## Problem Statement

Calculate the volumes of bodies bounded by the surfaces.

$$
\frac{x^{2}}{9}+\frac{y^{2}}{4}-\frac{z^{2}}{36}=-1, z=12
$$
## Solution

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

$$
\frac{x^{2}}{9}+\frac{y^{2}}{4}=\frac{z^{2}}{36}-1
$$

The area of the 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-07.jpg?height=768&width=1040&top_left_y=998&top_left_x=905)

$$
\begin{aligned}
& \frac{x^{2}}{9 \cdot \frac{z^{2}-36}{36}}+\frac{y^{2}}{4 \cdot \frac{z^{2}-36}{36}}=1 \rightarrow a=\frac{1}{2} \sqrt{z^{2}-36} ; b=\frac{1}{3} \sqrt{z^{2}-36} \\
& \Rightarrow S=\pi a b=\pi \cdot \frac{1}{2} \sqrt{z^{2}-36} \cdot \frac{1}{3} \sqrt{z^{2}-36}=\frac{\pi}{6} \cdot\left(z^{2}-36\right)
\end{aligned}
$$

$$
\begin{aligned}
& V=\int_{6}^{12} S(z) d z=\frac{\pi}{6} \int_{6}^{12}\left(z^{2}-36\right) d z=\left.\frac{\pi}{6}\left(\frac{z^{3}}{3}-36 z\right)\right|_{6} ^{12}= \\
& =\frac{\pi}{6}\left(\frac{12^{3}}{3}-36 \cdot 12-\frac{6^{3}}{3}+36 \cdot 6\right)=\frac{\pi}{6}(576-432-72+216)=48 \pi
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_20-4$ » Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals

- Last edited: 05:45, June 23, 2010.
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## Problem Kuznetsov Integrals 20-5

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