TITLE: How to prove $dxdy = r dr d \theta$?
QUESTION [10 upvotes]: $x = r \cos \theta$, $y = r \sin \theta$
I got $dx = \cos \theta dr - r \sin \theta d \theta $ 
$ dy = \sin \theta dr + r \cos \theta d \theta$
How to get $dx dy = r dr d \theta$??
I saw the same question Rigorous proof that $dx dy=r\ dr\ d\theta$.
But I am not getting where vectors are coming in to the picture thanks.

REPLY [3 votes]: A piece of an annulus swept out by a change of angle $\Delta \theta$ and a change of radius $\Delta r$, starting from a point given by $(r,\theta)$, has area $\Delta \theta \int_r^{r+\Delta r} s ds = \Delta \theta \frac{(r+\Delta r)^2-r^2}{2} = \Delta \theta \left ( r \Delta r + \frac{\Delta r^2}{2} \right )$. (This is computed by integrating the length of circular arcs.)
As $\Delta r \to 0$ the second term is asymptotically much smaller than the first, which heuristically justifies the change of variables formula. Showing that this procedure, which is equivalent to the more general procedure based on the Jacobian determinant, actually makes integrals do the correct thing takes some more work. The details can be found in a typical undergraduate real analysis text.