# Work Done to Pump Water out of a Tank

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

https://brainmass.com/math/calculus-and-analysis/work-done-pump-water-out-tank-11447

#### 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 ...

#### Solution Summary

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