## Task Condition

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

$$
r=2 \sin 4 \phi
$$
## Solution

Let's draw the graph of the function $r=2 \sin 4 \varphi:$

We will use

the formula

Integrals 16-26

for calculating

the area

of a region in polar

coordinates:

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-51.jpg?height=1405&width=1422&top_left_y=1134&top_left_x=520)

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

Obviously, to find the area of the entire figure, it is sufficient to calculate the area for $0 \leq \varphi \leq \frac{\pi}{4}$ and multiply it by 8.

$$
\begin{aligned}
& S=8 \cdot \frac{1}{2} \int_{0}^{\pi / 4} 4 \cdot \sin ^{2} 4 \varphi d \varphi= \\
& =16 \cdot \int_{0}^{\pi / 4} \sin ^{2} 4 \varphi d \varphi= \\
& =16 \cdot \int_{0}^{\pi / 4} \frac{1}{2}(1-\cos 8 \varphi) d \varphi= \\
& =\left.8 \cdot\left(\varphi-\frac{1}{8} \sin 8 \varphi\right)\right|_{0} ^{\pi / 4}= \\
& =8 \cdot\left(\frac{\pi}{4}-\frac{1}{8} \sin \left(8 \cdot \frac{\pi}{4}\right)\right)-8 \cdot\left(0-\frac{1}{8} \sin (8 \cdot 0)\right)= \\
& =2 \pi-\sin 2 \pi-0=2 \pi
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+16-26 \%$

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

- Last edited: 08:24, 13 July 2010.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 16-27

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