## Task Condition

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

$$
r=2 \cos 6 \phi
$$
## Solution

Let's draw the graph of the function $r=2 \cos 6 \phi$:

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-53.jpg?height=1237&width=1259&top_left_y=912&top_left_x=250)

We will use the formula for calculating the area of a region in polar coordinates:

$$
S=\frac{1}{2} \int_{\alpha}^{\beta} r^{2}(\phi) d \phi
$$

Obviously, to find the area of the entire figure, it is sufficient to calculate the area for $0 \leq \phi \leq \frac{\pi}{6}$ and multiply it by 12.

$$
S=12 \cdot \frac{1}{2} \int_{0}^{\frac{\pi}{6}} 4 \cos ^{2} 6 \phi d \phi=12 \cdot \frac{1}{2} \cdot 2 \cdot \int_{0}^{\frac{\pi}{6}} 2 \cos ^{2} 6 \phi d \phi=12 \cdot \int_{0}^{\frac{\pi}{6}}(1+\cos 12 \phi) d \phi=
$$

$$
\begin{aligned}
& =\left.12 \cdot\left(\phi+\frac{1}{12} \sin 12 \phi\right)\right|_{0} ^{\frac{\pi}{6}}=12 \cdot\left(\frac{\pi}{6}+\frac{1}{12} \sin \left(12 \cdot \frac{\pi}{6}\right)\right)-12 \cdot\left(0+\frac{1}{12} \sin (12 \cdot 0)\right)= \\
& =2 \pi+\sin 2 \pi-0=2 \pi
\end{aligned}
$$

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\%D1\%82 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_16-27$ "

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

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## Problem Kuznetsov Integrals 16-28

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