= Solution
Combining the electric-field equations gives the <Poisson equation> $\nabla^2\phi_i=-\rho_i/\varepsilon_0$. Apply the <product rule for divergence> to
$$
\mathbf F=\phi_1\nabla\phi_2-\phi_2\nabla\phi_1.
$$
The two cross terms in its <divergence> cancel, leaving
$$
\nabla\cdot\mathbf F=\phi_1\nabla^2\phi_2-\phi_2\nabla^2\phi_1
=\frac{\phi_2\rho_1-\phi_1\rho_2}{\varepsilon_0}.
$$
This <electrostatic reciprocity identity> follows from the <divergence theorem>, and is exactly <Green's second identity> for these potentials. Integrating and rearranging gives
$$
\boxed{\frac1{\varepsilon_0}\int_V\phi_1\rho_2\,dV
+\int_{\partial V}\phi_1\nabla\phi_2\cdot d\mathbf S
=\frac1{\varepsilon_0}\int_V\phi_2\rho_1\,dV
+\int_{\partial V}\phi_2\nabla\phi_1\cdot d\mathbf S.}
$$
The boundary orientation is outward from the chosen region. For smooth charge densities this follows directly from classical differentiation. For isolated <point charges>, one may excise small spheres and take their radii to zero, provided the other potential is regular at each charge.
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