Proof That the Volume Integral of the Curl of a Vector Normal to a Surface Everwhere is Zero

Theorem
If
$\mathbf{F}$
is a vector always normal to a surface
$S$
enclosing a volume
$V$
then
$\int \int \int_V \mathbf{\nabla} \times \mathbf{F} \: dV=0$
.
Proof
Let
$\mathbf{a}$
be a constant vector. The Divergence Theorem for the field
$\mathbf{F} \times \mathbf{a}$
states
$\int \int \int_V \mathbf{\nabla} \cdot ( \mathbf{F} \times \mathbf{a}) dV = \int \int_S ( \mathbf{F} \times \mathbf{a}) \cdot \mathbf{n} \: dS$

Since
$\mathbf{a}$
is a constant vector,
$\mathbf{\nabla} \cdot ( \mathbf{F} \times \mathbf{a}) = \mathbf{a} \cdot ( \mathbf{\nabla} \times \mathbf{f}) - \mathbf{F} \cdot ( \mathbf{\nabla} \times \mathbf{a}) = \mathbf{a} \cdot ( \mathbf{\nabla} \times \mathbf{f})$

and
$(\mathbf{F} \times \mathbf{a}) \cdot \mathbf{n} = \mathbf{F} \cdot (\mathbf{a} \times \mathbf{n} ) =(\mathbf{a} \times \mathbf{n} ) \cdot \mathbf{F} = \mathbf{a} \cdot ( \mathbf{n} \times \mathbf{F} )$

Hence
$\int \int \int_V \mathbf{a} \cdot ( \mathbf{\nabla} \times \mathbf{F}) \: dV = \int \int_S \mathbf{a} \cdot ( \mathbf{n} \times \mathbf{F}) \: dS$

or
$\mathbf{a} \cdot \int \int \int_V ( \mathbf{\nabla} \times \mathbf{F}) \: dV = \mathbf{a} \cdot \int \int_S ( \mathbf{n} \times \mathbf{F}) \: dS$

Hence
$\int \int \int_V ( \mathbf{\nabla} \times \mathbf{F}) \: dV = \int \int_S ( \mathbf{n} \times \mathbf{F}) \: dS$

$\mathbf{F}$
is normal to
$S$
so
$\mathbf{F} \times \mathbf{n} =0$
everywhere on
$S$

Therefore
$\int \int \int_V ( \mathbf{\nabla} \times \mathbf{F}) \: dV = 0$