## Problem Statement

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

$$
\rho=1-\sin \varphi, -\frac{\pi}{2} \leq \varphi \leq -\frac{\pi}{6}
$$
## Solution

The length of the arc of a curve given by an equation in polar coordinates is determined by the formula

$$
L=\int_{\phi_{1}}^{\phi_{2}} \sqrt{\rho^{2}+\left(\frac{d \rho}{d \phi}\right)^{2}} d \phi
$$

Let's find $\frac{d \rho}{d \phi}$:

$$
\frac{d \rho}{d \phi}=(-\cos \phi)
$$

We get:

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

Categories: Kuznetsov Problem Book Integrals Problem 19 | Integrals

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## Problem Kuznetsov Integrals 19-12

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