Polar Integration of $ r = 2\cos(\theta)$ $ r = 2\cos(\theta)$ has the graph
I want to know why the following integral to find area does not work $$\int_0^{2 \pi } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$
whereas this one does:
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$
Why do the limits of integration have to go from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$? Doesn't going from $0$ to $2\pi$ also sweep out the full circle?

Start with $\theta=0$ you get $r=2$ then move to $\theta=\frac{\pi}{2}$ you will get $r=0$ (then you have from here the upper half of the circle). From $\frac{\pi}{2}$ to $\pi$ you have a negative $r$ which don't have really much sense. Then you have to consider it from $\frac{3\pi}{2}$ to $2\pi$ to complete the lower half.
$$\int_{0}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta+\int_{\frac{3\pi}{2}}^{2\pi} \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta=\int_{0}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta+\int_{-\frac{\pi}{2}}^{0} \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$
$$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$
