## Task Condition

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

$$
y=(x-1)^{2}, x=0, x=2, y=0
$$
## 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-61.jpg?height=1099&width=1213&top_left_y=1004&top_left_x=750)

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

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

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

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

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

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

Now, let's find the volume of the left region:

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

In the end, we get:

$$
V=V_{1}+V_{2}=\frac{7 \pi}{6}+\frac{\pi}{6}=\frac{4 \pi}{3}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+21-31$ »

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

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