## Task Condition

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

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

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

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

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

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_16-16%"

- Last edited on this page: 12:52, 9 July 2010.
- Content is available under CC-BY-SA 3.0.


## Problem Kuznetsov Integrals $16-17$

## Material from PlusPi