## Problem Statement

Calculate the volumes of the bodies bounded by the surfaces.

$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}+\frac{z^{2}}{36}=1, z=3, z=0
$$
## Solution

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

$$
\begin{aligned}
& \frac{x^{2}}{16}+\frac{y^{2}}{9}=\frac{36-z^{2}}{36} \\
& \frac{x^{2}}{16 \cdot \frac{36-z^{2}}{36}}+\frac{y^{2}}{9 \cdot \frac{36-z^{2}}{36}}=1 \rightarrow a=\frac{4}{9} \sqrt{36-z^{2}} ; b=\frac{1}{4} \sqrt{36-z^{2}} \\
& \Rightarrow S=\pi a b=\pi \cdot \frac{4}{9} \sqrt{36-z^{2}} \cdot \frac{1}{4} \sqrt{36-z^{2}}=\frac{\pi}{9} \cdot\left(36-z^{2}\right) \\
& V=\int_{0}^{3} S(z) d z=\frac{\pi}{9} \int_{0}^{3}\left(36-z^{2}\right) d z=\left.\frac{\pi}{9}\left(36 z-\frac{z^{3}}{3}\right)\right|_{0} ^{3}= \\
& =\frac{\pi}{9}\left(36 \cdot 3-\frac{3^{3}}{3}-36 \cdot 0+\frac{0^{3}}{3}\right)=\frac{\pi}{9}(108-9)=\frac{99 \cdot \pi}{9}=11 \pi
\end{aligned}
$$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_20-16" Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals

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## Problem Kuznetsov Integrals 20-16

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