## Problem Statement

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

$$
r=2 \cos \phi, r=2 \sqrt{3} \sin \phi,\left(0 \leq \phi \leq \frac{\pi}{2}\right)
$$
## Solution

## Construct

 graphsIntegrals $16-5$

of the

functions:

Find the points

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-08.jpg?height=1422&width=1307&top_left_y=1082&top_left_x=503)

of intersection:

$$
2 \cos \varphi = 2 \sqrt{3} \sin \varphi
$$

$$
\begin{aligned}
& \operatorname{tg}(\varphi) = \frac{1}{\sqrt{3}} \\
& \varphi = \frac{\pi}{6} + \pi k, k \in \mathbb{Z}
\end{aligned}
$$

From the graph, it is clear that the desired area is:

$$
\begin{aligned}
& S = \frac{1}{2} \cdot \int_{0}^{\pi / 6} r_{1}^{2}(\varphi) d \varphi + \frac{1}{2} \cdot \int_{\pi / 6}^{\pi / 2} r_{2}^{2}(\varphi) d \varphi = \\
& = 6 \cdot \int_{0}^{\pi / 6} \sin ^{2} \varphi d \varphi + 2 \cdot \int_{\pi / 6}^{\pi / 2} \cos ^{2} \varphi d \varphi = \\
& = 3 \cdot \int_{0}^{\pi / 6}(1 - \cos 2 \varphi) d \varphi + \int_{\pi / 6}^{\pi / 2}(1 + \cos 2 \varphi) d \varphi = \\
& = \left.3 \cdot \left(\varphi - \frac{\sin 2 \varphi}{2}\right)\right|_{0}^{\pi / 6} + \left.\left(\varphi + \frac{\sin 2 \varphi}{2}\right)\right|_{\pi / 6}^{\pi / 2} = \\
& = \frac{\pi}{2} - \frac{3 \sqrt{3}}{4} + \frac{\pi}{3} - \frac{\sqrt{3}}{4} = \frac{5 \pi}{6} - \sqrt{3}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+16-5$ » Categories: Kuznetsov's Problem Book Integrals Problem 16 | Integrals

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## Problem Kuznetsov Integrals 16-6

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