## Problem Statement

Calculate the volumes of the bodies bounded by the surfaces.

$$
\frac{x^{2}}{27}+\frac{y^{2}}{25}=1, z=\frac{y}{\sqrt{3}}, z=0(y \geq 0)
$$
## Solution

The base of the considered area is a semi-ellipse, in which

$$
\begin{aligned}
& x=0 \text { when } y=5 \\
& y=0 \text { when } x=\sqrt{27}=3 \sqrt{3}
\end{aligned}
$$

That is,

$x \in[-3 \sqrt{3}, 3 \sqrt{3}], y \in[0,5]$

Consider the surface $z=\frac{y}{\sqrt{3}}:$

![](https://cdn.mathpix.com/cropped/2024_05_22_348f4289c7b5f46bc246g-36.jpg?height=859&width=985&top_left_y=1021&top_left_x=935)

$$
V_{z}=\int_{0}^{5} z d y=\int_{0}^{5} \frac{y}{\sqrt{3}} d y=\left.\frac{1}{\sqrt{3}} \cdot \frac{y^{2}}{2}\right|_{0} ^{5}=\frac{25}{2 \sqrt{3}}
$$

Now consider the area of the base and find the volume of the given body:

$$
\begin{aligned}
& \frac{x^{2}}{27}+\frac{y^{2}}{25}=1 \\
& y^{2}=25\left(1-\frac{x^{2}}{27}\right) \\
& V=\frac{25}{2 \sqrt{3}} \int_{-3 \sqrt{3}}^{3 \sqrt{3}} 25\left(1-\frac{x^{2}}{27}\right) d x=\left.\frac{625}{2 \sqrt{3}}\left(x-\frac{x^{3}}{3 \cdot 27}\right)\right|_{-3 \sqrt{3}} ^{3 \sqrt{3}} d x=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{625}{2 \sqrt{3}}\left(3 \sqrt{3}-\frac{(3 \sqrt{3})^{3}}{81}\right)-\frac{625}{2 \sqrt{3}}\left(-3 \sqrt{3}-\frac{(-3 \sqrt{3})^{3}}{81}\right)= \\
& =\frac{625}{2 \sqrt{3}}\left(3 \sqrt{3}-\frac{81 \sqrt{3}}{81}\right)+\frac{625}{2 \sqrt{3}}\left(3 \sqrt{3}-\frac{81 \sqrt{3}}{81}\right)= \\
& =\frac{625}{2 \sqrt{3}} \sqrt{3}(3-1)+\frac{625}{2 \sqrt{3}} \sqrt{3}(3-1)=625+625=1250
\end{aligned}
$$

Answer: $V=1250$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-21$ »

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

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