Proof of Formula for Volume Integrtion of Curl of a Vector Field

Theorem
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  
\[ \int \int \int_V ( \mathbf{\nabla} \times \mathbf{F}) \: dV = \int \int_S ( \mathbf{n} \times \mathbf{F}) \: dS\]