Calculate the work required to pump water out of a boiler

Calculate the work required to pump water out of a boiler shaped like a paraboloid of revolution, with depth $H=0.5$ m and base (opening) radius $R=0.4$ m, given that the boiler is completely full.

Step 1: Equation of the paraboloid

With the vertex at $y=0$ and the rim at $y=H$ with radius $R$, the paraboloid $y=cr^2$ (where $r^2=x^2+z^2$) must satisfy $H=cR^2$, so $c=H/R^2$ and

\[r^2(y)=\frac{R^2}{H}y.\]

Step 2: Cross-sectional area

\[A(y)=\pi r^2(y)=\frac{\pi R^2}{H}y.\]

Step 3: Work to lift a thin slice

A slice of thickness $dy$ at height $y$ has mass $dm=\rho A(y)\,dy$ and must rise $(H-y)$ to the rim:

\[dW=\rho g(H-y)A(y)\,dy=\frac{\rho\pi R^2g}{H}\,y(H-y)\,dy.\]

Step 4: Integrate

\[W=\frac{\rho\pi R^2g}{H}\int_0^H(Hy-y^2)\,dy=\frac{\rho\pi R^2g}{H}\left[\frac{Hy^2}{2}-\frac{y^3}{3}\right]_0^H=\frac{\rho\pi R^2g}{H}\cdot H^3\left(\frac12-\frac13\right).\]

\[\boxed{W=\frac{\rho\pi R^2gH^2}{6}}\]

Step 5: Numerical value

With $\rho=1000\ \text{kg/m}^3$, $g=9.8\ \text{m/s}^2$, $R=0.4$ m, $H=0.5$ m:

\[W=\frac{1000\cdot\pi\cdot0.16\cdot9.8\cdot0.25}{6}=\frac{392\pi}{6}\approx205\ \text{J}.\]

Step 6: Center-of-mass check

The water's volume is $V=\int_0^HA(y)\,dy=\pi R^2H/2$, giving mass $M=\rho\pi R^2H/2$; comparing, $W=\rho\pi R^2gH^2/6=(MgH)/3$, so the effective lifting height is $H/3$ — the known centroid height (from the vertex) of a solid paraboloid of revolution, confirming the setup.