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.


I agree with your integral setup for $A$, but I think you may have lost the $\frac{1}{2}$, since I get half your answer.

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