## Task Condition

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

$$
r=\frac{1}{2}+\sin \phi
$$
## Solution

Since the figure is symmetrical, we calculate the area in the I and IV quadrants (i.e., for $\left[-\frac{\pi}{2} ; \frac{\pi}{2}\right]$) and multiply by 2:

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-23.jpg?height=1025&width=1034&top_left_y=978&top_left_x=908)

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

$$
\begin{aligned}
& =\frac{1}{4} \cdot\left(3 \cdot \frac{\pi}{2}-4 \cos \frac{\pi}{2}-\sin \pi\right)-\frac{1}{4} \cdot\left(3 \cdot\left(-\frac{\pi}{2}\right)-4 \cos \left(-\frac{\pi}{2}\right)-\sin (-\pi)\right)= \\
& =\frac{3 \pi}{8}-0-\frac{1}{4} \cdot(-1)+\frac{3 \pi}{8}+0+\frac{1}{4} \cdot(-1)=\frac{3 \pi}{4}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+16-12 »$ Categories: Kuznetsov's Problem Book Integrals Problem 16 | Integrals

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- Last edited: 16:05, 26 June 2010.
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## Problem Kuznetsov Integrals 16-13

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