Let $x \in B_R(0) \subset \mathbb{R^3}$. To compute $$u(x)=\int_{B_R(0)} \frac{1}{|y-x|} dy$$
The integrand has singularity at $x$, so consider $$u_\epsilon(x)=\int_{\substack{|y|< R \\ |y- x|> \epsilon }} \frac{1}{|y-x|} dy$$ Clearly $$|u(x)-u_\epsilon(x)| \leq \int_{B_\epsilon(x)}\frac{1}{|y-x|}dy=\frac{1}{\epsilon} \cdot \frac{4\pi \epsilon^3}{3} \rightarrow 0 \qquad \text{as} \quad\epsilon \rightarrow 0$$
How can we compute $u_\epsilon(x)$ in spherical coordinates? We can show that $u(x)$ is a radial function so w.l.o.g we may assume $x=(0,0,x_3)$.