Given $$r_1(\theta)=2(1+\cos\theta) \\ r_2(\theta)=2(1-\cos\theta)$$ I want to find the area of the region resulting from the intersection of those curves. Is the following integral correct?
$$ 2A= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [r_2(\theta)]^2 ,\ d\theta + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} [r_1(\theta)]^2 \, d\theta $$
HINT....If you consider the graphs of the functions, the area enclosed by both of them is, by symmetry, $$4\times \frac 12\int_{0}^{\frac{\pi}{2}}4(1-\cos\theta)^2d\theta$$
Can you finish this?