## Problem Statement

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

$$
\rho=6(1+\sin \varphi),-\frac{\pi}{2} \leq \varphi \leq 0
$$
## 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 $\rho=6(1+\sin \varphi)$, we find: $\rho^{\prime}=6 \cos \varphi$

We obtain:

$$
\begin{aligned}
L & =\int_{-\pi / 2}^{0} \sqrt{(6(1+\sin \varphi))^{2}+(6 \cos \varphi)^{2}} d \varphi= \\
& =\int_{-\pi / 2}^{0} \sqrt{36(1+\sin \varphi)^{2}+36 \cos ^{2} \varphi} d \varphi= \\
& =6 \int_{-\pi / 2}^{0} \sqrt{1+2 \sin \varphi+\sin ^{2} \varphi+\cos ^{2} \varphi} d \varphi= \\
& =6 \int_{-\pi / 2}^{0} \sqrt{2+2 \sin \varphi} d \varphi= \\
& =6 \sqrt{2} \int_{-\pi / 2}^{0} \sqrt{1+\sin \varphi} d \varphi= \\
& =6 \sqrt{2} \int_{-\pi / 2}^{0} \sqrt{2 \cos { }^{2}\left(\frac{\pi}{4}-\frac{\varphi}{2}\right)} d \varphi= \\
& =12 \int_{-\pi / 2}^{0} \cos \left(\frac{\pi}{4}-\frac{\varphi}{2}\right) d \varphi= \\
& =12 \cdot-2 \int_{-\pi / 2}^{0} \cos \left(\frac{\pi}{4}-\frac{\varphi}{2}\right) d\left(\frac{\pi}{4}-\frac{\varphi}{2}\right)= \\
& =-\left.24 \sin \left(\frac{\pi}{4}-\frac{\varphi}{2}\right)\right|_{-\pi / 2} ^{0}=-24\left(\sin \frac{\pi}{4}-\sin \left(\frac{\pi}{4}-\frac{-\pi}{4}\right)\right)=-24\left(\frac{\sqrt{2}}{2}-1\right)=12(2-\sqrt{2})
\end{aligned}
$$

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\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0 $\% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+19-16$ »

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

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## Problem Kuznetsov Integrals 19-17

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