## Problem Statement

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

$$
r=\sin \phi, r=\sqrt{2} \cos \left(\phi-\frac{\pi}{4}\right),\left(0 \leq \phi \leq \frac{3 \pi}{4}\right)
$$
## Solution

Let's construct the graphs

of the given functions:

Let's find the points of intersection:

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-19.jpg?height=1111&width=1319&top_left_y=998&top_left_x=608)

$$
\begin{aligned}
& \sin \varphi=\sqrt{2} \cos \left(\varphi-\frac{\pi}{4}\right) \\
& \varphi=\frac{\pi}{2}+\pi k, k \in \mathbb{Z}
\end{aligned}
$$

From the graph, we can see that the desired area is:

$$
S=\frac{1}{2} \cdot \int_{0}^{\pi / 2} r_{1}^{2}(\varphi) d \varphi+\frac{1}{2} \cdot \int_{\pi / 2}^{3 \pi / 4} r_{2}^{2}(\varphi) d \varphi=
$$

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+16-10$ »

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

- Last edited: 04:29, 13 July 2010.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 16-11

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