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Work Done to Pump Water out of a Tank

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A tank is in the shape of an inverted cone (pointy at the top) 6 feet high and 8 feet across at the base. The tank is filled to a depth of 3 feet. How much work is done in emptying the tank through a hole at the top? (Weight density of water is 62.4 lb/ft^3).

*I found the distance to be 6-y and used the disk method to solve. I think volume of disk is pi(2/3y)^2 change in y. I set up the problem:
W= The integral (from 3 to 6) 62.4pi(2/3y)^2 (6-y)dy and found the answer to be approx. 1684.6pi

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Solution Summary

The work done to pump water out of a tank is calculated using an integral.

Solution Preview

First you calculate the position of the center of mass of the filled portion of cone:
y_c = (integral[0 to 3]{y*dV})/(integral[0 to 3]{dV})
where,
dV = volume of small strip of thickness dy at the height y from the bottom
=> dV = Ay*dy
Ay = area of cross section at height y from the ...

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