## Problem Statement

Calculate the volumes of the bodies bounded by the surfaces.

$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}+\frac{z^{2}}{196}=1, z=7, z=0
$$
## Solution

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

$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}=\frac{196-z^{2}}{196}
$$

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:

$$
\begin{aligned}
& \frac{x^{2}}{16 \cdot \frac{196-z^{2}}{196}}+\frac{y^{2}}{9 \cdot \frac{196-z^{2}}{196}}=1 \rightarrow a=\frac{2}{7} \sqrt{196-z^{2}} ; b=\frac{3}{14} \sqrt{196-z^{2}} \\
& \Rightarrow S=\pi a b=\pi \cdot \frac{2}{7} \sqrt{196-z^{2}} \cdot \frac{3}{14} \sqrt{196-z^{2}}=\frac{3 \pi}{49} \cdot\left(196-z^{2}\right) \\
& V=\int_{0}^{7} S(z) d z=\frac{3 \pi}{49} \int_{0}^{7}\left(196-z^{2}\right) d z=\left.\frac{3 \pi}{49}\left(196 z-\frac{z^{3}}{3}\right)\right|_{0} ^{7}= \\
& =\frac{3 \pi}{49}\left(196 \cdot 7-\frac{7^{3}}{3}-196 \cdot 0+\frac{0^{3}}{3}\right)=\frac{\pi}{49}\left(3 \cdot 1372-3 \cdot \frac{343}{3}\right)=\frac{3773 \cdot \pi}{49}=77 \pi
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-31$ » Categories: Kuznetsov's Problem Book Integrals Problem $20 \mid$ Integrals

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