## Problem Statement

Calculate the volumes of the bodies bounded by the surfaces.

$$
\frac{x^{2}}{3}+\frac{y^{2}}{4}=1, z=y \sqrt{3}, z=0(y \geq 0)
$$
## Solution

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

$$
\begin{aligned}
& x=0 \text { when } y=2 \\
& y=0 \text { when } x=\sqrt{3}
\end{aligned}
$$

That is, $x \in[-\sqrt{3}, \sqrt{3}], y \in[0,2]$

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

![](https://cdn.mathpix.com/cropped/2024_05_22_348f4289c7b5f46bc246g-19.jpg?height=856&width=985&top_left_y=1011&top_left_x=935)

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

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

$$
\begin{aligned}
& \frac{x^{2}}{3}+\frac{y^{2}}{4}=1 \\
& y^{2}=4\left(1-\frac{x^{2}}{3}\right) \\
& V=2 \sqrt{3} \int_{-\sqrt{3}}^{\sqrt{3}} 4\left(1-\frac{x^{2}}{3}\right) d x=\left.8 \sqrt{3}\left(x-\frac{x^{3}}{3 \cdot 3}\right)\right|_{-\sqrt{3}} ^{\sqrt{3}} d x=
\end{aligned}
$$

$$
\begin{aligned}
& =8 \sqrt{3}\left(\sqrt{3}-\frac{\sqrt{3}^{3}}{9}\right)-8 \sqrt{3}\left(-\sqrt{3}-\frac{-\sqrt{3}^{3}}{9}\right)= \\
& =8\left(3-\frac{3}{3}\right)+8\left(3-\frac{3}{3}\right)=16+16=32
\end{aligned}
$$

Answer: $V=32$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-11$ »

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

- Last modified: 08:29, 24 June 2010.
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