43

� �R2 � r2�3�2

� 2� ��R2 � r2�3�2 ��R2 � r2�3�2

3 �

� 2� ��R2 � r2�x �x3

3 ��R2�r2

0

� 2� ��R2�r2

0 �R2 � r2 � x2� dx

V � � ��R2�r2

��R2�r2

���R2 � x2 �2� r2 dx

r

R

RR2 2− r

66. (a) When represents a square.

When represents a circle.

(b)

To approximate the volume of the solid, form n slices, each of whose area is approximated by the integral above. Then sum the volumes of these n slices.

A � 2 �1

�1 �1 � �x�a�1�a dx � 4 �1

0 �1 � xa�1�a dx

�y� � �1 � �x�a�1�a

a � 2: �x�2 � �y�2 � 1

a � 1: �x� � �y� � 1

x

a = 2

a = 1

−1

−1

1

1

y

Page 16: 1

Section 6.3 Volume: The Shell Method

2.

� 2��1

0 �x � x2� dx � 2��x2

2�

x3

3 �1

0�

3

V � 2��1

0 x�1 � x� dx

h�x� � 1 � x

p�x� � x 4.

� 2��2x2 �x4

4 �2

0� 8�

� 2��2

0 �4x � x3� dx

V � 2��2

0 x�4 � x2� dx

h�x� � 8 � �x2 � 4� � 4 � x2

p�x� � x

6.

1−1−3

3

6

9

12

15

18

2 3 4 5 6

y

x

� ��x4

4 �6

0� 324�

V � 2��6

0

12

x3 dx

h�x� �12

x2

p�x� � x 8.

x21−1−2

1

2

3

y

� 2��2x2 �14

x4�2

0� 8�

V � 2��2

0 �4x � x3� dx

h�x� � 4 � x2

p�x� � x

278 Chapter 6 Applications of Integration

12.

� 2���

0 sin x dx � ��2� cos x�

0� 4�

V � 2���

0 x �sin x

x � dx

h�x� �sin x

x

1

−1

2

3

x

4π π

43π

yp�x� � x

10.

x

1

2

3

4

−1 1 2 3

y

� 2��2x2 �23

x3�2

0�

16�

3

� 2��2

0 �4x � 2x2� dx

V � 2��2

0 x�4 � 2x� dx

h�x� � 4 � 2x

p�x� � x

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 16.

� 2��128 � 64� � 128�

� 2��8y2 �y4

4 �4

0

� 2��4

0�16y � y3� dy

V � 2��4

0y�16 � y2� dy

1

−1

−2

−3

−4

2

3

4

4 8 12

y

x

h�y� � 16 � y2

p�y� � y

Page 17: 1

Section 6.3 Volume: The Shell Method 279

20.

x1 2 3 4

4

5

−2

−1

1

2

3

y

� 2��4x32 �25

x52�4

0�

192�

5

� 2� �4

0 �6x12 � x32� dx

V � 2� �4

0 �6 � x�x dx

h�x� � x

p�x� � 6 � x

22. (a) Disk

� �100�

3 � 1125

� 1� �49615

� 100��x�3

�3�5

1

� 100��5

1x�4 dx

V � ��5

1�10

x2 �2

dx

R�x� �10x2 , r�x� � 0

(b) Shell

� 20��ln x �5

1� 20� ln 5

� 20��5

1 1x dx

V � 2��5

1x�10

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

1−1−2

2

4

6

8

10

2 3 4 5

y

x

24. (a) Disk

� 2� �a3 �95

a3 �97

a3 �13

a3� �32� a3

105

� 2� �a2x �95

a43x53 �97

a23x73 �13

x3�a

0

� 2� �a

0 �a2 � 3a43x23 � 3a23x43 � x2� dx

V � � �a

�a

�a23 � x23�3 dx

r�x� � 0

R�x� � �a23 � x23�32

x

( , 0)a

(0, )a

( , 0)−a

y (b) Same as part a by symmetry

x

( , 0)a

(0, )a

(0, )−a

y

18.

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 � 2x2

p�x� � 2 � x

Page 18: 1

280 Chapter 6 Applications of Integration

26. (a)

x

y

2

2 5

−2

z (b)

xy

2

55

z (c)

a < c < b

xy

2

5 10

z

28.

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 � x2,

2� �4

0 x�x

2� dx 30. (a)

(b) V � 2� �1

0 x1 � x3 dx � 2.3222

x1

1

1

1

1

1

3

3

4

4

2

2

4

4

y

32. (a)

(b) V � 2� �3

1

2x1 � e1x dx � 19.0162

x1 2 3 4

4

3

2

1

y 34.

Matches (e)

x

2

1

y

Volume � 1

x ��

4x � 0,y � 0,y � tan x,

36. Total volume of the hemisphere is By the Shell Method,Find such that

Diameter: 29 � 1823 � 2.920

x0 � 9 � 1823 � 1.460.

�9 � x02� 32 � 18

� ��23

�9 � x2�32�x0

0� 18 �

23

�9 � x02� 32

6 � ��x0

0 �9 � x2�12��2x� dx

x

2

1

321−1−1

−3

−2

−3 −2

y 6� � 2� �x0

0 x9 � x2 dx

x0

h�x� � 9 � x2.p�x� � x,12�4

3��r3 �23� �3�3 � 18�.

Page 19: 1

Section 6.3 Volume: The Shell Method 281

38.

� 2� 2r2R

� 4� R�� r2

2 � � �2��23��r2 � x2�32�

r

�r

� 4� R �r

�r

r2 � x2 dx � 4� �r

�r

xr2 � x2 dx

V � 4� �r

�r

�R � x�r2 � x2 dx

42. (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

40. (a)

(b)

limn→�

�abn�b � �

limn→�

R1�n� � limn→�

n

n � 1� 1

R1�n� �

abn�1� nn � 1�

�abn�b �n

n � 1

� abn�1�1 �1

n � 1� � abn�1� nn � 1�

� abn�1 � abn�1

n � 1

� �abn x � 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� nn � 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

Page 20: 1

Section 6.4 Arc Length and Surfaces of Revolution

2.

(a)

(b)

s � �7

1 �1 � � 4

3�2

dx � �53

x�7

1� 10

y� �43

y �43

x �23

d � ��7 � 12 � �10 � 22 � 10

�1, 2, �7, 10 4.

�2

27�8232 � 1 � 54.929

� � 227

�1 � 9x32�9

0

s � �9

0

�1 � 9x dx

y� � 3x12, �0, 9

y � 2x32 � 3

6.

� 1032 � 232 � 28.794

� �32

�23

�x23 � 132�27

1

�32�

27

1

�x23 � 1� 23x13� dx

� �27

1�x23 � 1

x23 dx

s � �27

1�1 � � 1

x13�2

dx

y� � x�13, �1, 27

y �32

x23 � 4 8.

� � 110

x5 �1

6x3�2

1�

779240

� 3.246

� �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, 2

y� �12

x4 �1

2x4

y �x5

10�

16x3

10.

�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. (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

14. (a)

−1 2

−2

y = 1x + 1

5

y �1

1 � x, 0 ≤ x ≤ 1 (b)

L � �1

0 �1 �

1�1 � x4 dx

1 � �y�2 � 1 �1

�1 � x4

y� � �1

�1 � x2(c) L � 1.132

282 Chapter 6 Applications of Integration

Page 21: 1

16. (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

18. (a)

−1 6

−6

2

y � ln x, 1 ≤ x ≤ 5 (b)

L � �5

1 �1 �

1x2 dx

1 � �y�2 � 1 �1x2

y� �1x

(c) L � 4.367

20. (a)

−10

−5

10

10

y � �36 � x 2, 3�3 ≤ x ≤ 6

x � �36 � y2 , 0 ≤ y ≤ 3 (b)

� �3

0

6�36 � y2

dy

L � �3

0 �1 �

y2

36 � y2 dy

��y

�36 � y 2

dxdy

�12

�36 � y2�12��2y (c) L � 3.142 ��!

L � �6

3�3 �1 �

x2

36 � x2 dx � �6

3�3

6�36 � x2

dx

y� �dydx

� �x

�36 � x2

y � �36 � x2

Alternatively, you can convert to a function of x.

Although this integral is undefined at a graphing utility still gives L � 3.142.x � 0,

22.

Matches (e)

s � 1

��4

0 �1 � � d

dx�tan x�

2

dx

x

84

4

2

1

, 1( (

(0, 0)

y x= tan

π

π

y

24.

(a)

(b)

(c)

(d) 160.287

s � �4

0 �1 � �4x�x2 � 4 2 dx � 159.087

� 160.151

d � ��1 � 02 � �9 � 162 � ��2 � 12 � �0 � 92 � ��3 � 22 � �25 � 02 � ��4 � 32 � �144 � 252

d � ��4 � 02 � �144 � 162 � 128.062

f �x � �x2 � 42, �0, 4

Section 6.4 Arc Length and Surfaces of Revolution 283

Page 22: 1

26. 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

1 �1 �

1x2 dx.y� �

1x

1 ≤ x ≤ e,y � ln x,

28.

Thus, there are square feet of roofing on the barn.100�47 � 4700

�12

�20

�20 �ex20 � e�x20 dx � �10�ex20 � e�x20�

20

�20� 20�e �

1e� � 47 ft

s � �20

�20�� 1

2�ex20 � e�x20�

2

dx

1 � �y�2 � 1 �14

�ex10 � 2 � e�x10 � � 12

�ex20 � e�x20�2

y� � �12

�ex20 � e�x20

y � 31 � 10�ex20 � e�x20

30.

(Use Simpson’s Rule with or a graphing utility.)n � 100

s � �299.2239

�299.2239 �1 � ��0.6899619478 sinh 0.0100333x2 dx � 1480

y� � �0.6899619478 sinh 0.0100333x

y � 693.8597 � 68.7672 cosh 0.0100333x

32.

14

�2��5 � 7.8540 � s

� 5�arcsin 45

� arcsin ��35�� � 7.8540

� �5 arcsin x5�

4

�3

� �4

�3

5�25 � x2

dx

s � �4

�3 � 25

25 � x2 dx

1 � �y�2 �25

25 � x2

y� ��x

�25 � x2

y � �25 � x2 34.

�8�

3�1032 � 532 � 171.258

�83

��x � 132�9

4

� 4��9

4

�x � 1 dx

S � 2��9

42�x�1 �

1x dx

y� �1�x

, �4, 9

y � 2�x

284 Chapter 6 Applications of Integration

Page 23: 1

36.

� �2��58

x2�6

0� 9�5 �

S � 2� �6

0 x2�5

4 dx

1 � �y�2 �54

, �0, 6

y� �12

y �x2

38.

��

6�3732 � 1 � 117.319

� ��

6�1 � 4x232�

3

0

��

4�3

0�1 � 4x212 �8x dx

S � 2��3

0x�1 � 4x2 dx

y� � �2x

y � 9 � x2, �0, 3

40.

� 2� �e

1 �x2 � 1 dx � 22.943

S � 2� �e

1 x�x2 � 1

x2 dx

1 � �y�2 �x2 � 1

x2 , �1, e

y� �1x

y � ln x 42. The precalculus formula is the distance formula betweentwo points. The representative element is

���xi2 � ��yi2 ��1 � ��yi

�xi�2

�xi.

44. The surface of revolution given by will be larger. is larger for f1.r�xf1

46.

� 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 48. From Exercise 47 we have:

x

x y r2 2 2+ =

r

r h−

h

y

� 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 dx�r2 � x2

S � 2� �a

0

rx�r2 � x2

dx

Section 6.4 Arc Length and Surfaces of Revolution 285

Page 24: 1

50. (a) We approximate the volume by summing 6 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

�12

� �51 � 482 ��9 � � 3

2��2

�12

�70 � 662 ��9 � � 4

2��2

�12

� �66 � 582 ��9 � � 8

2��2

�12

S � �50 � 65.52 ��9 � �15.5

2� �2

�12

� �65.5 � 702 ��9 � �4.5

2��2

�12

� �50 � 65.52 ��9 � �65.5 � 50

2� �2

�12

.

r

R

sh

S1 � ��32 � �65.5 � 502� �

2

�12

� 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��y2 dy

V � �18

0 �r2 dy � 5275.9 cubic inches

52. Individual project, see Exercise 50, 51.

(c) You cannot evaluate this definite integral, since the integrand is not defined at Simpson’s Rule will not work for thesame reason. Also, the integrand does not have an elementary antiderivative.

x � 3.

56. Essay

54. (a)

Ellipse:

−5 5

−4 9x2 2y

4+ = 1

4

y2 � �2�1 �x2

9

y1 � 2�1 �x2

9

x2

9�

y2

4� 1 (b)

L � �3

0�1 �

4x2

81 � 9x2 dx

��2x

9�1 �x2

9

��2x

3�9 � x2

y� � 2�12��1 �

x2

9 ��12

��2x9 �

y � 2�1 �x2

9, 0 ≤ x ≤ 3

286 Chapter 6 Applications of Integration

Page 25: 1

Section 6.5 Work

Section 6.5 Work 287

2. W � Fd � �2800��4� � 11,200 ft � lb 4. W � Fd � �9�2000��� 12 �5280�� � 47,520,000 ft � lb

6. is the work done by a force F moving an object along a straight line from to x � b.x � aW � �b

a

F�x� dx

8. (a)

(b)

(c)

(d) W � �9

0 �x dx �

23

x3�29

0�

23

�27� � 18 ft � lbs

W � �9

0

127

x2 dx �x3

819

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 10.

� 40 in � lb 3.33 ft � lb

W � �10

0 54

x dx � �58

x210

6

12.

� 28000 n � cm � 280 Nm

W � �70

0 F�x� dx � �70

0 807

x dx �40x2

7 70

0

800 � k�70� ⇒ k �807

F�x� � kx 14.

� 240 ft � lb

W � 2 �4

0 15x dx � �15x2

4

0

15 � k�1� � k

F�x� � kx

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 18.

limh→�

W � 20,000 mi�ton 2.1 � 1011 ft � lb

��80,000,000

h� 20,000

W � �h

4000

80,000,000x2 dx � ��

80,000,000x

h

4000

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

Page 26: 1

22. The bottom half had to be pumped a greater distance then the top half.

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) W �49

�62.4�� �6

4 y3 dy � �4

9�62.4���1

4y4

6

4 7210.7� ft � lb

W �49

�62.4�� �2

0 y3 dy � �4

9�62.4���1

4y4

2

0 110.9� ft � lb

y

62.4��23

y 2

y

�� 23

y 2

y

x

y

4321−1−2−3−4

7

5

4

3

2

y

28. Volume of each layer:

Weight of each layer:

Distance:

� 3252.6 ft � lb

� 55.6�18y �3y2

2�

y3

3 3

0

W � �3

055.6�6 � y��y � 3� dy � 55.6�3

0�18 � 3y � y2� dy

6 � y

55.6�y � 3� y

x1 2 3 4

4

3

2

1

(2, 3)

y x= 3 3−

yy � 3

3�3� y � �y � 3� 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 .

� 1008�192

�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

V � 12�2���25�4� � y2 y

x

6

3

12

963−3−3

−6

−9 −6

192

− y

Ground

level

y

288 Chapter 6 Applications of Integration

Page 27: 1

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

34. The work required to lift the chain is (from Exercise 31). The work required to lift the 500-pound load isThe work required to lift the chain with a 100-pound load attached is

W � 337.5 � 7500 � 7837.5 ft � lbs

W � �500��15� � 7500.337.5 ft � lb

36. W � 3 �6

0 �12 � 2y� dy � ��3

4�12 � 2y�2

6

0�

34

�12�2 � 108 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 40.

� 2500 ln 3 2746.53 ft � lb

W � �3

1 2500

V dV � �2500 ln V

3

1

2500 �k1

⇒ k � 2500

p �kV

42. (a)

(b)

24,88.889 ft � lb

W 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

(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

44. W � �4

0 �ex2

� 1100 dx 11,494 ft � lb 46. W � �2

0 1000 sinh x dx 2762.2 ft � lb

Section 6.5 Work 289

Page 28: 1

Section 6.6 Moments, Centers of Mass, and Centroids

2. x �7��3� � 4��2� � 3�5� � 8�6�

7 � 4 � 3 � 8�

1711

4. x �12��6� � 1��4� � 6��2� � 3�0� � 11�8�

12 � 1 � 6 � 3 � 11� 0

6. The center of mass is translated k units as well. 8. (Person on left)

x � 323 feet

750x � 2750

200x � 2750 � 550x

200x � 550�5 � x�

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 12.

x

m1

m4

m3

m2

(2, 3)

(2, 2)−

(6, 8)

( 1, 5)−

−2−2

2 6 8

8

6

2

y �x, y� � �6227

, 6427�

y �

12�3� � 6�5� �152

�8� � 15��2�

12 � 6 �152

� 15�

9640.5

�6427

x �

12�2� � 6��1� �152

�6� � 15�2�

12 � 6 �152

� 15�

9340.5

�6227

14.

�x, y� � �32

, 35�

x �My

m�

2�

43

�3

2

My � ��2

0x�1

2x2� dx �

12

��2

0x3 dx � ��

8x4�

2

0� 2�

y �Mx

m�

45

43

�3

5

Mx � ��2

0

12 �

12

x2��12

x2� dx ��

8�2

0x4 dx � � �

40x5�

2

0�

3240

� �45

y

x1

1

2

2

32

35

,( )

m � ��2

0

12

x2 dx � ��x3

6 �2

0�

43

290 Chapter 6 Applications of Integration

Page 29: 1

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

m � � �1

0 �x � x� dx � ��2

3x32 �

x2

2�1

0�

6

x1

1

14

14

12

34

34

12

( , )x y

y

18.

2−2−1

1

2

3

4

5

4 6 8 10

y

x

185

52

,( )

(9, 4)

(0, 1)

�x, y� � �185

, 52�

x �My

m�

815

92

�18

5; y �

Mx

m�

454

92

�5

2

� � �4865

� 81� �815

My � ��9

0x�x � 1 �

13

x � 1� dx � ��9

0�x32 �

13

x2� dx � ��25

x52 �19

x3�9

0

��

2�x2

6�

x3

27�

43

x32�9

0�

2�272

� 27 � 36� �453

��

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

Section 6.6 Moments, Centers of Mass, and Centroids 291

Page 30: 1

x

−4

−4−8

12

4

8

8

( , )x y

y20.

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

m � 2� �8

0 �4 � x23� dx � 2��4x �

35

x53�8

0�

128�

5

x1 2

1

2

( , )x y

y22.

�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

m � � �2

0 �2y � y2� dy � ��y2 �

y3

3�2

0�

4�

3

1

1

−1

2

2

3

3 4x

( , )x y

y24.

�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

m � � �2

�1 ��y � 2� � y2� dy � ��y2

2� 2y �

y3

3 �2

�1�

9�

2

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

292 Chapter 6 Applications of Integration

Page 31: 1

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

30.

Therefore, the centroid is �2.2, 0.3�.

y �Mx

m� 0.3

x �My

m� 2.2

My � � �4

0 x2e�x2 dx � 5.1732�

Mx � � �4

0 �xe�x2

2 ��xe�x2� dx ��

2 �4

0 x2e�x dx � 0.7619�

m � � �4

0 xe�x2 dx � 2.3760�

−1

−1

5

( , )x y

3

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

3m � � �2

�2

8x2 � 4

dx � 6.2832�

34.

This is the point of intersection of the diagonals.

�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

y2 � a2y�c

0�

12ac

�abc � a2c� �12

�b � a�

�1

2ac �c

0 �2ab

cy � a2� dy

x �1ac

12

�c

0 ��b

cy � a�

2

� �bc

y�2

� dy

1A

�1ac

x

( , )x y

y = cb

x

cb ( )x a−y =

(0, 0)

( , )b c

( , 0)a

( + , )a b c

yA � bh � ac

Section 6.6 Moments, Centers of Mass, and Centroids 293

Page 32: 1

(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

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

36. by symmetry

�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

r

r

x

y

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

n 1 2 3 4

917 b7

13 b59 b3

5 by

294 Chapter 6 Applications of Integration

Page 33: 1

42. Let be the top curve, given by The bottom curve is

(a)

�x, y � � �0, 1.078�

y �Mx

A�

4.97974.62

� 1.078

�16

�29.878� � 4.9797

f

d

−2 20

3 �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

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

(b)

(c)

�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

44. Centroids of the given regions: and

Area:

�x, y � � �2514

, 1514�

y �3�32� � 2�12� � 2�1�

7�

1527

�1514

x �3�12� � 2�2� � 2�72�

7�

2527

�2514

A � 3 � 2 � 2 � 7

�72

, 1��12

, 32�, �2,

12�,

x

1

2

3

4

41 2 3

m1m3m2

y

46.

By symmetry,

�x, y � � �0, 55232128� � �0, 2.595�

y ��74��716� � �28764��5516�

�74� � �28764� �16,5696384

�55232128

x � 0.

m2 �78 �6 �

78� �

28764

, P2 � �0, 5516�

m1 �78

�2� �74

, P1 � �0, 7

16�

x1

2

3

4

5

−1 12

32

32

− 12

m1

m2

y

Section 6.6 Moments, Centers of Mass, and Centroids 295

Page 34: 1

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� 50. V � 2�rA � 2��3��4�� � 24�2

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

54. A planar lamina is a thin flat plate of constant density. The center of mass is the balancing point on the lamina.

�x, y�

56. Let R be a region in a plane and let L be a line such that L does not intersect the interiorof R. If r is the distance between the centroid of R and L, then the volume V of the solid of revolution formed by revolving R about L is

where A is the area of R.

V � 2�rA

58. The centroid of the circle is The distance traveled by the centroid is The arc length of the circle is also Therefore,

x

−2

−1

−1 1

1

3

2

d

C

y

S � �2���2�� � 4�2.2�.2�.�1, 0�.

296 Chapter 6 Applications of Integration

Page 35: 1

Section 6.7 Fluid Pressure and Fluid Force

Section 6.7 Fluid Pressure and Fluid Force 297

2. F � PA � �62.4�5���16� � 4992 lb 4.

� 62.4�4��48� � 11,980.8 lb

F � 62.4�h � 4��48� � �62.4��h��48�

6.

Force is one-third that of Exercise 5.

x1−1−2 2

4

2

1

y

�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

h�y� � 3 � y 8.

x1−1

−3

1

y

� �62.423�4 � y2�3�2�

0

�2� 332.8 lb

F � 62.4 �0

�2 ��y��2��4 � y2 dy

L�y� � 2�4 � y2

h�y� � �y

10.

� �62.449�9 � y2�3�2�

0

�3� 748.8 lb

� 62.423 �0

�3 �9 � y2�1�2��2y� dy

F � 62.4 �0

�3 ��y� 4

3�9 � y2 dy

L�y� �43�9 � y2

h�y� � �yx

1−1

−1

−2

−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�2�2

0� �3�2y � 18y �

y3

3�

6�2 � 12

y�3�2

3�2�2�

F � 2�9800���3�2�2

0�1 � 3�2 � y�y dy � �3�2

3�2�2�1 � 3�2 � y��3�2 � y� dy�

L2�y� � 2�3�2 � y� �upper part�

L1�y� � 2y �lower part�23

23

2 1+ 3

2

3

3

y

x1−1−2−3 2 3

h�y� � �1 � 3�2� � y

Page 36: 1

298 Chapter 6 Applications of Integration

14.

x−3 −2 −1 1 2 3

6

y

� 9800�6y �y2

2 �5

0� 171,500 Newtons

F � 9800 �5

0 1�6 � y� dy

L�y� � 1

h�y� � 6 � y 16.

x2−2

−6

−4

2

y

� ��140.7��4�3 2

3�9 � y2�3�2�0

�3� 3376.8 lb

��140.7��4�

3 �0

�3 �9 � y2 ��2y� dy

F � 140.7 �0

�3 ��y��2�4

3�9 � y2 dy

L�y� � 243�9 � y2

h�y� � �y

18.

� 140.7 �452

� 15� � 1055.25 lb

� 140.7 ��52

y2 �59

y3�0

�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

x2 3 4 6

−1

−2

−3

−4

−5

1

y

(5, −3)

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��9�4� � y2 �

32 .

F � 42 �3�2

�3�2 3

2� y�9 � 4y2 dy � 63 �3�2

�3�2 �9 � 4y2 dy �

214

�3�2

�3�2 �9 � 4y2 ��8y� dy

L�y� � 212�9 � 4y2

h�y� �32

� y

22. (a)

(b) F � wk�r2 � �62.4��5���32� � 2808� lbs

F � wk�r2 � �62.4��7���22� � 1747.2� lbs 24. (a)

(b) F � wkhb � �62.4�175 �5��10� � 10,608 lbs

F � wkhb � �62.4�112 �3��5� � 5148 lbs

Page 37: 1

Review Exercises for Chapter 6

26. From Exercise 21:

F � 64�15���12�

2

� 753.98 lb

28.

Solving for x, you obtain

x

2

4

1

5

321−1−1

−3 −2

y

� 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�

h�y� � 3 � y 30.

x−6 −4 −2 2 4 6

10

8

6

y

� 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�

h�y� � 12 � y

32. Fluid pressure is the force per unit of area exerted by afluid over the surface of a body.

34. The left window experiences the greater fluid forcebecause its centroid is lower.

Review Exercises for Chapter 6 299

2.

x1 2 3 4

1

2

3

5

6

6

(5, 4)

5, 125( )

12

, 4 )(

y

� 4x �1x�

5

12�

815

A � �5

12 �4 �

1x2� dx 4.

( 1, 1)−

x

1

−2

12

32

32

− 12

y

� �y � 1�3

3 �1

0�

13

� �1

0 �y � 1�2 dy

� �1

0 �y2 � 2y � 1� dy

A � �1

0 ��y2 � 2y� � ��1� dy

Page 38: 1

300 Chapter 6 Applications of Integration

6.

x

−1

−2

1

2

3

2 3 4 5

(2, 1)−

(5, 2)

y

� 2y �12

y2 �13

y3�2

�1�

92

� �2

�1 �2 � y � y2� dy

A � �2

�1 ��y � 3� � �y2 � 1� dy 8.

3

1

xπ2

π3

π6

2π3

5π6

π6 , 2( ) 5π

6, 2( )

y

� 22�

3� ln�2 � �3�� � 1.555

� 2��� � 0 � �

3� ln�2 � �3���

� 22x � ln�csc x � cot x���2

�6

A � 2 ���2

�6 �2 � csc x� dx

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.

14.

( 2, 8)−

(0, 0)

(2, 8)

−4 4

−2

10

� 243

x3 �15

x5�2

0�

12815

� 8.5333

� 2 �2

0 �4x2 � x4� dx

A � 2 �2

0 �2x2 � �x4 � 2x2� dx 16.

x2

2

3 4 5

1

−1

−2

3

(1, 0)

(5, 2)

y

� 23

�x � 1�32 �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 � 1

Page 39: 1

Review Exercises for Chapter 6 301

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

22. (a) Shell

x2

2

3

4

4

1

1

3

y

V � 2� �2

0 y3 dy � �

2y4�

2

0� 8�

(b) Shell

(d) Disk

x2

2

3

4

5

4

1

1

3

y

� �15

y5 �23

y3�2

0�

176�

15

� � �2

0 �y4 � 2y2� dy

V � � �2

0 ��y2 � 1�2 � 12 dy

x2 3

4

4

1

1

3

y

� 2�23

y3 �14

y4�2

0�

8�

3

� 2� �2

0 �2y2 � y3� dy

V � 2� �2

0 �2 � y�y2 dy

20. (a)

00

10

40

R1�t� � 5.2834�1.2701�t � 5.2834 e0.2391t (b)

Difference � �15

10 �R1�t� � R2�t� dt � 171.25 billion dollars

R2�t� � 10 � 5.28 e0.2t

(c) Disk

x2

2

3

4

4

1

1

3

y

� �

5y5�

2

0�

32�

5 V � � �2

0 y4 dy

Page 40: 1

302 Chapter 6 Applications of Integration

24. (a) Shell

(0, )b

( , 0)a

x

x2 y2

a2 b2+ = 1

y

� �4�b3a

�a2 � x2�32�a

0�

43

�a2b

��2�b

a �a

0 �a2 � x2�12��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

26. Disk

x1 2

2

3

−1−2

−1

y

��2

2

� 2���

4� 0�

� 2� arctan x�1

0

V � 2� �1

0 1�1 � x2�

2

dx

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

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

Page 41: 1

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,

� �3�4a3

3 �

V � �3 �a

�a

�a2 � x2� dx � �3a2x �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

2x2�2

y� �12

x2 �1

2x2

y �x3

6�

12x

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 � 0

f��x� � sec2 x,f �x� � tan x

Review Exercises for Chapter 6 303

38.

� 4��23��x � 1�32�

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�x

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 ftmin.

dVdt

��

9 �dhdt �

V � �r2h � ��19�h

dVdt

�4 galmin � 12 galmin

7.481 galft3� �

87.481

ft3min

Page 42: 1

304 Chapter 6 Applications of Integration

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

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� � ���29�x � 6,��43�x � 16,

0 ≤ x ≤ 9 9 ≤ x ≤ 12

W � �b

a

F�x� dx

48.

�x, y � � �1, 175 �

�3

649x � 6x2 �43

x3 �15

x5�3

�1�

175

( , )x y

−3 3 6

3

6

9

x

y

�3

64 �3

�1 �9 � 12x � 4x2 � x4� dx y � � 3

32�12

�3

�1 ��2x � 3�2 � x4 dx

�3

3232

x2 �23

x3 �14

x4�3

�1� 1 x �

332

�3

�1 x�2x � 3 � x2� dx �

332

�3

�1 �3x � 2x2 � x3� dx

1A

�3

32

A � �3

�1 ��2x � 3� � x2 dx � x2 � 3x �

13

x3�3

�1�

323

50.

�x, y � � �103

, 4021�

�12 �

516�

37

x73 �1

12x3�

8

0�

4021

y � � 516�

12

�8

0 �x43 �

14

x2� dx

� �516

38

x83 �16

x3�8

0�

103

x �5

16 �8

0 x�x23 �

12

x� dx

1A

�5

16

A � �8

0 �x23 �

12

x� dx � 35

x53 �14

x2�8

0�

165

x

−2

2

2

6

4

4

6 8

( , )x y

y