TITLE: Explain $\iint \mathrm dx\,\mathrm dy = \iint r \,\mathrm \,d\alpha\,\mathrm dr$
QUESTION [48 upvotes]: It is changing the coordinate from one coordinate to another. There is an angle and radius on the right side. What is it? And why?
I got:
$2\,\mathrm dy\,\mathrm dx = r(\cos^2\alpha-\sin^2\alpha)\,\mathrm d\alpha \,\mathrm dr$,
where $x = r \cos(\alpha)$ and $y = r \sin(\alpha)$.
but cannot understand and get the right side. The problem emerged when trying to integrate $\displaystyle \int_0^\infty e^{\frac{-x^2}{n^2}}\,\mathrm dx$ where I tried to change the problem knowing $r^2=x^2+y^2$ but stuck to this part. What is the change in the title called and why is it so?

REPLY [35 votes]: Let us approach this problem like some physicists would, that is, let us consider the differential $\mathrm{d}$ as an operator which obeys the rules of derivation, only with a sign. 
In the present case, $x=r\cos\alpha$ and $y=r\sin\alpha$ hence
$$
\mathrm{d}x=\cos\alpha\mathrm{d}r-r\sin\alpha\mathrm{d}\alpha,
\quad
\mathrm{d}y=\sin\alpha\mathrm{d}r+r\cos\alpha\mathrm{d}\alpha.
$$
Now, there are some magic rules which allow to multiply $\mathrm{d}r$ and $\mathrm{d}\alpha$ elements. These are
$$
\mathrm{d}r\mathrm{d}r=0,\quad
\mathrm{d}\alpha\mathrm{d}r=-\mathrm{d}r\mathrm{d}\alpha,\quad
\mathrm{d}\alpha\mathrm{d}\alpha=0.
$$
Hence,
$$
\mathrm{d}x\mathrm{d}y=(\cos\alpha\mathrm{d}r-r\sin\alpha\mathrm{d}\alpha)(\sin\alpha\mathrm{d}r+r\cos\alpha\mathrm{d}\alpha).
$$
The $\mathrm{d}r\mathrm{d}r$ and $\mathrm{d}\alpha\mathrm{d}\alpha$ terms disappear and the terms which interest us are the $\mathrm{d}r\mathrm{d}\alpha$ and $\mathrm{d}\alpha\mathrm{d}r$ ones. One gets
$$
\mathrm{d}x\mathrm{d}y=(\cos\alpha\cdot r\cos\alpha-r\sin\alpha\cdot (-1)\sin\alpha)\mathrm{d}r\mathrm{d}\alpha,
$$
that is,
$$
\color{green}{\mathrm{d}x\mathrm{d}y=r\mathrm{d}r\mathrm{d}\alpha},
$$
a formula which yields
$$
\color{red}{\iint f(x,y)\mathrm{d}x\mathrm{d}y=\iint f(r\cos\alpha,r\sin\alpha)r\mathrm{d}r\mathrm{d}\alpha}.
$$
One can also transform integrals the other way round, all there is to do is to compute a formula for $\mathrm{d}r\mathrm{d}\alpha$ as a multiple of $\mathrm{d}x\mathrm{d}y$. Our formula for $\mathrm{d}x\mathrm{d}y$ in terms of $\mathrm{d}r\mathrm{d}\alpha$ yields
$$
\mathrm{d}r\mathrm{d}\alpha=\frac1r\mathrm{d}x\mathrm{d}y=\frac1{\sqrt{x^2+y^2}}\mathrm{d}x\mathrm{d}y,
$$
hence
$$
\color{blue}{\iint f(r,\alpha)\mathrm{d}r\mathrm{d}\alpha=\iint f(x,y)\frac1{\sqrt{x^2+y^2}}\mathrm{d}x\mathrm{d}y}.
$$
Once again, this only describes the computational side of the story, but this recipe is supported by a well established theory of differential forms which we omitted.