## Task Condition

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

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

Construct the graph of the given curve. Since the polar radius is non-negative, i.e.,

$\mathrm{r} \geq 0$, then $\cos 2 \phi \geq 0, -\frac{\pi}{2} + 2 \pi \mathrm{k} \leq 2 \phi \leq \frac{\pi}{2} + 2 \pi \mathrm{k}; \mathrm{k} \in \mathrm{Z}_{0}$.

$-\frac{\pi}{4} + \pi \mathrm{k} \leq \phi \leq \frac{\pi}{4} + \pi \mathrm{k}, \mathrm{k} \in \mathrm{Z}_{0}$

$\mathrm{k}=0 \Rightarrow -\frac{\pi}{4} \leq \phi \leq \frac{\pi}{4}; \mathrm{k}=1 \Rightarrow \frac{3 \pi}{4} \leq \phi \leq \frac{5 \pi}{4}$, i.e., the graph will consist of two

parts

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-04.jpg?height=582&width=762&top_left_y=1108&top_left_x=110)

$\mathrm{S}=4 \cdot \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \cos ^{2} 2 \phi \mathrm{d} \phi=2 \cdot \frac{1}{2} \int_{0}^{\frac{\pi}{4}}(1+\cos 4 \phi) \mathrm{d} \phi=\left.\left(\phi+\frac{\sin 4 \phi}{4}\right)\right|_{0} ^{\frac{\pi}{4}}=\frac{\pi}{4}$.

Note: If the polar radius is arbitrary, i.e., $r \in R$, then the graph will consist of 4 petals, and the value of $\mathrm{S}$ will double, i.e., $S=\frac{\pi}{2}$.

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Categories: In the form of a scan | Kuznetsov's Problem Book Integrals Problem $16 \mid$ Integrals

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## Problem Kuznetsov Integrals 16-3

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