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This is problem 8.4.17. from Marsden Vector Calculus book.

Let $\rho$ be a continuous function which vanishes outside a 3D region $W$. Define

$\phi(\textbf{p})=\displaystyle\iiint_W\frac{\rho(\textbf{q})}{4\pi\lVert\textbf{p}-\textbf{q}\rVert}dV(\textbf{q})$.

How to show that

$\displaystyle\iint_{\partial W}\nabla\phi\cdot\textbf{dS}=-\iiint_W\rho\; dV$

without the knowledege that $\phi$ is the solution to the Poisson equation $\nabla^2\phi=-\rho$ (which will be a consequence of this)?

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I cannot hope to improve on the depth and rigor of the answer given in Show $\nabla^2g=-f$ almost everywhere by Abdullahi_A_Ibrahim. However I can provide a somewhat more accessible explanation, one that physicists may be accustomed to, which is less rigorous and may for that reason be more appropriate for some readers of this question. Using the definition $$ \phi(\textbf{p})=\displaystyle\iiint_W\frac{\rho(\textbf{q})}{4\pi\lVert\textbf{p}-\textbf{q}\rVert}dV(\textbf{q}), $$ we may take the gradient $$ \nabla\phi(\textbf{p})=\nabla_{\textbf{p}}\bigg[\displaystyle\iiint_W\frac{\rho(\textbf{q})}{4\pi}\frac{1}{\lVert\textbf{p}-\textbf{q}\rVert}dV(\textbf{q})\bigg], $$ which may be brought inside the integral using Leibniz's integral rule $$ \nabla\phi(\textbf{p})=\displaystyle\iiint_W\frac{\rho(\textbf{q})}{4\pi}\nabla_{\textbf{p}}\bigg[\frac{1}{\lVert\textbf{p}-\textbf{q}\rVert}\bigg]dV(\textbf{q}). $$ Substituting this integral into the LHS of the derised result renders $$ \text{LHS} \\ =\displaystyle\iint_{\partial W}\nabla\phi(\textbf{p})\cdot\textbf{dS}(\textbf{p}) \\ =\displaystyle\iint_{\partial W}\displaystyle\iiint_{W}\frac{\rho(\textbf{q})}{4\pi}\nabla_{\textbf{p}}\bigg[\frac{1}{\lVert\textbf{p}-\textbf{q}\rVert}\bigg]dV(\textbf{q})\cdot\textbf{dS}(\textbf{p}), \\ =\displaystyle\iiint_{W}\frac{\rho(\textbf{q})}{4\pi}\bigg[\displaystyle\iint_{\partial W}\nabla_{\textbf{p}}\bigg[\frac{1}{\lVert\textbf{p}-\textbf{q}\rVert}\bigg]\cdot\textbf{dS}(\textbf{p})\bigg]dV(\textbf{q}), $$ using a real version of Fubini's theorem. The inner bracketed integral may be evaluated explicitly $$ \nabla_{\textbf{p}}\bigg[\frac{1}{\lVert\textbf{p}-\textbf{q}\rVert}\bigg]=-\frac{\textbf{p}-\textbf{q}}{\lVert\textbf{p}-\textbf{q}\rVert^3} $$ $$ \displaystyle\iint_{\partial W}\nabla_{\textbf{p}}\bigg[\frac{1}{\lVert\textbf{p}-\textbf{q}\rVert}\bigg]\cdot\textbf{dS}(\textbf{p})=-\displaystyle\iint_{\partial W}\frac{\textbf{p}-\textbf{q}}{\lVert\textbf{p}-\textbf{q}\rVert^3}\cdot\textbf{dS}(\textbf{p}). $$ This integral is evaluated by shifting ($\textbf{p}\mapsto\textbf{p}+\textbf{q}$) and spherically parametrising $\partial W$ $$ \displaystyle\iint_{\partial W}\nabla_{\textbf{p}}\bigg[\frac{1}{\lVert\textbf{p}-\textbf{q}\rVert}\bigg]\cdot\textbf{dS}(\textbf{p}) \\ =-\displaystyle\iint_{\partial W}\frac{\textbf{p}\cdot\textbf{p}}{\lVert\textbf{p}\rVert^4}\text{dS}(\textbf{p})=-\displaystyle\int_{0}^{2\pi}\int_{0}^{\pi}\frac{\lVert\textbf{p}\rVert(\theta,\phi)^2}{\lVert\textbf{p}\rVert(\theta,\phi)^2}\sin(\theta)\text{d}\theta\text{d}\phi=-4\pi. $$ Substituting this result back into the LHS renders $$ \text{LHS}=\displaystyle\iint_{\partial W}\nabla\phi \cdot\textbf{dS}=-\displaystyle\iiint_{W}\rho dV=\text{RHS}, $$ as required. While it was not explored in this proof (for brevity) the application of Leibniz's integral rule and Fubini's theorem put constraints on $\rho$ similar to those described by Abdullahi_A_Ibrahim in Show $\nabla^2g=-f$ almost everywhere.

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