## Problem Statement

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

$$
\rho=2 \cos \varphi, 0 \leq \varphi \leq \frac{\pi}{6}
$$
## Solution

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

$$
L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi
$$

For the curve given by the equation $\rho=2 \cos \varphi$, we find: $\rho^{\prime}=2 \sin \varphi$

We get:

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+19-26 »$ Categories: Kuznetsov Problem Book Integrals Problem 19| Integrals

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## Problem Kuznetsov Integrals 19-27

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