 Solution

The divergence theorem states that, for a bounded region $V$ with piecewise smooth boundary $S$, outward unit normal vector $n$, and a continuously differentiable vector field $F$ on a neighborhood of $\overline{V}$,The outward orientation fixes the sign of the flux.

On the sphere of radius $R$, the outward unit normal vector is $\mathbf{r}/R$. Thus $r^n\mathbf{r}\cdot n=R^{n+1}$. Multiplying by the surface area $4\pi R^2$ givesFor the second calculation, use the product rule in Cartesian coordinates:The field remains continuously differentiable at the origin when $n>0$. Indeed, for $r>0$ its Jacobian matrix entries are $r^n{\delta }_{ij}+n{r}^{n-2}{x}_i{x}_j$, all tending to zero as $r\to 0$; its derivative at zero is also zero because $|r^n\mathbf{r}|/r=r^n\to 0$. The divergence theorem therefore applies to the entire ball. In spherical coordinates, it yieldsin agreement with the direct surface integral.