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?

Generally speaking, a double integral always has an area differential, so that you're integrating $\iint dA$.  Another way to view the question, then, is why $dA = dx dy$ in Cartesian coordinates and $dA = r dr d\theta$ in polar coordinates.  

An area element in Cartesian is a rectangle, as Qiaochu describes in his answer.  The area of the rectangle is the small change in $x$ times the small change in $y$, or $\Delta x \Delta y$.

An area element in polar, however, is a piece of a circle sector.  There's a nice picture below taken from here.  (The area element is the shaded part.)  



If the angle is measured in radians, we know that the area of a sector of angle $\theta$ of a circle of radius $r$ is $\frac{1}{2}r^2 \theta$.  So the area of the shaded piece in the picture is $$\Delta A = \frac{1}{2}(r + \Delta r)^2 \Delta \theta - \frac{1}{2}r^2 \Delta \theta = r \, \Delta r \, \Delta \theta + \frac{1}{2}(\Delta r)^2 \, \Delta \theta.$$  The quadratic factor of $\Delta r$ makes the second term negligible compared to the first term for small enough $\Delta r$ and $\Delta \theta$.  Thus in the limit we get $dA = r \, dr \, d\theta$.

As others have said, you can also use the multivariate change of variables formula involving the Jacobian of the transformation directly.  I like the geometric argument when first introducing the polar element in a calculus course, though.
