posted by COFFEE .
The hemispherical 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.
What is shown is just the tank (a hemisphere) with a radius of 5 ft.
First I calculated the Volume of the hemisphere, V = (2/3)*pi*r^3
V = (2/3)*pi*125 = (250/3)*pi
Then I took the integral of: Volume*5y*dy from 0 to 5.
Which equals: ((250/3)*pi)*(5/2)y^2 evaluated at 5 and 0.
I came up with 16362.5 ft*lb.
Am I using the wrong method?
The work required depends upon where the water is extracted. I assume you are pumping out the top.
I don't see why you claim that the energy is
Volume*5y*dy from 0 to 5
The work required is the weight of each differential slab of height dy, multiplied by the distance it must be lifted, 5 - y, integrated from 0 to 5. The area of each slab is different. It depends upon y.