## Problem Statement

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

$$
\rho=3 e^{3 \varphi / 4},-\frac{\pi}{2} \leq \varphi \leq \frac{\pi}{2}
$$
## 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 $\frac{d \rho}{d \phi}$:

$$
\frac{d \rho}{d \phi}=3 \cdot \frac{3}{4} e^{3 \phi / 4}=\frac{9}{4} e^{3 \phi / 4}
$$

We obtain:

$$
\begin{aligned}
L & =\int_{-\pi / 2}^{\pi / 2} \sqrt{\left(3 e^{3 \phi / 4}\right)^{2}+\left(\frac{9}{4} e^{3 \phi / 4}\right)^{2}} d \phi= \\
& =\int_{-\pi / 2}^{\pi / 2} \sqrt{\left(9+\frac{81}{16}\right) e^{3 \phi / 2}} d \phi= \\
& =\frac{15}{4} \int_{-\pi / 2}^{\pi / 2} e^{3 \phi / 4} d \phi=\left.\frac{15}{4} \cdot \frac{4}{3} e^{3 \phi / 4}\right|_{-\pi / 2} ^{\pi / 2}=5 \cdot\left(e^{3 \pi / 8}-e^{-3 \pi / 8}\right)=10 \cdot \operatorname{sh} \frac{3 \pi}{8}
\end{aligned}
$$

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_19-1» Categories: Kuznetsov's Problem Book Integrals Problem 19 | Integrals

- Last edited on this page: 12:00, 27 May 2010.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 19-2

## Material from PlusPi