Consider this problem.
Suppose I just want to find the area of a unit circle in polar coordinates. So let $r = 1$. Now I want to (for sake of theory of this question) integrate this over the shaded region (see picture)
Now I know that $\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta = \pi$ Because I am integrating over $\pi/2$ from below and to above.
But why is it that if I do $\frac{1}{2}\int_{-\frac{3\pi}{2}}^{\frac{\pi}{2}} d\theta = -\pi$ I thought that integrating from $\frac{-\pi}{2}$ is the same as integrating $\frac{3\pi}{2}$
Thank you for reading