Transform integral bounds when changing to polar coordinates

Question

When transforming an integral from euclidean to polar coordinates we have this for example: $$ \int^\infty_{-\infty}\int^\infty_{-\infty}f(x,y)dxdy = \int^\infty_{0}\int^{2\pi}_0f(r\cos\theta,r\sin\theta)rd\theta dr $$ I can plug in some functions and evaluate to prove this holds, but I can't think of a reason or intuition why the RHS is not $$ \int_\infty^{0}\int^{2\pi}_0f(r\cos\theta,r\sin\theta)rd\theta dr $$ or $$ \int^\infty_{0}\int_{2\pi}^0f(r\cos\theta,r\sin\theta)rd\theta dr $$ where one of the bounds is flipped, I know this would result a negation and break the equality, but why is it that integrating from the origin in polar coordinates corresponds to integrating from negative to positive infinity in the euclidean space?