## Problem Statement

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

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

Let's construct the graphs of the given functions:

Since the figure is symmetric, we will find the area of half, i.e., the upper limit will be $\frac{\pi}{4}$ instead of $\frac{\pi}{2}$.

Then we get:

![](https://cdn.mathpix.com/cropped/2024_05_22_17ecf51401f349d22271g-25.jpg?height=1114&width=1168&top_left_y=979&top_left_x=775)

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+16-13$ »

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

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## Problem Kuznetsov Integrals 16-14

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