## Problem Statement

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

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

$\sqrt{3} \cos \phi=\sin \phi ; \operatorname{tg} \phi=\sqrt{3} ; \phi=\frac{\pi}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-05.jpg?height=620&width=803&top_left_y=889&top_left_x=181)

$\mathrm{S}=\frac{1}{2} \int_{0}^{\frac{\pi}{3}} \sin ^{2} \phi \mathrm{d} \phi+\frac{3}{2} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cos ^{2} \phi \mathrm{d} \phi=\frac{1}{4} \int_{0}^{\frac{\pi}{3}}(1-\cos 2 \phi) \mathrm{d} \phi+\frac{3}{4} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}(1+\cos 2 \phi) \mathrm{d} \phi=$

$=\left.\frac{1}{4}\left(\phi-\frac{\sin 2 \phi}{2}\right)\right|_{0} ^{\frac{\pi}{3}}+\left.\frac{3}{4}\left(\phi+\frac{\sin 2 \phi}{2}\right)\right|_{\frac{\pi}{3}} ^{\frac{\pi}{2}}=\frac{\pi}{12}-\frac{\sqrt{3}}{16}+\frac{3 \pi}{8}-\frac{\pi}{4}-\frac{3 \sqrt{3}}{16}=\frac{5 \pi}{24}-\frac{\sqrt{3}}{4}$.

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Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� $\% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \ldots \% \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB}$ $\%$ D1\%8B_16-3"

Categories: In scan form | Kuznetsov Integral Problems 16 | Integrals

- Last edited: 10:33, May 11, 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 16-4

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