Answer:
1.338 * 10 ^ 9 lb-ft
Step-by-step explanation:
We have the following information:
Water density 62.4 lb / ft ^ 3
work equals force per distance, like this:
W = F * d
with the information in the statement we can deduce:
20/500 = r / y
r = y / 25
now the differential volume would be:
dv = pi * (r ^ 2) * dy
dv = pi * ((y / 25) ^ 2) * dy
dv = pi * (y ^ 2/625) * dy
weight of slice would be:
62.4 * pi * (y ^ 2/625) * dy
= 0.3136 * y ^ 2 * dy
work for the removal of slice:
(500 - y) * 0.3136 * y ^ 2 * dy
Now the total work is the integral of the previous expression that goes from y = 0 to y = 400
Total work = integral (from 0 to 400) {(500 - y) * 0.3136 * y ^ 2 * dy}
Total work = integral (from 0 to 400) {0.3136 * (500 * y ^ 2 - y ^ 3) dy}
We integrate and we have:
Total work = 0.3136 * ((500/3) * y ^ 3 - (y / 4) ^ 4)
Total work = 0.3136 * ((500/3) * 400 ^ 3 - (400/4) ^ 4) - 0.3136 * ((500/3) * 0 ^ 3 - (0/4) ^ 4)
Total work = 0.3136 * ((500/3) * 400 ^ 3 - (400 ^ 4) / 4)
Total work = 1.338 * 10 ^ 9 lb-ft
