## Problem Statement

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

$$
y=\arcsin \frac{x}{5}, y=\arcsin x, y=\frac{\pi}{2}
$$
## Solution

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

![](https://cdn.mathpix.com/cropped/2024_05_22_cfec74fe4e2360126fc5g-36.jpg?height=825&width=1356&top_left_y=1044&top_left_x=561)

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

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

$$
\begin{aligned}
& y=\arcsin \frac{x}{5} \Rightarrow x=5 \sin y \\
& y=\arcsin x \Rightarrow x=\sin y
\end{aligned}
$$

From the problem statement, we already have: $y_{2}=\frac{\pi}{2}$. Now, find the lower limit:

$$
5 \sin y=\sin y \Rightarrow y_{1}=0
$$

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

$$
\begin{aligned}
& V=\pi \cdot \int_{0}^{\pi / 2}(5 \sin y)^{2} d y-\pi \cdot \int_{0}^{\pi / 2} \sin ^{2} y d y= \\
& =\pi \cdot \int_{0}^{\pi / 2}\left(25 \sin ^{2} y-\sin y\right) d y= \\
& =12 \pi \cdot \int_{0}^{\pi / 2} 2 \sin ^{2} y d y= \\
& =12 \pi \cdot \int_{0}^{\pi / 2}\left(2 \sin ^{2} y-1+1\right) d y= \\
& =12 \pi \cdot \int_{0}^{\pi / 2}(-\cos 2 y+1) d y= \\
& =\left.12 \pi \cdot\left(y-\frac{\sin 2 y}{2}\right)\right|_{0} ^{\pi / 2}=12 \pi \cdot\left(\frac{\pi}{2}-0\right)-(0-0)=6 \pi^{2}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+21-18$ " Categories: Kuznetsov's Problem Book Integrals Problem 21 | Integrals | Problems for Checking

- Last edited: 05:25, June 11, 2010.
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## Problem Kuznetsov Integrals 21-19

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