Is Gaussian integral the only one that can be easily solved by this double integral trick? For a lot of people the favorite way of solving Gaussian integral $I=\int^{\infty}_{-\infty} e^{-x^2} dx$ is to find $I^2$ in polar coordinates and then take a root.
The trick may be useful in this case, but I struggle to find any other integral it can be applied to. The obvious condition for the integrated function is:
$$f(x) \cdot f(y)=g(x^2+y^2)=h(|r|)$$
I don't know any other function aside from $e^{bx^2}$ that meets this condition.
Moreover, the limits for the argument should be infinite. Otherwise we can't equate integration in the square $x,y \in (-a,a)$ with integration in the cirlce $r \in (0,a)$.

But maybe this method can be generalized? For example, there may be some functions that give elementary integrals in polar form when multiplied $f(x)f(y)$ even if their product depends on the angle too?

 A: From Nate Eldredge's answer here:

One well-known trick is a way to evaluate the Gaussian integral $G = \int_\mathbb{R} e^{-x^2}dx = \sqrt{\pi}$ by writing
  $$G^2 = \left(\int_\mathbb{R} e^{-x^2}dx\right)\left(\int_\mathbb{R} e^{-y^2}dy\right)
= \int_{\mathbb{R}^2} e^{-(x^2+y^2)}dxdy$$
  which when transformed to polar coordinates becomes
  $$G^2 = 2\pi \int_0^\infty e^{-r^2} r dr = \pi \int_0^\infty e^{-u} du = \pi$$
  via the substitution $u=r^2$.  It appears this idea is due to Poisson.
In a 2005 note in the American Mathematical MONTHLY, R. Dawson has observed that this is a trick that only works once; there are no other integrals that can be evaluated by this method.  Specifically:
Theorem. Any Riemann-integrable function $f$ on $\mathbb{R}$, such that $f(x)f(y) = g(\sqrt{x^2+y^2})$ for some $g$, is of the form $f(x)=ke^{ax^2}$. 
See: Dawson, Robert J. MacG.  On a "singular" integration technique of Poisson.  American Mathematical Monthly 112 (2005), 270-272.

A: See also my article, "Poisson's remarkable calculation --a method or a trick?", Elem. Math. 65 (2010), where a more general version of Dawson's result is proved. 
