Surface Area Using Divergence 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\[ \int \int \int_V ( \mathbf{\nabla} \cdot \mathbf{n}) \: dV = \int \int_S ( \mathbf{n} \cdot \mathbf{n}) \: dS = \int \int_S dS =A\]
