## Surface Area Using Divergence Theorem

Theorem
If
$S$
is a closed surface for a region
$V$
then
$\int \int \int_V \mathbf{\nabla} \cdot \mathbf{n} dV =A$
where
$\mathbf{n}$
is the outward normal and use t
$A$
is the area of t
$S$

Proof
The Divergence Theorem states
$\int \int \int_V ( \mathbf{\nabla} \cdot \mathbf{F}) \: dV = \int \int_S ( \mathbf{F} \cdot \mathbf{n}) \: dS$

where use t
$\mathbf{F}$
is a twice differentiable vector field. Substitute t
$\mathbf{F} = \mathbf{n}$
to get
$\int \int \int_V ( \mathbf{\nabla} \cdot \mathbf{n}) \: dV = \int \int_S ( \mathbf{n} \cdot \mathbf{n}) \: dS = \int \int_S dS =A$