## Problem Statement

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

$$
y=\sin ^{2} x, x=\frac{\pi}{2}, y=0
$$
## Solution

Since

the $O x$

axis

is

the

axis

of

rotation, the volume

is found

using Integrals $21-5$

by

the formula:

![](https://cdn.mathpix.com/cropped/2024_05_22_cfec74fe4e2360126fc5g-09.jpg?height=959&width=1516&top_left_y=1351&top_left_x=401)

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

Let's find the limits of integration: From the problem statement, we already have: $x_{2}=\frac{\pi}{2}$. Now, let's find the lower limit:

$$
y=\sin ^{2} x=0 \Rightarrow x_{1}=0
$$

Now, let's find the volume of the solid of revolution:

$$
\begin{aligned}
& V=\pi \cdot \int_{0}^{\pi / 2}\left(\sin ^{2} x\right)^{2} d x= \\
& =\pi \cdot \int_{0}^{\pi / 2}\left(\frac{1-\cos 2 x}{2}\right)^{2} d x= \\
& =\frac{\pi}{4} \cdot \int_{0}^{\pi / 2}\left(1-2 \cos 2 x+\cos ^{2} 2 x\right) d x= \\
& =\frac{\pi}{4} \cdot \int_{0}^{\pi / 2}(1-2 \cos 2 x) d x+\frac{\pi}{4} \cdot \int_{0}^{\pi / 2}\left(\cos ^{2} 2 x\right) d x= \\
& =\frac{\pi}{4} \cdot \int_{0}^{\pi / 2}(1-2 \cos 2 x) d x+\frac{\pi}{8} \cdot \int_{0}^{\pi / 2}(1+\cos 4 x) d x= \\
& =\left.\frac{\pi}{4} \cdot(x-\sin 2 x)\right|_{0} ^{\pi / 2}+\left.\frac{\pi}{8} \cdot\left(x+\frac{\sin 4 x}{4}\right)\right|_{0} ^{\pi / 2}= \\
& =\frac{\pi}{4} \cdot \frac{\pi}{2}+\frac{\pi}{8} \cdot \frac{\pi}{2}=\frac{3 \pi^{2}}{16}
\end{aligned}
$$

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

Categories: Kuznetsov's Problem Book Integrals Problem 21 | Integrals

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## Problem Kuznetsov Integrals 21-6

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