So how to find the potential energy of the water? Imagine the cylinder as a series of disks, each with a height dh. Each disk will have a volume of:
dV = (pi)r^2 * dh
Thus each disk will contain a mass of water equal to:
dm = (rho) * dV (where rho is the density of water.)
dm = (rho)*(pi)r^2 * dh
Thus the potenial energy of the disk will be:
d(PE) = dm * g * h, where h is the height of the disk.
d(PE) = g*(rho)*(pi)*r^2 * hdh
Now integrate over the height coordinate of the cylinder:
Int(d(PE)) = PE = Int(g*(rho)*(pi)*r^2 * hdh, h, 0, -H) where H is the height of the cylinder. (Recall that we require + to be upward, so we are integrating the cylinder from 0 to -H: we are integrating "upside down.")
PE = -(1/2)g*(rho)*(pi)*r^2*H^2
Thus W = -PE = -(1/2)g*(rho)*(pi)*r^2*H^2
b) This is the same idea as a) except that now we are pumping the water up 5 ft more. So add a Mg(5 ft) to the answer in a) where M is the total mass of the water in the cylinder.