## Task Condition

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

$$
r=1+\sqrt{2} \cos \phi
$$
## Solution

Since the figure is symmetrical, we calculate the area in the I and II quadrants (i.e., for $[0 ; \pi]$) and multiply by 2:

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-32.jpg?height=1003&width=1034&top_left_y=949&top_left_x=911)

\[
\begin{aligned}
& S=2 \cdot \frac{1}{2} \cdot \int_{0}^{\pi}(1+\sqrt{2} \cos \phi)^{2} d \phi=\int_{0}^{\pi}(1+\sqrt{2} \cos \phi)^{2} d \phi= \\
& =\int_{0}^{\pi}\left(1+2 \sqrt{2} \cos \phi+2 \cos ^{2} \phi\right) d \phi=\left|\cos ^{2} \phi=\frac{1}{2}(1+\cos 2 \phi)\right|= \\
& =\int_{0}^{\pi}(1+2 \sqrt{2} \cos \phi+(1+\cos 2 \phi)) d \phi= \\
& =\int_{0}^{\pi}(2+2 \sqrt{2} \cos \phi+\cos 2 \phi) d \phi=\left.\left(2 t+2 \sqrt{2} \sin \phi+\frac{1}{2} \sin 2 \phi\right)\right|_{0} ^{\pi}= \\
& =\left(2 \pi+2 \sqrt{2} \sin \pi+\frac{1}{2} \sin 2 \pi\right)-\left(2 \cdot 0+2 \sqrt{2} \sin 0+\frac{1}{2} \sin 0\right)=2 \pi
\end{aligned}
\]

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+16-17$ »

Categories: Kuznetsov's Problem Book Integrals Problem 16 | Integrals

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## Problem Kuznetsov Integrals 16-18

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