Finding area inside circle and outside another circle

Question

I was trying to find the area inside the circle $r=-2\cos(\theta)$ and outside $r=1$ and the upper bound was $2\pi/3$ while the lower bound is $0$. Is this correct? If not please help me set this up. Thanks.

Answer 1

You know that the curves intersect when $\cos(\theta)=-\dfrac{1}{2}$ which is at $\theta=\frac{2\pi}{3}$ and $\frac{4\pi}{3}$. But because of symmetry with respect to the horizontal axis you could just double the area between $\theta=\dfrac{2\pi}{3}$ and $\theta=\pi$. And as long as we are using symmetry, the area is the same as that lying outside $r=1$ and inside $r=2\cos\theta$ so why not just find and double the area of the region between the rays $\theta=0$ and $\theta=\dfrac{\pi}{3}$ and $\theta=0$ inside the curve $r=2\cos\theta$ and $r=1$? \begin{equation} A=2\int_0^{\pi/3}\frac{1}{2}\left(4\cos^2\theta-1\right)\,d\theta \end{equation}