I was working on a problem where I had to convert the region into polar equations in order to find the double integral. The region was
$$r = 2\cos(\theta)$$ and after solving the problem I realized that
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2 \cos \theta}(12-3 r \cos \theta-4 r \sin \theta) r d r d \theta$$ and $$ \int_{0}^{\pi} \int_{0}^{2 \cos \theta}(12-3 r \cos \theta-4 r \sin \theta) r d r d \theta$$ produce the same result. Intuitively, I understand that in the second expression the polar coordinate is flipped to the opposite side of the pole when
$ \frac{\pi}{2} < \theta < \pi $ because the value of cosine becomes negative in that range, making ‘r’ negative as well due to the region equation.
But purely mathematically speaking I don’t understand how getting the definite integral of any same function/expression (that is in terms of $\theta $ after integrating with respect to ‘r’) in two completely different bounds (I’m talking about the outer integral here, the 2nd integral in the iterated integral operation) always produce the same value as long as the polar coordinate is ‘flipped’ at some point to ultimately indicate the same region as the other integral.
I mean, normally integrating any same function for $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ vs. $\quad 0 < \theta < \pi$ should not produce the same value, right? Although I think I get it in terms of intuition, it doesn’t seem to make sense mathematically.
Can anyone who recognizes my confusion help me out? How can I thoroughly validate that the two double integrals above are equal without surface intuition?