= Solution
For a continuously differentiable <vector field> on a volume and its boundary, the <divergence theorem> equates $\int_V\nabla\cdot F\,dV$ to the outward flux $\int_{\partial V}F\cdot n\,dS$. For the given field, $\nabla\cdot F=5(x^2+y^2+z^2)=5r^2$, so
$$
\int_V\nabla\cdot F\,dV=4\pi\int_0^R5r^4\,dr=4\pi R^5.
$$
On the sphere, $n=(x,y,z)/R$, and
$$
F\cdot n=\frac{x^4+y^4+z^4+2x^2y^2+2y^2z^2+2z^2x^2}{R}=\frac{r^4}{R}=R^3.
$$
Thus the surface flux is also \b[$4\pi R^5$], verifying the theorem by explicit calculation.
Apply the <product rule> to $\nabla\cdot(\phi\nabla\psi)=\phi\Delta\psi+\nabla\phi\cdot\nabla\psi$, then apply the <divergence theorem>. This proves <Green's first identity>:
$$
\int_V(\phi\Delta\psi+\nabla\phi\cdot\nabla\psi)\,dV=\int_S\phi\,\partial_n\psi\,dS.
$$
Swap $\phi$ and $\psi$ and subtract. The gradient products cancel, giving <Green's second identity>:
$$
\int_V(\phi\Delta\psi-\psi\Delta\phi)\,dV=\int_S(\phi\,\partial_n\psi-\psi\,\partial_n\phi)\,dS.
$$
For uniqueness, set $w=\phi_1-\phi_2$. It is a <harmonic function> with zero <Dirichlet boundary conditions>. Under the stated smoothness assumptions, <Green's first identity> with $\phi=\psi=w$ gives $\int_V|\nabla w|^2dV=0$. Thus $w$ is constant on each connected component, and its boundary value makes every constant zero. \b[The two solutions agree throughout $V$.] This is <Dirichlet uniqueness for the Poisson equation>. For classical solutions merely continuous up to the boundary and twice continuously differentiable in the interior, the <maximum principle for harmonic functions> gives the same conclusion without requiring boundary derivatives.
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