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

Add comment

Security code
Refresh