## Problem Statement

Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.

$$
\rho=4(1-\sin \varphi), 0 \leq \varphi \leq \frac{\pi}{6}
$$
## Solution

The length of the arc of a curve given by an equation in polar coordinates is determined by the formula

$$
L=\int_{\phi_{1}}^{\phi_{2}} \sqrt{\rho^{2}+\left(\frac{d \rho}{d \phi}\right)^{2}} d \phi
$$

Let's find $\frac{d \rho}{d \phi}$:

$$
\frac{d \rho}{d \phi}=4(-\cos \phi)
$$

We get:

$$
\begin{aligned}
L & =\int_{0}^{\pi / 6} \sqrt{16(1-\sin \phi)^{2}+16(-\cos \phi)^{2}} d \phi= \\
& =4 \int_{0}^{\pi / 6} \sqrt{1-2 \sin \phi+\sin ^{2} \phi+\cos ^{2} \phi} d \phi= \\
& =4 \int_{0}^{\pi / 6} \sqrt{2-2 \sin \phi} d \phi= \\
& =4 \int_{0}^{\pi / 6} \sqrt{2-2 \cos \left(\frac{\pi}{2}-\phi\right)} d \phi= \\
& =4 \int_{0}^{\pi / 6} \sqrt{4 \sin ^{2}\left(\frac{\pi}{4}-\frac{\phi}{2}\right)} d \phi= \\
& =4 \int_{0}^{\pi / 6} 2 \sin \left(\frac{\pi}{4}-\frac{\phi}{2}\right) d \phi=\left.4 \cdot 4 \cos \left(\frac{\pi}{4}-\frac{\phi}{2}\right)\right|_{0} ^{\pi / 6}= \\
& =16\left(\cos \left(\frac{\pi}{4}-\frac{\pi}{12}\right)-\cos \left(\frac{\pi}{4}-0\right)\right)=16\left(\cos \left(\frac{\pi}{6}\right)-\cos \left(\frac{\pi}{4}\right)\right)=16\left(\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\right)=8(\sqrt{3}-\sqrt{2})
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� \%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\% $82 \% D 0 \% B 5 \% D 0 \% B 3 \% D 1 \% 80 \% D 0 \% B 0 \% D 0 \% B B$ $\%$ D1\%8B_19-14»

Categories: Kuznetsov's Problem Book Integrals Problem 19 | Integrals | Problems for Checking

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## Problem Kuznetsov Integrals $19-15$

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