## Problem Statement

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

$$
y=\arccos \frac{x}{3}, y=\arccos x, y=0
$$
## Solution

Since the $O y$ axis is the axis of rotation, the volume $\quad$ Integrals 21-17 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-34.jpg?height=954&width=1196&top_left_y=1136&top_left_x=744)

$$
\begin{aligned}
& y=\arccos \frac{x}{3} \Rightarrow x=3 \cos y \\
& y=\arccos x \Rightarrow x=\cos y
\end{aligned}
$$

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

$$
3 \cos y=\cos y \Rightarrow y_{2}=\frac{\pi}{2}
$$

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

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+21-17$ » Categories: Kuznetsov's Problem Book Integrals Problem 21 | Integrals

- Last edited: 04:40, June 2, 2010.
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## Problem Kuznetsov Integrals 21-18

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