 Solution

Write $r=|\mathbf{r}|$ and $q=\mathbf{p}\cdot \mathbf{r}$. Away from the origin, $\nabla q=\mathbf{p}$ and $\nabla r^{-m}=-m\mathbf{r}/r^{m+2}$. The product rule givesConsequently the divergence of $\mathbf{E}=3q\mathbf{r}{r}^{-5}-\mathbf{p}{r}^{-3}$ is zero. For any closed surface bounding a region that excludes the origin, the field is smooth throughout that region, and the divergence theorem provesFor the infinite cylinder, use $\mathbf{r}=(a\cos \varphi ,a\sin \varphi ,z)$ and outward unit normal $(\cos \varphi ,\sin \varphi ,0)$. With $\mathbf{p}=(0,0,1)$, $q=z$, $dS=ad\varphi dz$, and $\mathbf{E}\cdot \mathbf{n}=3az/(a^2+z^2{)}^{5/2}$. Hence the direct surface flux isThe integrand is odd in $z$ and absolutely integrable. This calculation treats the unbounded cylindrical side directly, without applying the bounded-region divergence theorem across the field's singularity.