I think I have it right now, if anyone could check to confirm will appreciate it. I want to solve $\int \int \int (x^2-y^2)^2 dV$, where the body is in the first 'eigth' bounded by: $z=\frac{1}{x^2+y^2}, x+y=2, x+y=1$. So in translating it to cylindrical triple integral (well, repeating one) I get:
$\int_{0}^\frac{pi}{2} \int_\frac{1}{cos\theta +sin \theta }^\frac{2}{cos\theta +sin\theta } \int_0^\frac{1}{r^2} (...) dz dr d\theta$?