## Problem Statement

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

$$
\rho=5(1-\cos \varphi),-\frac{\pi}{3} \leq \varphi \leq 0
$$
## Solution

As is known, the length of the arc of a curve given by the equation $\rho=\rho(\varphi)$ in polar coordinates, where $\varphi \in\left[\varphi_{0}, \varphi_{1}\right]$, is calculated using 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 $\rho=5(1-\cos \varphi)$, we find $\rho^{\prime}=5 \sin \varphi$, and according to formula (1), the length of the arc is calculated as follows:

$$
\begin{aligned}
& l=\int_{-\frac{\pi}{3}}^{0} \sqrt{(5(1-\cos \varphi))^{2}+(5 \sin \varphi)^{2}} d \varphi=\int_{-\frac{\pi}{3}}^{0} \sqrt{25(1-\cos \varphi)^{2}+25 \sin ^{2} \varphi} d \varphi= \\
& =\int_{-\frac{\pi}{3}}^{0} \sqrt{25\left((1-\cos \varphi)^{2}+\sin ^{2} \varphi\right)} d \varphi=\int_{-\frac{\pi}{3}}^{0} \sqrt{25} \sqrt{1-2 \cos \varphi+\cos ^{2} \varphi+\sin ^{2} \varphi} d \varphi
\end{aligned}
$$

Using the fundamental trigonometric identity, $\cos ^{2} \varphi+\sin ^{2} \varphi=1$, in (2):

$$
l=\int_{-\frac{\pi}{3}}^{0} \sqrt{25} \sqrt{1-2 \cos \varphi+1} d \varphi=\int_{-\frac{\pi}{3}}^{0} 5 \sqrt{2(1-\cos \varphi)} d \varphi=5 \sqrt{2} \int_{-\frac{\pi}{3}}^{0} \sqrt{1-\cos \varphi} d \varphi
$$

To compute the integral (3), we use the substitution method in definite integrals.

We apply the universal trigonometric substitution: $\tan \frac{\varphi}{2}=t$. Then $\cos \varphi=\frac{1-t^{2}}{1+t^{2}}, d \varphi=\frac{2 d t}{1+t^{2}}$.

Recalculate the limits according to the substitution: when $\varphi=0$, we get $t=\tan \left(\frac{0}{2}\right)=0$; when $\varphi=-\frac{\pi}{3}$, we get $t=\tan \left(-\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}}$. Therefore,

$l=5 \sqrt{2} \int_{-\frac{1}{\sqrt{3}}}^{0} \sqrt{1-\frac{1-t^{2}}{1+t^{2}}} \frac{2 d t}{1+t^{2}}=5 \sqrt{2} \int_{-\frac{1}{\sqrt{3}}}^{0} \sqrt{\frac{1+t^{2}}{1+t^{2}}-\frac{1-t^{2}}{1+t^{2}}} \frac{2 d t}{1+t^{2}}=$
$=5 \sqrt{2} \int_{-\frac{1}{\sqrt{3}}}^{0} \frac{\sqrt{2 t^{2}}}{\sqrt{1+t^{2}}} \frac{2 d t}{1+t^{2}}=5 \sqrt{2} \int_{-\frac{1}{\sqrt{3}}}^{0} \frac{2 \sqrt{2}|t| d t}{\left(1+t^{2}\right)^{\frac{3}{2}}}$

Since on the interval $\left[-\frac{1}{\sqrt{3}} ; 0\right]$ the inequality $t<0$ holds, we can open the modulus under the integral:

$l=-10 \int_{-\frac{1}{\sqrt{3}}}^{0} \frac{2 t d t}{\left(1+t^{2}\right)^{\frac{3}{2}}}$

In (5), we make another substitution: $1+t^{2}=y$. Then $d y=2 t d t$. Recalculate the limits according to the substitution: when $t=0$, we get $y=1+0^{2}=1$; when $t=-\frac{1}{\sqrt{3}}$, we get

$y=1+\left(-\frac{1}{\sqrt{3}}\right)=1+\frac{1}{3}=\frac{4}{3}$. Apply the substitution to (5):

$l=-10 \int_{\frac{4}{3}}^{1} \frac{d y}{(y)^{\frac{3}{2}}}=-10 \int_{\frac{4}{3}}^{1}(y)^{-\frac{3}{2}} d y=-\left.10 \cdot \frac{y^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}\right|_{\frac{4}{3}} ^{1}=20\left(1-\left(\frac{4}{3}\right)^{-\frac{1}{2}}\right)=20\left(1-\sqrt{\frac{3}{4}}\right)$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� $\% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} \% \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \% \% \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5$ $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+19-15$ "

Categories: Kuznetsov's Problem Book Integrals Problem 19 | Integrals | Problems for Checking

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