Proof of Integral Identity for Dot Product of Normal With Curl of a Vector Over a Surface
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
\[\int \int_S \mathbf{H} \cdot \mathbf{n} dS=0\]
