Which order to perform integration after coordinate change? Let $A$ be the unit circle. I know that:
$\iint_A (x^2 + y^2) \,dx\,dy$ = $\int_{0}^{1} \int_{0}^{2 \pi} r^2\ d\theta \ dr $
I'd like to know why wouldn't it be:
$\iint_A (x^2 + y^2) \,dx\,dy$ = $\int_{0}^{1} \int_{2 \pi}^{0} r^2\ d\theta \ dr $
My thinking is that the notation $\iint_A $ comes with an implicit assumption about the order of integration, though I fail to see how this carries over when we transform to polar coordinates or more generally $(u,v)$-coordinates.
Any pointers would be great. Thanks :)
 A: When you change to polar coordinates, you replace the integral over the region $A$ in the $xy$-plane with an integral over a corresponding region $B$ in the $r \theta$-plane, namely the rectangle given by $0 \le r \le 1$ and $0 \le \theta \le 2\pi$.
And when you integrate over a rectangle in the $xy$-plane, I don't think you would ever dream of putting the bounds of integration “backwards” (from the greater to the smaller), so why would you do it in the $r \theta$-plane?
A: Your choice of bounds on the integral depends on your parameterization of the unit circle in terms of $r, \theta$. Were it the case that you parameterize $\theta$ "backwards" like you showed above, the integral would indeed be the opposite. We parameterize $\theta$ forwards however, because that is  how we parameterize regular $1$ dimensional integrals, and because it gives us a positive area (anything else would be fairly nonsensical). 
A: The correct set up, by definition and convention, is the following one
$$\iint_A (x^2 + y^2) \,dx\,dy = \int_{0}^{1} \int_{0}^{2 \pi} r^3\ d\theta \ dr $$
Reversing the order for $\theta$ is equivalent to make a change of variable


*

*$\alpha=2\pi-\theta\implies d\theta=-d\alpha$


then
$$\int_{0}^{1} \int_0^{2 \pi} r^3\ d\theta \ dr=-\int_{0}^{1} \int_{2 \pi}^0 r^3\ d\alpha \ dr$$
