## Problem Statement

Calculate the volumes of the bodies bounded by the surfaces.

$$
\frac{x^{2}}{9}+y^{2}=1, z=y, z=0(y \geq 0)
$$
## Solution

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

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

That is, $x \in[-3,3], y \in[0,1]$

Consider the surface $z=y:$

![](https://cdn.mathpix.com/cropped/2024_05_22_348f4289c7b5f46bc246g-01.jpg?height=854&width=985&top_left_y=1012&top_left_x=932)

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

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

$$
\begin{aligned}
& \frac{x^{2}}{9}+y^{2}=1 \\
& y^{2}=1-\frac{x^{2}}{9} \\
& V=\frac{1}{2} \int_{-3}^{3}\left(1-\frac{x^{2}}{9}\right) d x=\left.\frac{1}{2}\left(x-\frac{x^{3}}{9 \cdot 3}\right)\right|_{-3} ^{3} d x=\frac{1}{2}\left(3-\frac{3^{3}}{27}\right)-\frac{1}{2}\left(-3-\frac{-3^{3}}{27}\right) \\
& =\frac{1}{2}(3-1)-\frac{1}{2}(-3+1)=1+1=2
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_20-1$ »

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

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