# Proof of Gauss’ Theorem for a Rectangular Prism

Recall Gauss’ theorem, $\int\int\int_V (\vec{\nabla} \cdot \vec{F}) dV = \int\int_{S} (\vec{F} \cdot \vec{e}_{n}) dS$. This theorem can be written more precisely. The following statement of the Divergence theorem is a copy from reference [1]. Definitions are provided first.

Volume $V$: Define $V$ as a region comprising three spatial dimensions. The volume has no holes in it.

Boundary $\partial V$: Define $\partial V$ as a once-differentiable surface surrounding the volume $V$. The boundary has a thickness of zero and no holes in it. Points not in $V$ are not in $\partial V,$ and the boundary $\partial V$ is comprised of points in $V$.

Area Vector $\vec{a}$: Define $\vec{a}$ as the vector that is perpendicular to the tangent plane at a point on the boundary $\partial V$. The area vector points away from the center of the volume $V$. The magnitude of $\vec{a}$ is equal to the area of $\partial V$ that has $\vec{a}$ perpendicular and outward-pointing.

Unit Area Vector $\vec{e}_a$: Define $\vec{e}_a \equiv \frac{\vec{a}}{a}$.

Dummy Variable $d \vec{a}$: Define $d\vec{a} \equiv \vec{e}_a da$.

Divergence Theorem: Let $V$ be a region in space with boundary $\partial V$. Then the volume integral of the divergence $\vec{\nabla} \cdot \vec{F}$ of $\vec{F}$ over $V$ and the surface integral of $\vec{F}$ over the boundary $\partial V$ of $V$ are related by $\int_V ( \vec{\nabla} \cdot \vec{F} ) dV = \int_{\partial V} \vec{F} \cdot d\vec{a}$.

Deriving this theorem for a generic surface is very difficult due to complications arising from the curvature of surfaces, so I will prove this theorem for the simplest possible case in which the volume $V$ is a rectangular prism centered at the origin of an arbitrary coordinate system. After all, why should one need an arbitrary shape? I hope that a rectangular prism will work for all the cases that will be encountered.

It helps to draw the situation, so the proof is handwritten.

References