Answer
See proof
Work Step by Step
From the Divergence Theorem: \[ \iint_{\sigma} \vec{F} \cdot \vec{n} \, dS = \iiint_V \nabla \cdot \vec{F} \, dV, \] Let $\vec{F}$ be a vector field such that $\vec{F} = \vec{a} \cdot \vec{a}$, where $\vec{a}$ is any constant vector. Then: \begin{align*} \iint_{\sigma} \vec{F} \cdot \vec{n} \, dS &= \iint_{\sigma} \left(\vec{a} \cdot \vec{a} \cdot \vec{F}\right) \cdot \vec{n} \, dS \\ &= \iint_{\sigma} \vec{a} \cdot \left(\vec{a} \cdot \vec{F}\right) \cdot \vec{n} \, dS \\ &= \iint_{\sigma} \vec{a} \cdot \left(\vec{F} \cdot \vec{a}\right) \cdot \vec{n} \, dS \\ &= \iint_{\sigma} \left(\vec{F} \cdot \vec{a}\right) \cdot \left(\vec{a} \cdot \vec{n}\right) \, dS \\ &= \iint_{\sigma} f(x, y, z) \left(\vec{a} \cdot \vec{n}\right) \, dS, \end{align*} and \begin{align*} \iiint_V \nabla \cdot \vec{F} \, dV &= \iiint_V \nabla \cdot \left(\vec{a} \cdot \vec{a}\right) \, dV \\ &= \iiint_V \nabla \cdot \left(\vec{a} \cdot \vec{F}\right) \, dV \\ &= \iiint_V \left(\nabla \cdot \vec{a} \cdot \vec{F} + \vec{a} \cdot \nabla \cdot \vec{F}\right) \, dV \\ &= \iiint_V \left(0 + \vec{a} \cdot \nabla \cdot \vec{F}\right) \, dV \\ &= \iiint_V \vec{a} \cdot \nabla \cdot \vec{F} \, dV. \end{align*} Since $\vec{a}$ is an arbitrary vector, we can conclude that: \[ \iint_{\sigma} \vec{F} \cdot \vec{n} \, dS = \iiint_V \nabla \cdot \vec{F} \, dV. \]
