I am trying to evaluate the following integral
$\int\frac{dxdy}{\sqrt{x^2+y^2+h^2}}$
One can clearly see, that this integral is invariant under change of x to y and y to x, so the answer must have same properties.
Transforming the integral to polar coordinates, one gets
$\int\frac{r}{\sqrt{r^2+h^2}}drd\theta$
This integral is trivial to solve
$\theta\sqrt{r^2+h^2}$
Substituing back to cartesian coordinates
$arctg\left(\frac{y}{x}\right)\sqrt{x^2+y^2+h^2}$
Which is clearly not invariant under change of y to x and x to y. But it is really important for that integral to be invariant under such transformation in the problem I am solving.
P.S.
For anyone wondering the bounds of the definite integrals I need to substitute are
- $\left(-d/2,d/2\right)$ and $\left(-d/2,d/2\right)$
- $\left(0,d\right)$ and $\left(-d/2,d/2\right)$
- $\left(0,d\right)$ and $\left(0,d\right)$
For the second case, symmetry is super important
Assuming that the domain is star shaped around the origin with a boundary given as $r\le R(\theta)$, after solving the inner integral you get $$ \int_0^{2\pi}(\sqrt{h^2+R(\theta)^2}-|h|)d\theta. $$ Whether that can be further simplified now depends on what the boundary function is. For a circle with $R(\theta)=R=const$, one gets the value $$ 2\pi(\sqrt{h^2+R^2}-|h|)=\frac{2\pi R^2}{\sqrt{h^2+R^2}+|h|} $$