## Task Condition

Calculate the volumes of bodies bounded by surfaces.

$$
z=2 x^{2}+8 y^{2}, z=4
$$
## Solution

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

$$
2 x^{2}+8 y^{2}=z
$$

The area of an 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-21.jpg?height=926&width=1105&top_left_y=979&top_left_x=818)

$$
\begin{aligned}
& \frac{2 x^{2}}{z}+\frac{8 y^{2}}{z}=1 \rightarrow a=\frac{\sqrt{z}}{\sqrt{2}} ; b=\frac{\sqrt{z}}{\sqrt{8}} \\
& \Rightarrow S=\pi a b=\pi \cdot \frac{\sqrt{z}}{\sqrt{2}} \cdot \frac{\sqrt{z}}{\sqrt{8}}=\frac{\pi \cdot z}{4}
\end{aligned}
$$

$$
\begin{aligned}
& V=\int_{0}^{4} S(z) d z=\frac{\pi}{4} \int_{0}^{4} z d z=\left.\frac{\pi}{4}\left(\frac{z^{2}}{2}\right)\right|_{0} ^{4}= \\
& =\frac{\pi}{8}\left(4^{2}-0^{2}\right)=2 \pi
\end{aligned}
$$

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}+20-12 »

Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals | Problems for Checking

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