Finding a Volume Using the Divergence Theorem

Theorem
Let  
\[V\]
  be any region of space with a surface  
\[S\]

The volume enclosed by  
\[S\]
  is  
\[V= \frac{1}{3} \int \int_S \mathbf{r} \cdot \mathbf{n} \]
  where  
\[\mathbf{n}\]
  is the outward normal to  
\[S\]

Proof
Use the Divergence Theorem  
\[\int \int \int_V \mathbf{\nabla} \cdot \mathbf{F} dV = \int \int_S \mathbf{F} \cdot \mathbf{n} dV\]

Let  
\[\mathbf{F} = \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}\]

Then
\[\begin{equation} \begin{aligned}\mathbf{\nabla} \cdot \mathbf{r} &= (\frac{\partial}{\partial x} \mathbf{i} + \frac{\partial}{\partial y} \mathbf{j} + \frac{\partial}{\partial z} \mathbf{k}) \cdot (x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \\ &= 1 + 1+1 =3 \end{aligned} \end{equation}\]

Then  
\[\int \int \int_V 3 dV = \int \int_S \mathbf{r} \cdot \mathbf{n} dS \rightarrow V = \frac{1}{3} \int \int_S \mathbf{r} \cdot \mathbf{n} dS\]