## Problem Statement

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

$$
\rho=7(1-\sin \varphi),-\frac{\pi}{6} \leq \varphi \leq \frac{\pi}{6}
$$
## 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 $\frac{d \rho}{d \phi}:$

$$
\frac{d \rho}{d \phi}=-7 \cos \phi
$$

We get:

$$
\begin{aligned}
L & =\int_{-\pi / 6}^{\pi / 6} \sqrt{49(1-\sin \phi)^{2}+49(-\cos \phi)^{2}} d \phi= \\
& =7 \int_{-\pi / 6}^{\pi / 6} \sqrt{1-2 \sin \phi+\sin ^{2} \phi+\cos ^{2} \phi} d \phi= \\
& =7 \int_{-\pi / 6}^{\pi / 6} \sqrt{2-2 \sin \phi} d \phi= \\
& =7 \int_{-\pi / 6}^{\pi / 6} \sqrt{2-2 \cos \left(\frac{\pi}{2}-\phi\right)} d \phi= \\
& =7 \int_{-\pi / 6}^{\pi / 6} \sqrt{4 \sin ^{2}\left(\frac{\pi}{4}-\frac{\phi}{2}\right)} d \phi= \\
& =7 \int_{-\pi / 6}^{\pi / 6} 2 \sin \left(\frac{\pi}{4}-\frac{\phi}{2}\right) d \phi=\left.7 \cdot 4 \cdot \cos \left(\frac{\pi}{4}-\frac{\phi}{2}\right)\right|_{-\pi / 6} ^{\pi / 6}= \\
& =28\left(\cos \left(\frac{\pi}{4}-\frac{1}{2} \cdot \frac{\pi}{6}\right)-\cos \left(\frac{\pi}{4}-\frac{1}{2} \cdot \frac{-\pi}{6}\right)\right)=28\left(\cos \frac{\pi}{6}-\cos \frac{\pi}{3}\right)=28\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)=14(\sqrt{3}-1)
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� \%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB $\% \mathrm{D} 1 \% 8 \mathrm{~B} \_19-17$ »

Categories: Kuznetsov's Problem Book Integrals Problem $19 \mid$ Integrals

## Problem Kuznetsov Integrals 19-18

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