## Problem Statement

Calculate the volumes of the bodies bounded by the surfaces.

$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}+\frac{z^{2}}{64}=1, z=4, z=0
$$
## Solution

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

$$
\begin{aligned}
& \frac{x^{2}}{16}+\frac{y^{2}}{9}=\frac{64-z^{2}}{64} \\
& \frac{x^{2}}{16 \cdot \frac{64-z^{2}}{64}}+\frac{y^{2}}{9 \cdot \frac{64-z^{2}}{64}}=1 \rightarrow a=\frac{1}{2} \sqrt{64-z^{2}} ; b=\frac{3}{8} \sqrt{64-z^{2}} \\
& \Rightarrow S=\pi a b=\pi \cdot \frac{1}{2} \sqrt{64-z^{2}} \cdot \frac{3}{8} \sqrt{64-z^{2}}=\frac{3 \pi}{16} \cdot\left(64-z^{2}\right) \\
& V=\int_{0}^{4} S(z) d z=\frac{3 \pi}{16} \int_{0}^{4}\left(64-z^{2}\right) d z=\left.\frac{3 \pi}{16}\left(64 z-\frac{z^{3}}{3}\right)\right|_{0} ^{4}= \\
& =\frac{3 \pi}{16}\left(64 \cdot 4-\frac{4^{3}}{3}-64 \cdot 0+\frac{0^{3}}{3}\right)=\frac{\pi}{16}(3 \cdot 256-64)=\frac{704 \cdot \pi}{16}=44 \pi
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-20$ » Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals

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## Problem Kuznetsov Integrals 20-21

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