Consider the unit open ball $ B_1(0) = \{x \in \mathbb R^n : \|x\|_2 < 1\}$ in $\mathbb R^n $, $n ≥ 3$. Let $Ω := B_1(0)$ and let the function $u: Ω \rightarrow \mathbb R$ be defined as $ u(x) := \frac{1}{\|x\|_2^q}$ for $x \ne 0; x(0) = 0$.
Find conditions on the real numbers $q$ and $p$ such that $u \in L^p(Ω)$.
I try to convert into spherical coordinates: $ u(r) := \frac{1}{r^q}$. Should I find integral $ \int_0^1 \frac{1}{r^q} \, dr$?
That's almost the integral you want to look at, but the power is not quite right.
If you transform into polar coordinates in $\mathbb{R}^n$, you'll get $$\int\limits_{B_1(0)}\frac{1}{|x|^q}\, dx=\int\limits_{S^{n-1}}\int\limits_0^1\frac{1}{r^q}r^{n-1}\, drd\sigma(\omega),$$ where $d\sigma$ denotes the surface measure on the $S^{n-1}$ and $\omega\in S^{n-1}$. Since the integrand is only dependent on $r$, you get that this equals $$\text{vol}\left(S^{n-1}\right)\int\limits_0^1r^{n-1-q}\, dr.$$ So, all you need to do is find when this integral is finite, which is a simple calculus $2$ exercise. You should get that this is finite when $n>q.$ Hence, $|x|^{-q}$ is integrable if $n>q,$ meaning that it'll be $L^p$ if $n>pq.$