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Proof of Identity For Integral of Cross Product of Vector Field With Tangent Around Closed Curve

Theorem
If  
\[\mathbf{F}\]
  is a vector field with differentiable components then  
\[\oint_C d \mathbf{r} \times \mathbf{F}= \int \int_S (\mathbf{n} \times \mathbf{\nabla} ) \times \mathbf{F} dS\]
  and  
\[C\]
  is the boundary of  
\[S\]

Proof
Stoke's Theorem states  
\[\oint_C \mathbf{A} d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times \mathbf{A}) \cdot \mathbf{n} dS\]

Let  
\[\mathbf{A} = \mathbf{F} \times \mathbf{c}\]
  (1) where  
\[\mathbf{c}\]
  is a constant vector, then,br>
\[\oint_C (\mathbf{F} \times \mathbf{c}) \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times ( \mathbf{F} \times \mathbf{c})) \cdot \mathbf{n} dS\]

Now use the identity  
\[\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) =(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}\]

and the properties of the dot product to rewrite this as
\[\oint_C (\mathbf{F} \times \mathbf{c}) \cdot d \mathbf{r} = \oint_C d \mathbf{r} \cdot (\mathbf{F} \times \mathbf{c}) = \oint_C (d \mathbf{r} \times \mathbf{F}) \cdot \mathbf{c} = \oint_C \mathbf{c} \cdot ( d \mathbf{r} \times \mathbf{F}) = \mathbf{c} \cdot \oint_C ( d \mathbf{r} \times \mathbf{F})\]
  (2)
Npw use the identity  
\[\mathbf{\nabla} \times (\mathbf{a} \times \mathbf{b})=(\mathbf{b} \cdot \mathbf{\nabla} ) \mathbf{a} - \mathbf{b} ( \mathbf{\nabla} \cdot \mathbf{a}) - (\mathbf{a} \cdot \mathbf{\nabla} ) \mathbf{b} + \mathbf{a} ( \mathbf{\nabla} \cdot \mathbf{b})\]

and note that  
\[\mathbf{\nabla} \cdot \mathbf{c} = 0\]
  since  
\[\mathbf{c}\]
  is a constant vector to get
\[\begin{equation} \begin{aligned} \int \int_S (\mathbf{\nabla} \times (\mathbf{F} \times \mathbf{c})) \cdot \mathbf{n} dS &= \in \int_S ((\mathbf{c} \cdot \mathbf{\nabla}) \mathbf{F} - \mathbf{c} \mathbf{\nabla} \cdot \mathbf{F})) \cdot \mathbf{n} dS \\ &= \int \int)S (\mathbf{c} \cdot \mathbf{\nabla})\mathbf{F} \cdot \mathbf{n} dS - \int \int_S ( \mathbf{c} (\mathbf{\nabla} \cdot \mathbf{F})) \cdot \mathbf{n} dS \\ &= \int \int_S \mathbf{c} \cdot (\mathbf{\nabla} (\mathbf{F} \cdot \mathbf{n}))dS - \int \int_S \mathbf{c} \cdot (\mathbf{n} (\mathbf{\nabla} \cdot \mathbf{F})) dS \end{aligned} \end{equation}\]

Substitute this and (2) into (1) to get
\[\begin{equation} \begin{aligned} \mathbf{c} \cdot \oint_C d \mathbf{r} \cdot \mathbf{F} &= \int \int_S \mathbf{c} \cdot (\mathbf{\nabla} (\mathbf{F} \cdot \mathbf{n}))dS - \int \int_S \mathbf{c} \cdot (\mathbf{n} (\mathbf{\nabla} \cdot \mathbf{F})) dS \\ &= \mathbf{c} \cdot \int \int_S (\mathbf{\nabla} (\mathbf{F} \cdot \mathbf{n}))dS - \int \int_S (\mathbf{n} (\mathbf{\nabla} \cdot \mathbf{F})) dS \\ &= \mathbf{c} \cdot \int \int (\mathbf{n} \times \mathbf{\nabla} ) \times \mathbf{F} dS \end{aligned} \end{equation}\]

Removing  
\[\mathbf{c} \cdot {}\]
  gives
\[ \oint_C d \mathbf{r} \cdot \mathbf{F} = \int \int (\mathbf{n} \times \mathbf{\nabla} ) \times \mathbf{F} dS \]

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