## Problem Statement

Calculate the volumes of bodies bounded by the surfaces.

$$
\frac{x^{2}}{9}+\frac{y^{2}}{16}-\frac{z^{2}}{64}=-1, z=16
$$
## Solution

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

$$
\frac{x^{2}}{9}+\frac{y^{2}}{16}=\frac{z^{2}}{64}-1
$$

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-16.jpg?height=705&width=962&top_left_y=927&top_left_x=978)

$$
\begin{aligned}
& \frac{x^{2}}{9 \cdot \frac{z^{2}-64}{64}}+\frac{y^{2}}{16 \cdot \frac{z^{2}-64}{64}}=1 \rightarrow a=\frac{3}{8} \sqrt{z^{2}-64} ; b=\frac{1}{2} \sqrt{z^{2}-64} \\
& \Rightarrow S=\pi a b=\pi \cdot \frac{3}{8} \sqrt{z^{2}-64} \cdot \frac{1}{2} \sqrt{z^{2}-64}=\frac{3 \pi}{16} \cdot\left(z^{2}-64\right)
\end{aligned}
$$

$$
\begin{aligned}
& V=\int_{8}^{16} S(z) d z=\frac{3 \pi}{16} \int_{8}^{16}\left(z^{2}-64\right) d z=\left.\frac{3 \pi}{16}\left(\frac{z^{3}}{3}-64 z\right)\right|_{8} ^{16}= \\
& =\frac{3 \pi}{16}\left(\frac{16^{3}}{3}-64 \cdot 16-\frac{8^{3}}{3}+64 \cdot 8\right)=\frac{\pi}{16}\left(3 \cdot \frac{4096}{3}-3 \cdot 1024-3 \cdot \frac{512}{3}+3 \cdot 512\right)= \\
& =\frac{\pi}{16}(4096-3072-512+1536)=128 \pi
\end{aligned}
$$

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\%D1\%82 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_20-9$ »

Categories: Kuznetsov's Integral Problems 20 | Integrals | Problems for Checking

- Last modified: 06:02, 23 June 2010.
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## Problem Kuznetsov Integrals 20-10

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