Page 322 - Calculus Demystified

Chapter 8
6
√ 309
2
2
(e) π[y + 2] dy
2

π/2
2
(f) π[sin x] dx
0
4
#
4. (a) 2π · x 2/3 · 1 +[(2/3)x −1/3 2
] dx
0
√

3 #
2
2
(b) 2π · y · 1 +[2y] dy
0
√

3
2
2
(c) 2π ·[x − 2]· 1 +[2x] dx
0

π
2
(d) 2π · sin x · 1 +[cos x] dx
0
2 #

2
2
(e) 2π ·[y + 2]· 1 +[2y] dy
1

1
3
2 2
(f) 2π · x · 1 +[3x ] dx
0
5. The depth of points in the window ranges from 6 to 10 feet. At depth x
√
2
in this range, the window has chord of length 2 16x − x − 60. Thus the
total pressure on the lower half of the window is
10

2
P = 62.4 · x · 2 16x − x − 60 dx .
8
√
6. At depth x, the corresponding subtriangle has side-length 2(5 − x/ 3).
Therefore the total pressure on one end of the pool is
√
5 3 √

P = 62.4 · x · 2(5 − x/ 3)dx.
0
7. Let t = 0 be the moment when the climb begins. The weight of the sack at
time t is then 100−t pounds. Then the work performed during the climb is
8

W = (100 − t) · 5 dt.
0
Thus
2 8
5t 320 0
W = 500t − = 4000 − − 100 · 0 − = 3840 ft lbs.
2 0 2 2