## Problem Statement

Calculate the volumes of the bodies bounded by the surfaces.

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

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

$$
\begin{aligned}
& x=0 \text { when } y=1 \\
& y=0 \text { when } x=3 \sqrt{3}
\end{aligned}
$$

That is,

$x \in[-3 \sqrt{3}, 3 \sqrt{3}], y \in[0,1]$

Therefore, the volume will be

![](https://cdn.mathpix.com/cropped/2024_05_22_348f4289c7b5f46bc246g-44.jpg?height=859&width=985&top_left_y=1018&top_left_x=932)

$$
V=\int_{-3 \sqrt{3}}^{3 \sqrt{3}} d x \int_{0}^{\sqrt{1-x^{2} / 27}} d y \int_{0}^{y / \sqrt{3}} d z
$$

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

$$
V_{z}=\int_{0}^{\sqrt{1-x^{2} / 27}} z d y=\int_{0}^{\sqrt{1-x^{2} / 27}} \frac{y}{\sqrt{3}} d y=\left.\frac{y^{2}}{2 \sqrt{3}}\right|_{0} ^{\sqrt{1-x^{2} / 27}}=\frac{1}{2 \sqrt{3}} \cdot\left(1-\frac{x^{2}}{27}\right)
$$

Now the volume of the given body:

$$
\begin{aligned}
& V=\frac{1}{2 \sqrt{3}} \int_{-3 \sqrt{3}}^{3 \sqrt{3}}\left(1-\frac{x^{2}}{27}\right) d x=\left.\frac{1}{2 \sqrt{3}}\left(x-\frac{x^{3}}{27 \cdot 3}\right)\right|_{-3 \sqrt{3}} ^{3 \sqrt{3}} d x= \\
& =\frac{1}{2 \sqrt{3}}\left(3 \sqrt{3}-\frac{(3 \sqrt{3})^{3}}{81}\right)-\frac{1}{2 \sqrt{3}}\left(-3 \sqrt{3}-\frac{(-3 \sqrt{3})^{3}}{81}\right)= \\
& =\frac{\sqrt{3}}{2 \sqrt{3}}\left(3-\frac{27 \cdot 3}{81}\right)+\frac{\sqrt{3}}{2 \sqrt{3}}\left(3-\frac{27 \cdot 3}{81}\right)= \\
& =\frac{1}{2}(3-1)+\frac{1}{2}(3-1)=1+1=2
\end{aligned}
$$

Answer: $V=2$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-26$ »

Categories: Kuznetsov's Problem Book Integrals Problem $20 \mid$ Integrals

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## Problem Kuznetsov Integrals 20-27

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