Thomas Calculus 13th [Solutions]

472 Chapter 6 Applications of Definite Integrals
15. The slab is a disk of area
work to pump the oil in this slab, W , is 57
2
2 y
x , thickness y , and height below the top of the tank (10 y ). So the
2
3 4 10
y
2
(10 y ) . The work to pump all the oil to top of the tank is
2
10
57 2 3 57 10y
y
W 10 y y dy
11,875 ft lb 37,306 ft-lb
0 4 4 3 4
0
y
2
16. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is (14 y )( ) and since
2
2 2 3
the tank is half full and the volume of the original cone is V 1 r h 1 5 (10) 250 ft , half the
3 3 3
volume
250
6
3
ft , and with half the volume the cone is filled to a height y, 250 1 y
y
6 3 4
3
3 3 4 500
500 2 3
14
So
57 57 y y
W 14 y y dy
60,042 ft-lb.
0 4 4 3 4
0
17. The typical slab between the planes at y and y y has a volume of
V
2
y
2
(radius) (thickness)
20
2
3
y 100 y ft . The force F required to lift the slab is equal to its weight:
2
F 51.2 V 51.2 100 y lb F 5120 y lb The distance through which F must act is about
3
500 ft.
30 30
(30 y ) ft. The work it takes to lift all the kerosene is approximately W W 5120 (30 y) y ft-lb
0 0
which is a Riemann sum. The work to pump the tank dry is the limit of these sums:
30
2 30
900
0
5120 (30 ) 5120 30 y
W y dy y
2 5120 2
(5120)(450 ) 7,238,229.48 ft-lb
0
18. (a) Follow all the steps of Example 5 but make the substitution of 64.5 lb for 57 lb . Then,
3
3
ft ft
8
8
64.5 2 64.5 10 y y 64.5 10 8 8 64.5 3
W (10 y) y dy
8 10 2 64.5 8
0 4 4 3 4 4 3 4 4 3 3
0
3
21.5 8 34,582.65 ft-lb
3 4 3 4 3
(b) Exactly as done in Example 5 but change the distance through which F acts to distance (13 y ) ft.
3 4 8
8 2 13
3 4
3
3
Then 57 57 y y
W (13 y) y dy
57 13 8 8 57 8 13 2 57 8 7
0 4 4 3 4 4 3 4 4 3 3 4
0
2
(19 )(8 )(7)(2) 53,482.5 ft-lb
19. The typical slab between the planes at y and y y has a volume of about
2
3
y y ft . The force F( y ) required to lift this slab is equal to its weight: F ( y) 73
V
2
(radius) (thickness)
2
73 y y 73 y y lb. The distance through which F( y ) must act to lift the slab to the top of the
reservoir is about (4 y ) ft, so the work done is approximately W 73 y(4 y) y ft-lb. The work done
n
lifting all the slabs from y 0 ft to y 4 ft is approximately W 73 yk
4 yk
y ft-lb. Taking the limit
k 0
V
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