Preparing to enter graduate schools in mathematics, I have come across this volume integral
\begin{array}{c} \iiint_{V} \frac{1}{\left(1+x^{2}+y^{2}\right) z^{3 / 2}} d x d y d z \\ V=\left\{(x, y, z) \in \mathbb{R}^{3} \mid x^{2}+y^{2} \leq z\right\}. \end{array}
It definitely appears as if prepared for a simplification through a surface integral, integrating the volume over z between two surfaces. That is, using the theory behind Gauss' (Divergence) Theorem. However, I am not sure as how to approach it. I.e. how exactly to fit it into said theorem or a coordinate transformation. Neither a polar transformation nor the theorem directly seem to reasonably simplify it.
It also seems that this type of integral can be solved through a clever substitution, that is other substitution than the standard.
Anybody up for help? Thanks.
Rewrite in cylindrical coordinates: let $x = r\cos\theta, y = r\sin\theta, z = z$. Our one bound for $z$ turns into $z \geq r^2$.
So we have:
$$\iiint_V \frac{1}{(1 + x^2 + y^2)z^{\frac{3}{2}}} dV = \int_0^{2\pi}\int_0^{\infty}\int_{r^2}^{\infty}\frac{1}{(1 + r^2)z^{\frac{3}{2}}} r dz dr d\theta$$
$$=2\pi\int_0^{\infty} \frac{r}{1+r^2} \Bigg[-\frac{2}{z^{\frac{1}{2}}}\Bigg]^{\infty}_{r^2}dr = 4\pi\int_0^{\infty}\frac{1}{1+r^2} dr = 4\pi\cdot\frac{\pi}{2} = 2\pi^2.$$