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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
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
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
2π
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
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
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
4π
2π
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�.
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
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
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,
(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
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
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
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
Section 6.6 Moments, Centers of Mass, and Centroids 295
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
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
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
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
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
5π
37π
π3
,
,
,
(
(
(
)
)
)
y
��
3� 2�3
� y2
� sin y�5�3
�3� sin y �
y2�
7�3
5�3
A � �5�3
�3 �1
2� cos y� dy � �7�3
5�3 �cos y �
12� dy 12. Point of intersection is given by:
(.7832, .4804)
−1 3
−2
4
� 1.189
� 3x � 2x2 �13
x3 �14
x4�0.783
0
A � �0.783
0 �3 � 4x � x2 � x3� dx
x3 � x2 � 4x � 3 � 0 ⇒ x � 0.783.
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
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
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
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
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