Let
\[\mathbf{F}\]
be a continuously differentiable vector field on a region \[V\]
with surface \[S\]
. Then\[\int \int \int_V \mathbf{\nabla} \times \mathbf{F} \: dV = \int \int_S \mathbf{n} \times \mathbf{F} \: dS\]
Proof
Let
\[\mathbf{a}\]
be a constant vector. Apply the Divergence Theorem to \[\mathbf{F} \times \mathbf{a}\]
.
\[\int \int \int_V \mathbf{\nabla} \cdot ( \mathbf{F} \times \mathbf{a}) dV = \int \int_S ( \mathbf{F} \times \mathbf{a}) \cdot \mathbf{n} \: dS\]
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
\[ \int \int \int_V ( \mathbf{\nabla} \times \mathbf{F}) \: dV = \int \int_S ( \mathbf{n} \times \mathbf{F}) \: dS\]