Calculate the work required to pump all the water out

Calculate the work required to pump all the water out of a cylindrical barrel with base radius $R$ and height $H$, filled to the top.

Step 1: Mass of a thin horizontal slice

A slice at height $y$ (measured from the base) with thickness $dy$ has volume $dV=\pi R^2\,dy$ and mass $dm=\rho\pi R^2\,dy$, where $\rho$ is water's density.

Step 2: Work to lift the slice

It must rise a distance $(H-y)$ to reach the rim, so $dW=(dm)g(H-y)=\rho\pi R^2g(H-y)\,dy$.

Step 3: Integrate over the barrel

\[W=\rho\pi R^2g\int_0^H(H-y)\,dy=\rho\pi R^2g\left[Hy-\frac{y^2}{2}\right]_0^H=\rho\pi R^2g\cdot\frac{H^2}{2}.\]

\[\boxed{W=\frac{\rho\pi R^2gH^2}{2}}\]

Step 4: Interpretation via the center of mass

The total mass is $M=\rho\pi R^2H$, and the water's center of mass sits at height $H/2$ (by symmetry), a distance $H/2$ below the rim, so $W=Mg\cdot(H/2)$, matching Step 3 exactly — pumping out the barrel does exactly the work of lifting all the water's mass through the height of its own center of mass.