## Problem Statement

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

$$
\rho=2 \varphi, 0 \leq \varphi \leq \frac{12}{5}
$$
## Solution

The length of an 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 obtain:

$$
\begin{aligned}
L & =\int_{0}^{12 / 5} \sqrt{(2 \varphi)^{2}+2^{2}} d \varphi= \\
& =2 \int_{0}^{12 / 5} \sqrt{\varphi^{2}+1} d \varphi={ }^{(1)} \\
& =2\left(\left.\frac{\varphi}{2} \sqrt{\varphi^{2}+1}\right|_{0} ^{12 / 5}+\left.\frac{1}{2} \ln \left|\varphi+\sqrt{\varphi^{2}+1}\right|\right|_{0} ^{12 / 5}\right)= \\
& =\frac{12}{5} \cdot \sqrt{\frac{144}{25}+1}-0+\ln \left|\frac{12}{5}+\sqrt{\frac{144}{25}+1}\right|-\ln |0+\sqrt{0+1}|= \\
& =\frac{12}{5} \cdot \sqrt{\frac{169}{25}}+\left(\ln \left|\frac{12}{5}+\sqrt{\frac{169}{25}}\right|-0\right)= \\
& =\frac{12}{5} \cdot \frac{13}{5}+\ln \left|\frac{12}{5}+\frac{13}{5}\right|=\frac{156}{25}+\ln \left|\frac{25}{5}\right|=\frac{156}{25}+\ln 5
\end{aligned}
$$

In $(1)$, we used the formula:

$$
\begin{aligned}
\int \sqrt{\varphi^{2}+1} d \varphi & =\left|\begin{array}{c}
t=\sqrt{\varphi^{2}+1} \\
\varphi=\sqrt{t^{2}-1} \\
d \varphi=\frac{2 t}{2 \sqrt{t^{2}-1}} d t
\end{array}\right|=\int \frac{t^{2}}{\sqrt{t^{2}-1}} d t=\left|\begin{array}{c}
d v=\frac{t}{\sqrt{t^{2}-1}} d t \\
v=\sqrt{t^{2}-1} \\
u=t \\
d u=d t
\end{array}\right|= \\
& =t \sqrt{t^{2}-1}-\int \sqrt{t^{2}-1} d t= \\
& =t \sqrt{t^{2}-1}-\int \frac{t^{2}-1}{\sqrt{t^{2}-1}} d t= \\
& =t \sqrt{t^{2}-1}-\int \frac{t^{2}}{\sqrt{t^{2}-1}} d t+\int \frac{1}{\sqrt{t^{2}-1}} d t
\end{aligned}
$$

Thus:

$$
\begin{aligned}
& \int \frac{t^{2}}{\sqrt{t^{2}-1}} d t=t \sqrt{t^{2}-1}-\int \frac{t^{2}}{\sqrt{t^{2}-1}} d t+\ln \left|t+\sqrt{t^{2}-1}\right| \\
& \int \frac{t^{2}}{\sqrt{t^{2}-1}} d t=\frac{t}{2} \sqrt{t^{2}-1}+\frac{1}{2} \ln \left|t+\sqrt{t^{2}-1}\right|
\end{aligned}
$$

Therefore:

$$
\int \sqrt{\varphi^{2}+1} d \varphi=\frac{\varphi}{2} \sqrt{\varphi^{2}+1}+\frac{1}{2} \ln \left|\varphi+\sqrt{\varphi^{2}+1}\right|
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+19-22$ »

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

- Last edited: 11:58, July 5, 2010.
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## Problem Kuznetsov Integrals 19-23

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