## Problem Statement

Calculate the volumes of the bodies bounded by the surfaces.

$$
\frac{x^{2}}{3}+\frac{y^{2}}{16}=1, z=y \sqrt{3}, z=0(y \geq 0)
$$
## Solution

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

$$
\begin{aligned}
& x=0 \text { when } y=4 \\
& y=0 \text { when } x=\sqrt{3}
\end{aligned}
$$

That is, $x$ belongs to the interval $[-\sqrt{3}, \sqrt{3}]$, and $y \in [0,4]$

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

$$
V_{z}=\int_{0}^{4} z d y=\int_{0}^{4} y \sqrt{3} d y=\sqrt{3} \int_{0}^{4} y d y=\left.\sqrt{3} \cdot \frac{y^{2}}{2}\right|_{0} ^{4}=\sqrt{3} \cdot \frac{16}{2}=8 \sqrt{3}
$$

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

$$
\begin{aligned}
& \frac{x^{2}}{3}+\frac{y^{2}}{16}=1 \\
& \frac{y^{2}}{16}=1-\frac{x^{2}}{3} \\
& V=8 \sqrt{3} \int_{-\sqrt{3}}^{\sqrt{3}}\left(1-\frac{x^{2}}{3}\right) d x=\left.8 \sqrt{3}\left(x-\frac{x^{3}}{9}\right)\right|_{-\sqrt{3}} ^{\sqrt{3}} d x=16 \sqrt{3}\left(\sqrt{3}-\frac{\sqrt{3}^{2}}{9}\right)=32
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-16 \%$

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

Kuznetsov Integrals 20-16 — PlusPi

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## Problem Kuznetsov Integrals 20-17

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