A tank has the shape of an elliptic cylinder lying on its

A tank has the shape of an elliptic cylinder lying on its side: the base is an ellipse with horizontal semi-axis $a$ and vertical semi-axis $b$, and the cylinder's length is $L$. The tank is filled halfway (by volume) with liquid of specific gravity $\gamma=1$. Find the work needed to pump all the liquid out over the highest point of the tank.

Step 1: Width of the tank at height $y$

The ellipse is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, so at height $y$ the tank spans $x\in\left[-a\sqrt{1-y^2/b^2},\,a\sqrt{1-y^2/b^2}\right]$, a width $w(y)=2a\sqrt{1-y^2/b^2}$.

Step 2: Mass of a thin horizontal layer

A layer at height $y$, thickness $dy$, running the tank's full length $L$, has volume $dV=w(y)L\,dy$ and (with density $\rho=1000\ \text{kg/m}^3$ for $\gamma=1$) mass

\[dm=2000aL\sqrt{1-\frac{y^2}{b^2}}\,dy.\]

Step 3: Work to lift the layer

The water fills $-b\le y\le0$ and must be raised to the highest point $y=b$, a distance $(b-y)$:

\[dW=2000aLg\sqrt{1-\frac{y^2}{b^2}}\,(b-y)\,dy.\]

Step 4: Substitute $y=b\sin\theta$

Then $\sqrt{1-y^2/b^2}=\cos\theta$, $dy=b\cos\theta\,d\theta$; $y=-b\Rightarrow\theta=-\pi/2$, $y=0\Rightarrow\theta=0$:

\[W=2000aLgb^2\int_{-\pi/2}^{0}(1-\sin\theta)\cos^2\theta\,d\theta.\]

Step 5: Evaluate the two pieces

\[\int_{-\pi/2}^{0}\cos^2\theta\,d\theta=\frac{\pi}{4},\qquad \int_{-\pi/2}^{0}\sin\theta\cos^2\theta\,d\theta=\left[-\frac{\cos^3\theta}{3}\right]_{-\pi/2}^{0}=-\frac13.\]

\[\int_{-\pi/2}^{0}(1-\sin\theta)\cos^2\theta\,d\theta=\frac{\pi}{4}-\left(-\frac13\right)=\frac{\pi}{4}+\frac13.\]

Step 6: Combine

\[W=2000aLgb^2\left(\frac{\pi}{4}+\frac13\right)=2000aLgb^2\cdot\frac{3\pi+4}{12}=\frac{500aLgb^2(3\pi+4)}{3}.\]

\[\boxed{W=\frac{500aLgb^2(3\pi+4)}{3}}\]

With $g=9.8\ \text{m/s}^2$, numerically $W\approx21{,}900\,aLb^2$ joules (lengths in meters).