## Task Condition

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

$$
r=\cos \phi-\sin \phi
$$
## Solution

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-55.jpg?height=1696&width=1422&top_left_y=942&top_left_x=451)

Since $r \geq 0$, we get $\sin \left(\frac{\pi}{4}-\varphi\right) \geq 0$, hence

$$
-\frac{3 \pi}{4}+2 \pi k \leq \varphi \leq \frac{\pi}{4}+2 \pi k ; k \in Z
$$

Then the desired area is:

$$
\begin{aligned}
& S=\int_{-3 \pi / 4}^{\pi / 4} \frac{1}{2} r^{2}(\varphi) d \varphi= \\
& =\frac{1}{2} \int_{-3 \pi / 4}^{\pi / 4}(\cos \varphi-\sin \varphi)^{2} d \varphi= \\
& =\frac{1}{2} \int_{-3 \pi / 4}^{\pi / 4}\left(\cos ^{2} \varphi-2 \cos \varphi \sin \varphi+\sin ^{2} \varphi\right) d \varphi= \\
& =\frac{1}{2} \int_{-3 \pi / 4}^{\pi / 4}(1-2 \cos \varphi \sin \varphi) d \varphi= \\
& =\frac{1}{2} \int_{-3 \pi / 4}^{\pi / 4}(1-\sin 2 \varphi) d \varphi= \\
& =\left.\frac{1}{2}\left(\varphi+\frac{\cos 2 \varphi}{2}\right)\right|_{-3 \pi / 4} ^{\pi / 4}= \\
& =\frac{1}{2}\left(\frac{\pi}{4}+\frac{\cos 2 \frac{\pi}{4}}{2}\right)-\frac{1}{2}\left(-\frac{3 \pi}{4}+\frac{\cos 2 \cdot\left(-\frac{3 \pi}{4}\right)}{2}\right)= \\
& =\frac{\pi}{8}+\frac{\cos \frac{\pi}{2}}{4}+\frac{3 \pi}{8}-\frac{\cos \left(\pi+\frac{\pi}{2}\right)}{4}= \\
& =\frac{\pi}{8}-0+\frac{3 \pi}{8}+0=\frac{\pi}{2}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+16-28$ »

Categories: Kuznetsov's Problem Book Integrals Problem 16 | Integrals | Problems for Checking

- Last modified: 08:15, 9 July 2010.
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## Problem Kuznetsov Integrals 16-29

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