Question

Найти площадь фигуры ограниченной линией p=5cos6x.

Answer 1

$\rho =5\, cos6\varphi \\\\\rho \geq 0\; \; \to \; \; cos6\varphi \geq 0\; ,\; \; -\frac{\pi}{2}+2\pi n\leq 6\varphi \leq \frac{\pi }{2}+2\pi n\; ,\; n\in Z\\\\-\frac{\pi }{12}+\frac{\pi n}{3} \leq \varphi \leq \frac{\pi }{12}+\frac{\pi n}{3}\; ,\; n\in Z\\\\S_1= \frac{1}{2}\int \limits _{ \alpha }^{ \beta }\rho^2(\varphi )d\varphi = \frac{1}{2} \, \int\limits^{\frac{\pi}{12}}_{-\frac{\pi}{12}}\, 25\, cos^2\, 6\varphi \, d\varphi = \frac{25}{4}\; \int\limits^{\frac{\pi}{12}}_{-\frac{\pi }{12}} \, (1+cos12\varphi )\, d\varphi =$

$= \frac{25}{4}\cdot 2\cdot (\varphi +\frac{1}{12}\, sin12\varphi )\Big |_0^{\frac{\pi}{12}}= \frac{25}{2}\cdot ( \frac{\pi }{12}+0)= \frac{25}{24}\, \pi \\\\S=6\, S_1=\frac{25}{4}\, \pi =6,25\, \pi$