## Task Condition

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

$$
r=\cos 3 \phi
$$
## Solution

## Integral Calculus. Option 8

## Problem 16.

Calculate the area of the figure bounded by the lines given by the equations in polar coordinates

$\mathrm{r}=\cos 3 \varphi$

Let's plot the graph of the given function in the polar coordinate system (see fig.)

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-15.jpg?height=697&width=943&top_left_y=688&top_left_x=291)

Since $r \geq 0$, we get $\cos 3 \varphi \geq 0$, hence $-\frac{\pi}{6}+\pi k \leq \varphi \leq \frac{\pi}{6}+\pi k$

From the figure, it is clear that due to symmetry, the desired area \( S \) is

$$
\begin{aligned}
& S=6 \int_{0}^{\pi / 6} \frac{1}{2} r^{2}(\varphi) d \varphi=3 \int_{0}^{\pi / 6} \cos ^{2} 3 \varphi d \varphi=\frac{3}{2} \int_{0}^{\pi / 6}(1+\cos 6 \varphi) d \varphi= \\
& =\left.\frac{3}{2}\left(\varphi+\frac{1}{6} \sin 6 \varphi\right)\right|_{0} ^{\pi / 6}=\pi / 4
\end{aligned}
$$

Answer: \( \mathrm{S}=\pi / 4 \approx 0.78 \)

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## Problem Kuznetsov Integrals $16-9$

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