From a similar posted problem posted here:
http://math.stackexchange.com/questions/1185952/how-much-work-is-done-in-pumping-water-out-over-the-top-edge-in-order-to-empty
Specific weight of water (given) = 62.4 lbf/ft^3
The force required to move the water will be the specific weight x the volume
Weight = mg = F = 62.4 x pi x r^2 x Δy = 3136.57 Δy where y = height
Work is F x d or 3136.57 (6-y) Δy where y varies from the top of the tank (6) to the bottom (o)
so integrate this with respect to Δy from 0 to 6
\(\int_{0}^{6}\) 3136.57 (6-y) Δy = |6==>0 18819.42y - 1568.29 y^2 = 56458.08 j
(IF I did this correctly..There are no promises given......I hope someone can 2nd it !)