A tank is in the shape of an inverted circular cone with height 500 ft and base radius 20 ft. It is filled with water to a height of 400 ft. Find the work required to empty the tank by pumping all of the water to the top of the tank. (Weight density of water rhog = 62.4 lb/ft3 .)

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

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