Example 2.76. Calculate the work required to pump out a liquid of density $\rho$ from a fully filled tank of height $H$, which has the shape of a paraboloid of revolution defined by the equation $z=$ $=a^{2}\left(x^{2}+y^{2}\right)$.
Solution. Let us mentally slice the paraboloid with planes parallel to the XY plane (Fig. 2.14). Suppose at a depth of $H-z$, the thickness of the layer between two adjacent parallel planes is $d z$, then the volume element of the liquid between the specified planes is calculated by the formula $d V=\pi x^{2} d z$.

![](https://cdn.mathpix.com/cropped/2024_05_22_0f84e17422d7c0df8d68g-100.jpg?height=378&width=472&top_left_y=1255&top_left_x=627)

Fig. 2.14. From the equation of the parabola, we get $x^{2}=\frac{z}{a^{2}}$, hence $d V=\frac{\pi z d z}{a^{2}}$.

The weight of the elementary layer of liquid $d m=\rho g d V=\rho g \pi z \frac{d z}{a^{2}}$. To pump this mass $d m$ from a depth of $H-z$, the elementary work required is $d A=\pi \rho g z(H-z) \frac{d z}{a^{2}}$. The total work required to pump all the liquid is

$$
\begin{gathered}
A=\int_{0}^{H} \pi \rho g z(H-z) \frac{d z}{a^{2}}=\frac{\pi \rho g}{a^{2}} \int_{0}^{H} z(H-z) d z= \\
=\frac{\pi \rho g}{a^{2}} \int_{0}^{H}\left(z H-z^{2}\right) d z=\left.\frac{\pi \rho g}{a^{2}}\left(\frac{H z^{2}}{2}-\frac{z^{3}}{3}\right)\right|_{0} ^{H}=\frac{\pi \rho g H^{3}}{6 a^{2}}
\end{gathered}
$$