Consider the following well-known theorem: If $f$ is continuous on the region $R$ in the $xy$-plane expressed in polar coordinates as $$ R=\{(r,\theta):0\leqslant g(\theta)\leqslant r\leqslant h(\theta),\alpha\leqslant\theta\leqslant\beta\} $$ where $0<\beta-\alpha\leqslant 2\pi$, then $$ \iint\limits_R f(x,y)\;dA=\int_\alpha^\beta\int_{g(\theta)}^{h(\theta)} f(r\cos\theta,r\sin\theta)\;r\;dr\;d\theta $$ In particular, converting a double integral to polar coordinates requires that the bounds satisfy $r\geqslant 0$. To give a simple example: $$ \int_0^1\int_{-\sqrt{x-x^2}}^{\sqrt{x-x^2}}dy\;dx=\dfrac{\pi}{4} $$ can be converted to: $$ \int_{-\pi/2}^{\pi/2}\int_0^{\cos\theta}r\;dr\;d\theta=\dfrac{\pi}{4} $$ However, if we ignore the requirement that $r\geqslant 0$, then we might be tempted to instead compute: $$ \int_0^\pi\int_0^{\cos\theta}r\;dr\;d\theta=\dfrac{\pi}{4} $$ which of course produces the same result, even though it is not guaranteed by the theorem above.
This seems like more than a coincidence. We can't just ignore the requirement, but is there maybe some alternative requirement that can replace $r\geqslant 0$ in the theorem?
Thanks!