## Task Condition

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

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

## Since

the $O x$ 

axis 

is 

the 

axis 

of 

rotation for Integrals $21-8$, the volume is 

found 

by 

the 

formula:

![](https://cdn.mathpix.com/cropped/2024_05_22_cfec74fe4e2360126fc5g-15.jpg?height=1268&width=1514&top_left_y=1165&top_left_x=405)

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

Determine the limits of integration: From the problem statement, we already have: $x_{1}=0$. Now, find the upper limit:

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

Find the volume of the solid as the difference in volumes of two solids of revolution:

$$
\begin{aligned}
& V=\pi \cdot \int_{0}^{1}\left(\left(2 x-x^{2}\right)^{2}-(-x+2)^{2}\right) d x= \\
& =\pi \cdot \int_{0}^{1}\left(-x^{4}+4 x^{3}-3 x^{2}-4 x+4\right) d x= \\
& =\left.\pi \cdot\left(-\frac{x^{5}}{5}+4 \frac{x^{4}}{4}-3 \frac{x^{3}}{3}-4 \frac{x^{2}}{2}+4 x\right)\right|_{0} ^{1}= \\
& =\pi \cdot\left(-\frac{1}{5}+1-1-2+4\right)=\frac{9}{5} \pi
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 84 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_21-8$ »

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

- Last edited on this page: 05:57, 8 July 2010.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 21-9

## Material from Plusi