Integrating a Function About a Curve

Theorem
If  
\[\phi\]
  is a function and  
\[S\]
  is a surface with boundary  
\[C\]
  then  
\[\oint_C \phi d \mathbf{r} = \int \int d \mathbf{S} \times (\mathbf{\nabla} \phi)\]
, 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  
\[\mathbf{F} = \phi \mathbf{a}\]
  where  
\[\mathbf{a}\]
  is a constant vector then  
\[\oint_C ( \phi \mathbf{a} ) \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times ( \phi \mathbf{a})) \cdot \mathbf{n} dS \]

We can take the dot product out as a factor from the left hand side  
\[\oint_C (\mathbf{a} \phi ) \cdot d \mathbf{r} = \mathbf{a} \cdot \oint_C \phi d \mathbf{r} \]

On the right hand side, since  
\[\mathbf{a}\]
  is a constant vector,  
\[\mathbf{\nabla} \times (\phi \mathbf{a})= (\mathbf{\nabla} \phi ) \times \mathbf{a}\]
  so the surface integral becomes  
\[ \int \int_S (\mathbf{\nabla} \times ( \phi \mathbf{a})) \cdot \mathbf{n} dS = \int \int_S (\mathbf{\nabla} \phi) \times ( \mathbf{a})) \cdot \mathbf{n} dS \]

Now use the identity  
\[(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})= \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b} )\]
  to get
\[ \int \int_S (\mathbf{\nabla} \phi) \times ( \mathbf{a})) \cdot \mathbf{n} dS = \int \int_S \mathbf{a} \cdot (\mathbf{n} \phi ( \mathbf{\nabla} \phi)) dS = \mathbf{a} \cdot \int \int_S d \mathbf{S} \times (\mathbf{\nabla} \phi)\]

Since  
\[\mathbf{a}\]
  is arbitrary,  
\[\oint_C \phi d \mathbf{r} = \int \int d \mathbf{S} \times (\mathbf{\nabla} \phi)\]

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