## Problem Statement

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

$$
2 x - x^{2} - y = 0, \quad 2 x^{2} - 4 x + y = 0
$$
## Solution

Since the $O x$ axis is the axis of rotation, the volume is found using the formula:

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

Let's find the limits of integration:

![](https://cdn.mathpix.com/cropped/2024_05_22_cfec74fe4e2360126fc5g-03.jpg?height=1217&width=860&top_left_y=985&top_left_x=1095)

$$
y=2 x-x^{2}=4 x-2 x^{2} \Rightarrow x_{1}=0 ; x_{2}=2
$$

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

$$
\begin{aligned}
& V=\pi \cdot \int_{0}^{2}\left(4 x-2 x^{2}\right)^{2}-\left(2 x-x^{2}\right)^{2} d x= \\
& =\pi \cdot \int_{0}^{2}\left(3 x^{4}-12 x^{3}+12 x^{2}\right) d x=
\end{aligned}
$$

$$
\begin{aligned}
& =\left.\pi \cdot\left(\frac{3 x^{5}}{5}-\frac{12 x^{4}}{4}+\frac{12 x^{3}}{3}\right)\right|_{0} ^{2}= \\
& =\pi \cdot\left(\frac{3 \cdot 32}{5}-3 \cdot 16+4 \cdot 8\right)= \\
& =\pi \cdot\left(\frac{96}{5}-48+32\right)=\pi \cdot \frac{96-80}{5}=\frac{16 \pi}{5}
\end{aligned}
$$

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} \_21-2 »

Categories: Kuznetsov's Problem Book Integrals Problem 21 | Integrals | Problems for Checking

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## Problem Kuznetsov Integrals 21-3

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