## Problem Statement

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

$$
\rho=2 \varphi, 0 \leq \varphi \leq \frac{4}{3}
$$
## 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 $\rho=2 \varphi$, we find: $\rho^{\prime}=2$

We get:

$$
\begin{aligned}
L & =\int_{0}^{4 / 3} \sqrt{(2 \varphi)^{2}+2^{2}} d \varphi= \\
& =2 \int_{0}^{4 / 3} \sqrt{\varphi^{2}+1} d \varphi={ }^{(1)} \\
& =2\left(\left.\frac{\varphi}{2} \sqrt{\varphi^{2}+1}\right|_{0} ^{4 / 3}+\left.\frac{1}{2} \ln \left|\varphi+\sqrt{\varphi^{2}+1}\right|\right|_{0} ^{4 / 3}\right)= \\
& =2\left(\frac{1}{2} \cdot \frac{4}{3} \sqrt{\frac{16}{9}+1}-0\right)+2\left(\frac{1}{2} \ln \left|\frac{4}{3}+\sqrt{\frac{16}{9}+1}\right|-\ln |0+\sqrt{0+1}|\right)= \\
& =\frac{4}{3} \cdot \sqrt{\frac{25}{9}}+2\left(\frac{1}{2} \ln \left|\frac{4}{3}+\sqrt{\frac{25}{9}}\right|-0\right)= \\
& =\frac{4}{3} \cdot \frac{5}{3}+\ln \left|\frac{4}{3}+\frac{5}{3}\right|=\frac{20}{9}+\ln 3
\end{aligned}
$$

In (1) we used the formula: $\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \ln \left|x+\sqrt{x^{2}+a^{2}}\right|$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+19-20$ »

Categories: Kuznetsov Problem Book Integrals Problem 19| Integrals

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## Problem Kuznetsov Integrals 19-21

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