A 10-ft trough filled with water has semicircular cross section of diameter 4 ft. How much work is done in pumping all the water over the edge of the trough? Assume that water weighs $\displaystyle 62.5 lb/ft^3$.

$\displaystyle \int_{0}^{2}10*62.5*2x*dy = \int_{0}^{2}1250*\sqrt{1 -y^2}dy$$\displaystyle =\frac{1250(1-y^2)^(3/2)}{3y}=5356 $lb.ft.