## Problem Statement

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

$$
r=\sqrt{2} \cos \left(\phi-\frac{\pi}{4}\right), r=\sqrt{2} \sin \left(\phi-\frac{\pi}{4}\right),\left(\frac{\pi}{4} \leq \phi \leq \frac{3 \pi}{4}\right)
$$
## Solution

Let's find the points of intersection:

$$
\begin{aligned}
& \sqrt{2} \cos \left(\phi-\frac{\pi}{4}\right)=\sqrt{2} \sin \left(\phi-\frac{\pi}{4}\right) \\
& \operatorname{tg}\left(\phi-\frac{\pi}{4}\right)=1 \\
& \phi-\frac{\pi}{4}=\frac{\pi}{4}+\pi n, n \in \mathbb{Z} \\
& \phi=\frac{\pi}{2}+\pi n, n \in \mathbb{Z}
\end{aligned}
$$

From the graph, the desired area is:

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-27.jpg?height=926&width=925&top_left_y=919&top_left_x=1002)

$$
S=\frac{1}{2} \cdot \int_{\pi / 4}^{\pi / 2}\left(\sqrt{2} \cos \left(\phi-\frac{\pi}{4}\right)\right)^{2} d \phi+\frac{1}{2} \cdot \int_{\pi / 2}^{3 \pi / 4}\left(\sqrt{2} \sin \left(\phi-\frac{\pi}{4}\right)\right)^{2} d \phi=
$$

Substitution:

$$
\begin{aligned}
t & =\phi-\frac{\pi}{4} \\
d t & =d \phi \\
\phi & =\frac{\pi}{4} \Rightarrow t=0 \\
\phi & =\frac{\pi}{2} \Rightarrow t=\frac{\pi}{4} \\
\phi & =\frac{3 \pi}{4} \Rightarrow t=\frac{\pi}{2}
\end{aligned}
$$

We get:

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

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+16-14$ »

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

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## Problem Kuznetsov Integrals 16-15

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