## Task Condition

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

$$
y=(x-1)^{2}, y=1
$$
## Solution

Since the Oy axis is the axis of rotation, the volume is found using the formula:

![](https://cdn.mathpix.com/cropped/2024_05_22_cfec74fe4e2360126fc5g-46.jpg?height=1082&width=1451&top_left_y=995&top_left_x=468)

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

Express \( x \) in terms of \( y \) and find the limits of integration:

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

From the problem statement, we already have: \( y_{2}=1 \). Now, let's find the lower limit:

$$
y=(x-1)^{2} ; x=1 \Rightarrow y_{1}=(1-1)^{2}=0
$$

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

$$
\begin{aligned}
& V=\pi \cdot \int_{0}^{1}(1+\sqrt{y})^{2}-(1-\sqrt{y})^{2} d y= \\
& =\pi \cdot \int_{0}^{1}(1+2 \sqrt{y}+y)-(1-2 \sqrt{y}+y) d y= \\
& =\pi \cdot \int_{0}^{1}(1+2 \sqrt{y}+y-1+2 \sqrt{y}-y) d y= \\
& =\pi \cdot \int_{0}^{1} 4 \sqrt{y} d y=4 \pi \cdot \int_{0}^{1} y^{1 / 2} d y= \\
& =\left.4 \pi \cdot\left(\frac{2}{3} \cdot y^{3 / 2}\right)\right|_{0} ^{1}=4 \pi \cdot\left(\frac{2}{3} \cdot 1-\frac{2}{3} \cdot 0\right)=\frac{8 \pi}{3}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_21-23$ »

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

Ukrainian Banner Network

- Last edited on this page: 07:53, 5 July 2010.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 21-24

## Material from PlusPi