## Problem Statement

Calculate the volumes of the bodies bounded by the surfaces.

$$
\frac{x^{2}}{4}+y^{2}-z^{2}=1, z=0, z=3
$$
## Solution

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

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

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-14.jpg?height=922&width=1111&top_left_y=995&top_left_x=815)

$$
\begin{aligned}
& \frac{x^{2}}{4 \cdot\left(z^{2}+1\right)}+\frac{y^{2}}{z^{2}+1}=1 \rightarrow a=2 \sqrt{z^{2}+1} ; b=\sqrt{z^{2}+1} \\
& \Rightarrow S=\pi a b=\pi \cdot 2 \sqrt{z^{2}+1} \cdot \sqrt{z^{2}+1}=2 \pi \cdot\left(z^{2}+1\right) \\
& V=\int_{0}^{3} S(z) d z=2 \pi \int_{0}^{3}\left(z^{2}+1\right) d z=\left.2 \pi\left(\frac{z^{3}}{3}+z\right)\right|_{0} ^{3}= \\
& =2 \pi\left(\frac{3^{3}}{3}+3-\frac{0^{3}}{3}-0\right)=2 \pi(9+3-0-0)=24 \pi
\end{aligned}
$$

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

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

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## Problem Kuznetsov Integrals 20-9

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