How to verify a conversion to spherical coordinates? Is it possible to verify if a conversion of an integral in Cartesian coordinates to spherical coordinates was done correctly other than revising it looking for mistakes? I mean, is there some kind of test?
I'm working with integrals that are (much) better to solve in spherical coordinates than in Cartesian coordinates - of course I could integrate the function both in Cartesian and spherical coordinates to verify the conversion, but the whole point of converting it is not having to integrate in Cartesian coordinates. So, is there any other way to tell? I am interested only in the specific case below since I don't have the result of the integral and I cannot use a software to convert from one coordinate system to the other, but I figured it'd be nice to know a general way of testing.

$$\int^2_{-2} \int^\sqrt{4-x^2}_0 \int^\sqrt{4-x^2-y^2}_0 e^{-(x^2+y^2+z^2)^\frac{3}{2}} dzdydx$$
Considering the axis:
  y
  |_ x
 /
z

(1) $0 \le z \le \sqrt{4-x^2-y^2}$ The front half of a sphere with radius 2, centered in the origin.
(2) $-2 \le x \le 2$ Coincides with the fact that the sphere goes from -2 to 2 on x.
(3) $0 \le y \le \sqrt{4-x^2}$ The upper half of a circumference with radius 2, centered in the origin, limiting the sphere to the upper front 4th of it (x from -2 to 2, y from 0 to 2, z from 0 to 2).
I then concluded that 
$0 \le \theta \le \pi$ Since it is the angle around the z axis and (3).
$0 \le \rho \le 2$ From (1).
$0 \le \varphi \le \frac{\pi}{2}$ From the fact that it is the angle around the x axis and (1).
Giving off (which I'm not sure if it's correct)
$$\int^\frac{\pi}{2}_{0} \int^\pi_0 \int^2_0 e^{-(p^2)^\frac{3}{2}} \rho^2 sin(\varphi) d\rho d\theta d\varphi$$

A useful approach to check for large errors is Monte Carlo integration.  Your integral is contained in a cube $4 \times 4 \times 4$ centered on the origin.  Generate a bunch of 3D random points in that volume and add up your integrand for those that fall within the limits.  Then do the same with your spherical expression.  If they disagree badly, you have an error.  If not, you might be right.
