## Problem Statement

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

$$
\rho=3(1+\sin \varphi),-\frac{\pi}{6} \leq \varphi \leq 0
$$
## 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 $\rho=3(1+\sin \varphi)$, we find: $\rho^{\prime}=3 \cos \varphi$

We get:

$$
\begin{aligned}
L & =\int_{-\pi / 6}^{0} \sqrt{(3(1+\sin \varphi))^{2}+(3 \cos \varphi)^{2}} d \varphi= \\
& =\int_{-\pi / 6}^{0} \sqrt{9(1+\sin \varphi)^{2}+9 \cos ^{2} \varphi} d \varphi= \\
& =3 \int_{-\pi / 6}^{0} \sqrt{1+2 \sin \varphi+\sin ^{2} \varphi+\cos ^{2} \varphi} d \varphi= \\
& =3 \int_{-\pi / 6}^{0} \sqrt{2+2 \sin \varphi} d \varphi= \\
& =3 \sqrt{2} \int_{-\pi / 6}^{0} \sqrt{1-\cos \left(\frac{\pi}{2}+\varphi\right)} d \varphi= \\
& =3 \sqrt{2} \int_{-\pi / 6}^{0} \sqrt{2 \sin { }^{2}\left(\frac{\pi}{4}+\frac{\varphi}{2}\right)} d \varphi= \\
& =6 \int_{-\pi / 6}^{0} \sin \left(\frac{\pi}{4}+\frac{\varphi}{2}\right) d \varphi= \\
& =6 \cdot 2 \int_{-\pi / 6}^{0} \sin \left(\frac{\pi}{4}+\frac{\varphi}{2}\right) d\left(\frac{\pi}{4}+\frac{\varphi}{2}\right)= \\
& =-\left.12 \cos \left(\frac{\pi}{4}+\frac{\varphi}{2}\right)\right|_{-\pi / 6} ^{0}=-12\left(\cos \frac{\pi}{4}-\cos \left(\frac{\pi}{4}-\frac{\pi}{12}\right)\right)=-12\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\right)=6(\sqrt{3}-\sqrt{2})
\end{aligned}
$$

Source — "http://pluspi.org/wiki/index.php/%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%BD%D0%B8%D0%BA_%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_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_19-13"

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

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## Problem Kuznetsov Integrals $19-14$

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