## Task Condition

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

$$
r=\sin 6 \phi
$$
## Solution

We will consider one petal.

Obviously, the total area will be 12 times larger.

$$
\begin{aligned}
& \sin 6 \phi=0 \\
& 6 \phi=\pi k, k \in \mathbb{Z} \\
& \phi=\frac{\pi k}{6}, k \in \mathbb{Z}
\end{aligned}
$$

For the first petal (see figure):

$$
\frac{\pi}{6} \geq \phi \geq 0
$$

Then the desired area is:

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-44.jpg?height=1013&width=1022&top_left_y=924&top_left_x=911)

$$
\begin{aligned}
& S=12 \cdot \frac{1}{2} \cdot \int_{0}^{\pi / 6} \sin ^{2} 6 \phi d \phi=\left|\sin ^{2} 6 \phi=\frac{1}{2}(1-\cos 12 \phi)\right|= \\
& =6 \cdot \int_{0}^{\pi / 6} \frac{1}{2}(1-\cos 12 \phi) d \phi=\left.3\left(t-\frac{1}{12} \sin 12 \phi\right)\right|_{0} ^{\pi / 6}= \\
& =3\left(\frac{\pi}{6}-\frac{1}{12} \sin 2 \phi\right)-3\left(0-\frac{1}{12} \sin 0\right)=\frac{\pi}{2}
\end{aligned}
$$

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\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB\%D1\%8B_16-23» Categories: Kuznetsov's Problem Book Integrals Problem 16 | Integrals

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## Problem Kuznetsov Integrals 16-24

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