## Task Condition

Calculate the volumes of bodies bounded by surfaces.

$$
z=x^{2}+5 y^{2}, z=5
$$
## Solution

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

$$
x^{2}+5 y^{2}=z
$$

The area of the 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-30.jpg?height=920&width=1091&top_left_y=982&top_left_x=834)

$$
\begin{aligned}
& \frac{x^{2}}{z}+\frac{5 y^{2}}{z}=1 \rightarrow a=\sqrt{z} ; b=\frac{\sqrt{z}}{\sqrt{5}} \\
& \Rightarrow S=\pi a b=\pi \cdot \sqrt{z} \cdot \frac{\sqrt{z}}{\sqrt{5}}=\frac{\pi \cdot z}{\sqrt{5}}
\end{aligned}
$$

$$
\begin{aligned}
& V=\int_{0}^{5} S(z) d z=\frac{\pi}{\sqrt{5}} \int_{0}^{5} z d z=\left.\frac{\pi}{\sqrt{5}}\left(\frac{z^{2}}{2}\right)\right|_{0} ^{5}= \\
& =\frac{\pi}{2 \sqrt{5}}\left(5^{2}-0^{2}\right)=\pi \cdot \frac{5 \sqrt{5}}{2}
\end{aligned}
$$

Kuznetsov Integral 20-17

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

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## Problem Kuznetsov Integrals 20-18

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