## Task Condition

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

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

## Integrals $16-15$

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-29.jpg?height=1502&width=1516&top_left_y=1042&top_left_x=356)

Let's construct the graphs of the given functions in polar coordinates.

The area under the curve in polar coordinates can be found as an integral:

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+16-15$ »

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

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## Problem Kuznetsov Integrals 16-16

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