## Task Condition

Calculate the volumes of bodies bounded by surfaces.

$$
z=4 x^{2}+9 y^{2}, z=6
$$
## Solution

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

$$
4 x^{2}+9 y^{2}=z
$$

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-38.jpg?height=922&width=1105&top_left_y=978&top_left_x=818)

$$
\begin{aligned}
& \frac{4 x^{2}}{z}+\frac{9 y^{2}}{z}=1 \rightarrow a=\frac{\sqrt{z}}{2} ; b=\frac{\sqrt{z}}{3} \\
& \Rightarrow S=\pi a b=\pi \cdot \frac{\sqrt{z}}{2} \cdot \frac{\sqrt{z}}{3}=\frac{\pi \cdot z}{6}
\end{aligned}
$$

$$
\begin{aligned}
& V=\int_{0}^{6} S(z) d z=\frac{\pi}{6} \int_{0}^{6} z d z=\left.\frac{\pi}{6}\left(\frac{z^{2}}{2}\right)\right|_{0} ^{6}= \\
& =\frac{\pi}{12}\left(6^{2}-0^{2}\right)=3 \pi
\end{aligned}
$$

%tegraly_20-22»

Categories: Kuznetsov Integral Problems 20 | Integrals | Problems for Checking

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## Problem Kuznetsov Integrals 20-23

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