# Calculus

posted by
**COFFEE** on
.

The tank shown is full of water. Given that water weighs 62.5 lb/ft3, find the work required to pump the water out of the tank. The tank shown is a hemisphere with r = 5 ft. The water is to be pumped out at the top.

First I solved for ri, ri / (5 - xi) = 5 / 5

ri = 5 - xi

Vi = pi*(ri)^2*delta x

Vi = pi*(5 - xi)^2*delta x

mi = density x volume

mi = 62.5*pi*(5 - xi)^2*delta x

Fi = mi x g, g = 9.8 m/s^2 or 32.152231 ft/s^2

Fi = (32.152231 ft/s^2)*(62.5*pi* (5 - xi)^2*delta x)

Fi = 2009.5*pi*(5 - xi)^2*delta x

Wi = (Fi)*(xi) = (2009.5*pi)(xi)(5 - xi)^2*delta x

W = INTEGRAL from 0 to 5 of: 2009.5*pi*(xi)(5 - xi)^2*dx

W = 2009.5*pi INTEGRAL xi*(5 - xi)^2*dx, from 0 to 5.

W = 2009.5*[12.5x^2 - (10x^3)/3 + (x^4)/4] evaluated at 5 and 0

W = 2009.5*[12.5*(5)^2 - (10*(5^3))/3 + (5^4)/4]

W = 100220 ft-lb

I am not sure which part I did wrong. Could you please point me in the right direction? Thanks!

Your calculation of force included g. But you used lbs for mass. Lbs is a force unit (it really isnt mass). So delete the 32.15 ft/s^2 from the force.

Work=NTEGRAL from 0 to 5 of: 62.5*pi*(xi)(5 - xi)^2*dx