## Proof of Integral Identity for Dot Product of Normal With Curl of a Vector Over a Surface

Theorem
If
$\mathbf{A}$
is a differentiable vector field defined on a volume
$V$
with surface
$S$
and
$\mathbf{H} = \mathbf{\nabla} \times \mathbf{A}$
then
$\int \int_S \mathbf{H} \cdot \mathbf{n} dS=0$

Proof
Apply the Divergence Theorem to
$\mathbf{H}$
.
$\int \int_S \mathbf{H} \cdot \mathbf{n} dS = \int \int \int_V \mathbf{\nabla} \cdot \mathbf{H} dV$

For any twice differentiable vector field
$\mathbf{F}$
,
$\mathbf{\nabla} \cdot ( \mathbf{\nabla} \times \mathbf{F}) =0$
,
$\int \int_S (\mathbf{\nabla} \times \mathbf{A}) \cdot \mathbf{n} dS = \int \int \int_V \mathbf{\nabla} \cdot (\mathbf{\nabla} \times \mathbf{A}) dV= \int \int \int_V 0 dV =0$

Hence
$\int \int_S \mathbf{H} \cdot \mathbf{n} dS=0$