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
\[ \int \int \int_V ( \mathbf{\nabla} \times \mathbf{F}) \: dV = 0\]