## Problem Statement

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

$$
\rho=8(1-\cos \varphi),-\frac{2 \pi}{3} \leq \varphi \leq 0
$$
## 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 $\rho=8(1-\cos \varphi)$, we find: $\rho^{\prime}=8 \sin \varphi$

We get:

$$
\begin{aligned}
L & =\int_{-2 \pi / 3}^{0} \sqrt{(8(1-\cos \varphi))^{2}+(8 \sin \varphi)^{2}} d \varphi= \\
& =\int_{-2 \pi / 3}^{0} \sqrt{64(1-\cos \varphi)^{2}+64 \sin ^{2} \varphi} d \varphi= \\
& =8 \int_{-2 \pi / 3}^{0} \sqrt{1-2 \cos \varphi+\cos ^{2} \varphi+\sin ^{2} \varphi} d \varphi= \\
& =8 \int_{-2 \pi / 3}^{0} \sqrt{2-2 \cos \varphi} d \varphi= \\
& =8 \sqrt{2} \int_{-2 \pi / 3}^{0} \sqrt{1-\cos \varphi} d \varphi= \\
& =8 \sqrt{2} \int_{-2 \pi / 3}^{0} \sqrt{2 \sin ^{2} \frac{\varphi}{2}} d \varphi= \\
& =16 \int_{-2 \pi / 3}^{0} \sin ^{0} \frac{\varphi}{2} d \varphi= \\
& =16 \cdot 2 \int_{-2 \pi / 3}^{0} \sin \frac{\varphi}{2} d \frac{\varphi}{2}= \\
& =-\left.32 \cos \frac{\varphi}{2}\right|_{-2 \pi / 3} ^{0}=-32\left(\cos 0-\cos \frac{-\pi}{3}\right)=-32\left(1-\frac{1}{2}\right)=16
\end{aligned}
$$

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_19-18 »$ Categories: Kuznetsov's Problem Book Integrals Problem 19 | Integrals | Problems for Checking

- Last edited on this page: 06:01, 28 May 2010.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 19-19

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