## Problem Statement

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

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

$$
\frac{d \rho}{d \phi}=2 \cdot \frac{4}{3} e^{4 \phi / 3}=\frac{8}{3} e^{4 \phi / 3}
$$

We obtain:

$$
\begin{aligned}
L & =\int_{-\pi / 2}^{\pi / 2} \sqrt{\left(2 e^{4 \phi / 3}\right)^{2}+\left(\frac{8}{3} e^{4 \phi / 3}\right)^{2}} d \phi= \\
& =\int_{-\pi / 2}^{\pi / 2} \sqrt{\left(4+\frac{64}{9}\right)\left(e^{4 \phi / 3}\right)^{2}} d \phi= \\
& =\frac{10}{3} \int_{-\pi / 2}^{\pi / 2} e^{4 \phi / 3} d \phi=\left.\frac{10}{3} \cdot \frac{3}{4} e^{4 \phi / 3}\right|_{-\pi / 2} ^{\pi / 2}=\frac{5}{2} \cdot\left(e^{2 \pi / 3}-e^{-2 \pi / 3}\right)=5 \cdot \operatorname{sh} \frac{2 \pi}{3}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_19-2$ » Categories: Kuznetsov's Problem Book Integrals Problem 19 | Integrals

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## Problem Kuznetsov Integrals 19-3

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