Find the area of the region lying inside the polar curve $r=1+\cos\theta$ and outside the polar curve $r=2\cos\theta$.
Let
$$A_1 = \frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2d\theta = \frac{3\pi}{2}$$ and $$A_2=\pi(1)^2 = \pi$$ because the radius of the curve $r=2\cos\theta$ is $1$. $$A_1 – A_2 = \frac{\pi}{2}$$ which is the final answer that I got.
However, when I tried to solve the problem on Wolfram Alpha with the input $$\frac{1}{2}\int_0^{2\pi}((1+\cos\theta)^2-(2\cos\theta)^2)d\theta$$ it gave me $-\frac{\pi}{2}$ even though the curve $r=1+\cos\theta$ looks bigger than $r=2\cos\theta$.
How do I reconcile these two answers? Any help would be appreciated!
Best Answer
I see where your problem lies, in the integral around $2\cos t$ you integrate from $0$ to $2\pi$ but this goes around the circle twice, double counting the area, the right bounds should be from $0$ to $\pi$.