## Problem Statement

Calculate the volumes of bodies bounded by the surfaces.

$$
\frac{x^{2}}{25}+\frac{y^{2}}{9}-\frac{z^{2}}{100}=-1, z=20
$$
## Solution

In the section of the given figure by the plane $z=$ const, there is an ellipse:

$$
\frac{x^{2}}{25}+\frac{y^{2}}{9}=\frac{z^{2}}{100}-1
$$

The area of an ellipse described by the formula: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

is $\pi \cdot a \cdot b$

Let's find the radii of the ellipse:

![](https://cdn.mathpix.com/cropped/2024_05_22_348f4289c7b5f46bc246g-42.jpg?height=617&width=831&top_left_y=822&top_left_x=1109)

$$
\begin{aligned}
& \frac{x^{2}}{25 \cdot \frac{z^{2}-100}{100}}+\frac{y^{2}}{9 \cdot \frac{z^{2}-100}{100}}=1 \rightarrow a=\frac{1}{2} \sqrt{z^{2}-100} ; b=\frac{3}{10} \sqrt{z^{2}-100} \\
& \Rightarrow S=\pi a b=\pi \cdot \frac{1}{2} \sqrt{z^{2}-100} \cdot \frac{3}{10} \sqrt{z^{2}-100}=\frac{3 \pi}{20} \cdot\left(z^{2}-100\right)
\end{aligned}
$$

$$
\begin{aligned}
& V=\int_{10}^{20} S(z) d z=\frac{3 \pi}{20} \int_{10}^{20}\left(z^{2}-100\right) d z=\left.\frac{3 \pi}{20}\left(\frac{z^{3}}{3}-100 z\right)\right|_{10} ^{20}= \\
& =\frac{3 \pi}{20}\left(\frac{20^{3}}{3}-100 \cdot 20-\frac{10^{3}}{3}+100 \cdot 10\right)=\frac{\pi}{20}\left(3 \cdot \frac{8000}{3}-3 \cdot 2000-3 \cdot \frac{1000}{3}+3 \cdot 1000\right)=200 \pi
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\� \%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0 $\% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_20-24$ »

Categories: Kuznetsov's Problem Book Integrals Problem $20 \mid$ Integrals

- Last modified: 06:56, 23 June 2010.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals $20-25$

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