A tank is full of water. Find the work required to pump the water out of the spout. (Use 9.8 m/s^2 for g. Use 1000 kg/m^3 as the density of water. Assume r = 9 m and h = 3 m.)

How high is the top of the spout from the center of gravity of the tank ???
radius + 3 = 9 + 3 = 12 meters

So you have to lift the weight of the water 12 meters.
mass of water in tank = m = (4/3) pi r^3 * 1000 Kg
weight = w = m g = 9.8 * m
work = weight * distance lifted = (4/3) pi 9^3 * 1000 Kg * 9.8 * 12 Joules

Note, You will need more because the water has velocity when it exits but we do not have data for that
(1/2) m v^2 :)

For now, assuming that the weight of water is 1 kg/m^2 (so we just have to worry about figuring the volume * distance)

moving the center of mass to the height of the top of the sphere (forget the spout -- assume the water just runs out the top) requires

4/3 π * 9^3 * 9 = 8748π J

Doing the calculus, and figuring that the cross-section area of a slice of water at a distance of y units above the bottom of the sphere is π√(81-(9-y)^2) (this is correct, because even though 9-y goes from 9..-9, it is squared)

∫[0..18] πr^2h dy = ∫[0..18] π(81-(9-y)^2)(18-y) dy = 8748π J

Now just multiply that by the correct density (N/m^3) and increase the height by the length of the spout, and you should get Damon's answer.