## Problem Statement

Calculate the areas of figures bounded by lines given in polar coordinates.

$$
r=6 \sin \phi, r=4 \sin \phi
$$
## Solution

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-61.jpg?height=1416&width=1156&top_left_y=1031&top_left_x=584)

$$
S=\frac{1}{2} \int_{\alpha}^{\beta}\left(r_{1}^{2}(\varphi)-r_{2}^{2}(\varphi)\right) d \varphi
$$

Where $\alpha=-\frac{\pi}{2} ; \beta=\frac{\pi}{2}$, and then we get:

$$
\begin{aligned}
& S=\frac{1}{2} \int_{-\pi / 2}^{\pi / 2}\left(36 \sin ^{2} \varphi-16 \sin ^{2} \varphi\right) d \varphi= \\
& =10 \int_{-\pi / 2}^{\pi / 2} \sin ^{2} \varphi d \varphi= \\
& =10 \int_{-\pi / 2}^{\pi / 2} \frac{1-\cos 2 \varphi}{2} d \varphi= \\
& =\left.5\left(\varphi-\frac{\sin 2 \varphi}{2}\right)\right|_{-\pi / 2} ^{\pi / 2}= \\
& =5\left(\frac{\pi}{2}-\frac{1}{2} \sin \pi\right)-5\left(-\frac{\pi}{2}-\frac{1}{2} \sin (-\pi)\right)= \\
& =5 \cdot \frac{\pi}{2}+5 \cdot \frac{\pi}{2}=5 \pi
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+16-31$ »

Categories: Kuznetsov's Problem Book Integrals Problem 16 | Integrals | Problems for Checking

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