## Problem Statement

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

$$
\rho=5 e^{5 \varphi / 12},-\frac{\pi}{2} \leq \varphi \leq \frac{\pi}{2}
$$
## Solution

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

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

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

$$
\frac{d \rho}{d \phi}=5 \cdot \frac{5}{12} e^{5 \phi / 12}=\frac{25}{12} e^{5 \phi / 12}
$$

We obtain:

$$
\begin{aligned}
L & =\int_{-\pi / 2}^{\pi / 2} \sqrt{\left(5 e^{5 \phi / 12}\right)^{2}+\left(\frac{25}{12} e^{5 \phi / 12}\right)^{2}} d \phi= \\
& =\int_{-\pi / 2}^{\pi / 2} \sqrt{\left(25+\frac{625}{144}\right) e^{5 \phi / 6}} d \phi= \\
& =\frac{65}{12} \int_{-\pi / 2}^{\pi / 2} e^{5 \phi / 12} d \phi=\left.\frac{65}{12} \cdot \frac{12}{5} e^{5 \phi / 12}\right|_{-\pi / 2} ^{\pi / 2}=13 \cdot\left(e^{5 \pi / 24}-e^{-5 \pi / 24}\right)=26 \cdot \operatorname{sh} \frac{5 \pi}{24}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+19-4$ »

Categories: Kuznetsov's Problem Book Integrals Problem 19 | Integrals

- Last modified: 11:54, 27 May 2010.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo  佂
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## Problem Kuznetsov Integrals $19-5$

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