A boiler has the shape of a hemisphere of radius R

A boiler has the shape of a hemisphere of radius $R$, sitting on a flat circular base with a domed top, and is completely filled with water. How much work is required to pump the water out through an opening at the top of the dome?

Step 1: Set up coordinates

Put the flat circular base of the boiler at $y=0$ (its center at the origin) and the apex of the dome, where the outlet sits, at $y=R$. The domed surface is the part of the sphere $x^2+y^2+z^2=R^2$ with $0\le y\le R$.

Step 2: Cross-sectional area at height $y$

From the sphere's equation, the horizontal cross-section at height $y$ is a circle of radius $\sqrt{R^2-y^2}$, so

\[A(y)=\pi(R^2-y^2).\]

Step 3: Mass of a thin slice

A slice of thickness $dy$ has $dV=\pi(R^2-y^2)\,dy$ and mass $dm=\rho\pi(R^2-y^2)\,dy$.

Step 4: Work to lift the slice

It must rise a distance $(R-y)$ to the apex, so

\[dW=(dm)g(R-y)=\rho\pi g(R^2-y^2)(R-y)\,dy=\rho\pi g\left(R^3-R^2y-Ry^2+y^3\right)dy.\]

Step 5: Integrate over the boiler

\[W=\rho\pi g\int_0^R\left(R^3-R^2y-Ry^2+y^3\right)dy=\rho\pi g\left[R^3y-\frac{R^2y^2}{2}-\frac{Ry^3}{3}+\frac{y^4}{4}\right]_0^R.\]

\[=\rho\pi gR^4\left(1-\frac12-\frac13+\frac14\right)=\rho\pi gR^4\cdot\frac{12-6-4+3}{12}=\frac{5\rho\pi gR^4}{12}.\]

\[\boxed{W=\frac{5\rho\pi gR^4}{12}}\]

Step 6: Interpretation via the center of mass

The total mass of water is $M=\rho\cdot\tfrac23\pi R^3$, and the centroid of a solid hemisphere sits at height $\tfrac38R$ above its flat face, a distance $R-\tfrac38R=\tfrac58R$ below the apex. Then $Mg\cdot\tfrac58R=\tfrac23\pi\rho R^3g\cdot\tfrac58R=\tfrac{5}{12}\pi\rho gR^4$, exactly matching Step 5.