## Task Condition

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

$$
r=1+\sqrt{2} \sin \phi
$$
## 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-36.jpg?height=980&width=991&top_left_y=909&top_left_x=949)

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

$$
=\pi-0-\frac{1}{2} \cdot(-1)+\pi+0+\frac{1}{2} \cdot(-1)=2 \pi
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+16-19$ » Categories: Kuznetsov's Problem Book Integrals Problem $16 \mid$ Integrals

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- Last edited: 16:44, 26 June 2010.
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## Problem Kuznetsov Integrals 16-20

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