Proof of Identity for Intefral of Product of a Function With a Gradient Around a Curve

Theorem
If  
\[\phi , \: \psi\]
  are functions and  
\[S\]
  is a surface with boundary  
\[C\]
  then  
\[\oint_C \phi \mathbf{\nabla} \psi \cdot d \mathbf{r} = \int \int (\mathbf{\nabla} \phi \times \mathbf{\nabla} \phi ) \cdot d \mathbf{n} \]
, with  
\[C\]
  taken anticlockwise about  
\[S\]
. Proof
Stoke's Theorem states  
\[\oint_C \mathbf{F} \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS \]

Let  
\[\mathbf{F} = \phi \mathbf{\nabla} \psi \]
  then  
\[\oint_C (\phi \mathbf{\nabla} \psi) \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times (\phi \mathbf{\nabla} \psi)) \cdot \mathbf{n} dS \]

Now use the identity  
\[(\mathbf{\nabla} \times (\phi \mathbf{\nabla \psi}) = \phi \mathbf{\nabla} \times \mathbf{\nabla \psi} + (\mathbf{\nabla} \phi \times \mathbf{\nabla \psi} ) = (\mathbf{\nabla} \phi \times \mathbf{\nabla \psi} )\]
 
Since  
\[ \phi \mathbf{\nabla} \times \mathbf{\nabla \psi} =0\]

We get  
\[\oint_C (\phi \mathbf{\nabla} \psi) \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \phi \times \mathbf{\nabla \psi} ) \cdot \mathbf{n} dS \]

Substitute the last identity into  
\[\oint_C (\phi \mathbf{\nabla} \psi) \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times (\phi \mathbf{\nabla} \psi)) \cdot \mathbf{n} dS \]

to get  
\[\oint_C \phi \mathbf{\nabla} \psi \cdot d \mathbf{r} = \int \int (\mathbf{\nabla} \phi \times \mathbf{\nabla} \phi ) \cdot d \mathbf{n} \]
.

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