## Task Condition

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

$$
y=x^{2}-2 x+1, x=2, y=0
$$
## Solution

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

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

Express $x$ in terms of $y$ and find the limits of integration:

![](https://cdn.mathpix.com/cropped/2024_05_22_cfec74fe4e2360126fc5g-55.jpg?height=1108&width=1222&top_left_y=1005&top_left_x=745)

$$
y=x^{2}-2 x+1=(x-1)^{2} \Rightarrow x=1 \pm \sqrt{y}
$$

From the problem statement, we already have: $y_{1}=0$. Now, let's find the lower limit:

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

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

$$
V=\pi \cdot \int_{0}^{1}(2)^{2}-(1+\sqrt{y})^{2} d y=
$$

$$
\begin{aligned}
& =\pi \cdot \int_{0}^{1} 4-(1+2 \sqrt{y}+y) d y= \\
& =\pi \cdot \int_{0}^{1}\left(3-2 y^{1 / 2}-y\right) d y= \\
& =\left.\pi \cdot\left(3 y-2 \cdot \frac{2}{3} \cdot y^{3 / 2}-\frac{y^{2}}{2}\right)\right|_{0} ^{1}=\pi \cdot\left(3-\frac{4}{3} \cdot 1-\frac{1}{2}\right)=\frac{7 \pi}{6}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+21-28$ »

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

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## Problem Kuznetsov Integrals 21-29

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