## Problem Statement

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

$$
\rho=2(1-\cos \varphi),-\pi \leq \varphi \leq-\frac{\pi}{2}
$$
## 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 $\rho=2(1-\cos \varphi)$, we find: $\rho^{\prime}=2 \sin \varphi$

## We obtain:

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

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_19-12 »$ Categories: Kuznetsov's Problem Book Integrals Problem 19 | Integrals | Popular Problems | Problems for Checking

- Last edited on this page: 12:04, 23 June 2010.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 19-13

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