Let $a>0,$ $$f_a(x,y,z)= \frac{|xy|^a}{nz^3}$$ on $$A_n=\{(x,y,z):\sqrt{x^2+y^2}<z<\sqrt{n^2-x^2-y^2}\}$$
I want to find all $a>0$ such that $f_a \in L^1(A_n)$ for $n \geq1.$
Looking at the domain of integration I see that $z>0,$ (thus $f_a$ is continuous and hence measurable), and also $n>\sqrt{x^2+y^2}.$
Besides this, I have no idea how to visualize $A_n.$
I think this looks like a case where I should pass to cylindrical coordinates, so
$$\int_A f_a=\frac{1}{n}\int_0^{2\pi}\ |\cos^at||\sin^at |\ dt\int_0^nr^{2a+1}\ dr\int_r^{\sqrt{(n^2-r^2)}}\frac{1}{z^3} \ dz=...$$ $ |\cos^at||\sin^at|$is bounded on $[0,2\pi]$ for $a>0$ so we do not need to worry about it.
The integral in $z$ gives $$\frac{1}{2(n^2-r^2)}-\frac{1}{2r^2}$$ and so we have $$\frac{1}{2}\int_0^n\frac{r^{2a+1}(r^2-(n-r)(n+r))}{(n-r)(n+r)r^2}=\frac{1}{2}\int_0^n\frac{r^{2a+1}-r^{2a-1}(n-r)(n+r)}{(n-r)(n+r)}=$$$$=\frac{1}{2}\int_0^n\frac{r^{2a+1}}{(n-r)(n+r)}-\frac{1}{2}\int_0^nr^{2a-1}$$ Now $$\frac{1}{2}\int_0^nr^{2a-1}<+\infty \iff 2a-1\geq-1 \iff a\geq0 $$ so this one is fine; the other one function is unbounded for $r \to n$ and I have no idea how to treat it.
I think maybe I made a mistake in choosing the bounds of integration? Because that $n$ both in the function and in the upper bound of integration is causing me problems.
I would be grateful for both a solution or to know what am I doing wrong with my approach