## Problem Statement

Calculate the volumes of the bodies bounded by the surfaces.

$$
\frac{x^{2}}{25}+\frac{y^{2}}{9}-z^{2}=1, z=0, z=2
$$
## Solution

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

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

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-47.jpg?height=923&width=1108&top_left_y=992&top_left_x=817)

$$
\begin{aligned}
& \frac{x^{2}}{25 \cdot\left(z^{2}+1\right)}+\frac{y^{2}}{9 \cdot\left(z^{2}+1\right)}=1 \rightarrow a=5 \sqrt{z^{2}+1} ; b=3 \sqrt{z^{2}+1} \\
& \Rightarrow S=\pi a b=\pi \cdot 5 \sqrt{z^{2}+1} \cdot 3 \sqrt{z^{2}+1}=15 \pi \cdot\left(z^{2}+1\right) \\
& V=\int_{0}^{2} S(z) d z=15 \pi \int_{0}^{2}\left(z^{2}+1\right) d z=\left.15 \pi\left(\frac{z^{3}}{3}+z\right)\right|_{0} ^{2}= \\
& =15 \pi\left(\frac{2^{3}}{3}+2-\frac{0^{3}}{3}-0\right)=5 \pi\left(3 \cdot \frac{8}{3}+3 \cdot 2-0-0\right)=70 \pi
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-28$ "

Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals | Problems for Checking

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## Problem Kuznetsov Integrals 20-29

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