## Problem Statement

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

$$
r=2 \cos \phi, r=3 \cos \phi
$$
## Solution

Integrals $16-24$

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-46.jpg?height=1299&width=1350&top_left_y=1001&top_left_x=590)

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

| $\varphi$ | $\frac{3 \pi}{2}$ | $\frac{5 \pi}{3}$ | $\frac{11 \pi}{6}$ | 0 | $\frac{\pi}{6}$ | $\frac{\pi}{3}$ | $\frac{\pi}{2}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $r_{1}$ | 0 | 1 | $\sqrt{3}$ | 2 | $\sqrt{3}$ | 1 | 0 |
| $r_{2}$ | 0 | $\frac{3}{2}$ | $\frac{3 \sqrt{3}}{2}$ | 3 | $\frac{3 \sqrt{3}}{2}$ | $\frac{3}{2}$ | 0 |

Where $\alpha=\frac{3 \pi}{2} ; \beta=\frac{\pi}{2}$ and then we get:

$$
\begin{aligned}
& S=\frac{1}{2} \int_{3 \pi / 2}^{\pi / 2}\left(4 \cos ^{2} \varphi-9 \cos ^{2} \varphi\right) d \varphi= \\
& =-\frac{5}{2} \int_{3 \pi / 2}^{\pi / 2} \cos ^{2} \varphi d \varphi= \\
& =-\frac{5}{2} \int_{3 \pi / 2}^{\pi / 2} \frac{1+\cos 2 \varphi}{2} d \varphi= \\
& =-\left.\frac{5}{4}\left(\varphi+\frac{1}{2} \sin 2 \varphi\right)\right|_{3 \pi / 2} ^{\pi / 2}= \\
& =-\frac{5}{4}\left(\frac{\pi}{2}+\frac{1}{2} \sin \pi\right)+\frac{5}{4}\left(\frac{3 \pi}{2}+\frac{1}{2} \sin (3 \pi)\right)=-\frac{5 \pi}{8}+\frac{15 \pi}{8}=\frac{5 \pi}{4}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+16-24$ »

Categories: Kuznetsov's Problem Book Integrals Problem 16 | Integrals

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## Problem Kuznetsov Integrals 16-25

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