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