# Triple integral. Spherical coordinates. Too much calculations

I have troubles with integral:

$$\iiint \limits_{S} g(x;y;z)dxdydz\ \label{orig} \tag{1}$$

Where $g(x;y;z) = \frac{xyz}{(a^2 + x^2 + y^2 + z^2)^3}$ and area is given by inequalities: $$(x^2 + y^2 + z^2)^{3/2} \leqslant 4xy, \\ x \geqslant 0, y \geqslant 0, z\geqslant 0$$

I know one way to solve this. I used conversion to spherical coordinate system and got the following integral:

$$\int\limits_{0}^{\pi / 2}d\varphi \int\limits_{0}^{\pi/2}d\theta \int\limits_{0}^{\sin2\varphi(1 - \cos2\theta)} \frac{r^5 \sin2\varphi \sin2\theta(1 - \cos2\theta)}{8(a^2 + r^2)^3} dr$$

This integral is solvable, but there are a lot of calculations in the process. My question is: Is it possible to solve original integral $\eqref{orig}$ differently? is there more elegant solution? Maybe I'm making some mistakes?

• you can proceed along the lines given in this other post – G Cab Feb 1 '18 at 12:08
• Sometimes things suck...and this seems to be one of those times. – DonAntonio Feb 1 '18 at 12:25