## Problem Statement

Calculate the volumes of the bodies bounded by the surfaces.

$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}+\frac{z^{2}}{4}=1, z=1, z=0
$$
## Solution

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

$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}=\frac{4-z^{2}}{4}
$$

The area of an ellipse with radii $a$ and $b$ is $\pi \cdot a \cdot b$

By the definition of the radius of an ellipse:

$$
\begin{aligned}
& \frac{x^{2}}{16 \cdot \frac{4-z^{2}}{4}}+\frac{y^{2}}{9 \cdot \frac{4-z^{2}}{4}}=1 \rightarrow a=2 \sqrt{4-z^{2}} ; b=\frac{3}{2} \sqrt{4-z^{2}} \\
& \Rightarrow S=\pi a b=\pi \cdot 2 \sqrt{4-z^{2}} \cdot \frac{3}{2} \sqrt{4-z^{2}}=3 \pi \cdot\left(4-z^{2}\right) \\
& V=\int_{0}^{1} S(z) d z=3 \pi \int_{0}^{1}\left(4-z^{2}\right) d z=\left.3 \pi\left(4 z-\frac{z^{3}}{3}\right)\right|_{0} ^{1}= \\
& =3 \pi\left(4 \cdot 1-\frac{1^{3}}{3}-4 \cdot 0+\frac{0^{3}}{3}\right)=\pi\left(3 \cdot 4-3 \cdot \frac{1}{3}\right)=\pi \cdot(12-1)=11 \pi
\end{aligned}
$$

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

Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals

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## Problem Kuznetsov Integrals 20-6

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