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For a sufficiently regular vector field $F$ on a bounded region $V$ with piecewise smooth boundary $S$, the divergence theorem stateswhere $n$ is the outward unit normal vector. Apply it to $F=fk$ for an arbitrary constant vector $k$. Since $\nabla \cdot (fk)=k\cdot \nabla f$, it givesEquality for every constant $k$ proves the vector gradient form of the divergence theorem:

Applying the same divergence theorem directly to $G$, with $\nabla \cdot G=\rho$, proves the flux lawThis is the Gauss flux law for the prescribed source density, obtained directly from the divergence theorem. For the piecewise source in this problem the divergence relation is understood away from the interface, or almost everywhere; applying the theorem separately to the two sides gives the same law because the normal field is continuous across the interface and the internal boundary fluxes cancel.

For the radial field, its normal component on a sphere of radius $r$ is the constant $G(r)$, so its total flux is $4\pi r^{2}G(r)$. For $0<r\le a$ the enclosed source is $4\pi {\rho }_0{r}^3/3$; for $r>a$ it is $4\pi {\rho }_0{a}^3/3$. Thus the origin-regular spherical Poisson flux law givesThe value at $r=0$ is the continuous value zero. A singular inverse-square radial addition is excluded by the source equation throughout space, including the origin: it would add a point source there.

Since $G=\nabla f$, radial integration gives $f_r={\rho }_{0}r/3$ inside and $f_r={\rho }_0{a}^3/(3r^2)$ outside. The decay condition fixes the exterior constant, giving $f=-{\rho }_0{a}^3/(3r)$ for $r\ge a$. Integrating inside gives $f={\rho }_0{r}^2/6+C_0$, and matching at $r=a$ yields $C_0=-{\rho }_0{a}^2/2$. Hence the potential of a uniform spherical source isBoth $f$ and its radial derivative are continuous at $a$; there is no surface source hidden in the matching.

For any ball centred at the origin, the gradient $G(x)$ is odd under $x\mapsto -x$, so its volume integral vanishes by symmetry. On the bounding sphere, $f$ is constant and outward normals at antipodal points are opposite, so ${\int }_{S}fndS=0$ as well. Both sides of the vector gradient identity are therefore zero, for radii below, above, or equal to $a$.

Rotational symmetry of the vector field means $\mathbf{F}(r\mathbf{n})=h(r)\mathbf{n}$: rotations fixing $\mathbf{n}$ force the field to be parallel to $\mathbf{n}$, and rotations between directions make its magnitude depend only on $r$. Its flux integral over the radius-$r$ sphere is $4\pi r^{2}h(r)$. Applying the flux identity from part (i) to the ball, with the spherically symmetric source, givesTherefore the origin-regular spherical Poisson flux law isOnly source values at $s\le |\mathbf{x}|$ occur. Changing the exterior source can change the scalar potential by an interior constant, but cannot change this radial gradient. Regularity on all of ${\mathbb{R}}^3$, inherited from the Poisson equation setting, matters: on a punctured domain alone an additional $C\mathbf{x}/r^3$ would have zero divergence away from the origin and would not be fixed by the local source. It is excluded here by the origin regularity needed for the ball flux identity.