## Problem Statement

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

$$
\rho=2 \varphi, 0 \leq \varphi \leq \frac{3}{4}
$$
## 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 obtain:

$$
\begin{aligned}
L & =\int_{0}^{3 / 4} \sqrt{(2 \varphi)^{2}+2^{2}} d \varphi= \\
& =2 \int_{0}^{3 / 4} \sqrt{\varphi^{2}+1} d \varphi={ }^{(1)} \\
& =2\left(\left.\frac{\varphi}{2} \sqrt{\varphi^{2}+1}\right|_{0} ^{3 / 4}+\left.\frac{1}{2} \ln \left|\varphi+\sqrt{\varphi^{2}+1}\right|\right|_{0} ^{3 / 4}\right)= \\
& =2\left(\frac{1}{2} \cdot \frac{3}{4} \sqrt{\frac{9}{16}+1}-0\right)+2\left(\frac{1}{2} \ln \left|\frac{3}{4}+\sqrt{\frac{9}{16}+1}\right|-\ln |0+\sqrt{0+1}|\right)= \\
& =\frac{3}{4} \cdot \sqrt{\frac{25}{16}}+2\left(\frac{1}{2} \ln \left|\frac{3}{4}+\sqrt{\frac{25}{16}}\right|-0\right)= \\
& =\frac{3}{4} \cdot \frac{5}{4}+\ln \left|\frac{3}{4}+\frac{5}{4}\right|=\frac{15}{16}+\ln 2
\end{aligned}
$$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_19-19"

Categories: Kuznetsov Problem Book Integrals Problem 19 | Integrals

- Last edited: 08:37, 2 July 2010.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 19-20

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