## Problem Statement

Calculate the volumes of solids formed by rotating the figures bounded by the graphs of the functions. The axis of rotation is $O x$.

$$
y=2 x-x^{2}, y=-x+2
$$
## Solution

Since the $O x$ axis

is the axis of rotation, the volume is

found using the formula:

![](https://cdn.mathpix.com/cropped/2024_05_22_cfec74fe4e2360126fc5g-17.jpg?height=1270&width=1514&top_left_y=1164&top_left_x=405)

$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$

Let's find the limits of integration:

$$
y=2 x-x^{2}=-x+2 \Rightarrow x_{1}=1 ; x_{2}=2
$$

We will find the volume of the solid as the difference in volumes of two solids of revolution:

$$
V=V_{1}-V_{2}
$$

Where:

$$
\begin{aligned}
& V_{1}=\pi \cdot \int_{1}^{2}\left(2 x-x^{2}\right)^{2} d x= \\
& =\pi \cdot \int_{1}^{2}\left(4 x^{2}-4 x^{3}+x^{4}\right) d x= \\
& =\left.\pi \cdot\left(4 \frac{x^{3}}{3}-4 \frac{x^{4}}{4}+\frac{x^{5}}{5}\right)\right|_{1} ^{2}= \\
& =\pi \cdot\left(\left(4 \frac{8}{3}-4 \frac{16}{4}+\frac{32}{5}\right)-\left(4 \frac{1}{3}-4 \frac{1}{4}+\frac{1}{5}\right)\right)= \\
& =\pi \cdot\left(\frac{32}{3}-16+\frac{32}{5}-\frac{4}{3}+1-\frac{1}{5}\right)= \\
& =\pi \cdot\left(\frac{28}{3}-15+\frac{31}{5}\right)=\pi \cdot \frac{28 \cdot 5-15 \cdot 15+31 \cdot 3}{15}=\pi \cdot \frac{8}{15} \\
& V_{2}=\pi \cdot \int_{1}^{2}(-x+2)^{2} d x= \\
& =\pi \cdot \int_{1}^{2}\left(4-4 x+x^{2}\right) d x= \\
& =\left.\pi \cdot\left(4 x-4 \frac{x^{2}}{2}+\frac{x^{3}}{3}\right)\right|_{1} ^{2}= \\
& =\pi \cdot\left(\left(8-8+\frac{8}{3}\right)-\left(4-2+\frac{1}{3}\right)\right)=\pi \cdot\left(\frac{8}{3}-2-\frac{1}{3}\right)=\frac{1}{3} \pi
\end{aligned}
$$

Then we get:

$$
V=V_{1}-V_{2}=\pi \cdot \frac{8}{15}-\pi \cdot \frac{1}{3}=\pi \cdot \frac{8-5}{15}=\frac{3}{15} \pi=\frac{1}{5} \pi
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+21-9$ »

Categories: Kuznetsov's Problem Book Integrals Problem 21 | Integrals

- Last edited: 08:03, July 8, 2010.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 21-10

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