I'm trying to find the area in the curve $r^2=2\cos \theta$ and out of $r=2(1-\cos \theta)$
The intersections are at $\theta=\frac{\pi}{3}$ and $\theta=\frac{-\pi}{3}$, then, the integral to find the area is:
$A=\frac{1}{2} \int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} (\sqrt{2 \cos{\theta}})^2-(2-2\cos{\theta})^2 d\theta=9\sqrt{3}-4\pi$
Using the result that the area of a region in polar coordinates is given by:
$\frac{1}{2} \int_{\theta_1}^{\theta_2} (f(\theta))^2 d\theta$
Is this correct?
Thanks for your help.