setting up the equation of area between two polar curves

Question

I have to find the area inside $r=2\cos x$ and outside $r=1.$ Can someone please check if I set up my equation correctly? I did:

$$2\left(\frac12\int\limits_0^\pi 2\cos^2 x\ dx-\frac12\int\limits_{-\pi/3}^{\pi/3} 1^2\ dx\right)$$ I have a gut feeling that my bounds for the second part are incorrect but that's were the radio lines hit on the graph.

Answer 1

Indeed, two polar curves intersect themselves at:

$$2\cos(t)=1\to t=\pi/3$$ Note that i consider the symmetric which rules the whole story so, $$0\leq t\leq\pi/3$$and so $$A=2\times\left(\frac{1}2\int_0^{\pi/3}(4\cos^2(t)-1)dt\right)$$