Stoke's Theorem and 1 - Forms

We can derive Stoke's formula from the statement that for a 1 - form  
\[\omega^1 =f_1 dx_1 + f_2 dx_2 + f_3 dx_3\]
  with domain  
\[D\]
  and domain  
\[\partial D\]
,
\[\int_D d \omega^1 = \int_{\partial D} \omega^1 \]
.
The exterior derivative of  
\[\omega^1\]
  is
\[d \omega^1 =( \frac{\partial f_3}{\partial 2} - \frac{\partial f_2}{\partial 3}) dx_2 \wedge dx_3 +( \frac{\partial f_1}{\partial 3} - \frac{\partial f_1}{\partial 3}) dx_3 \wedge dx_1 + ( \frac{\partial f_2}{\partial 1} - \frac{\partial f_1}{\partial 2}) dx_2 \wedge dx_1\]

(1) becomes
\[\int_D \frac{\partial f_3}{\partial 2} - \frac{\partial f_2}{\partial 3}) dx_2 \wedge dx_3 +( \frac{\partial f_1}{\partial 3} - \frac{\partial f_1}{\partial 3}) dx_3 \wedge dx_1 + ( \frac{\partial f_2}{\partial 1} - \frac{\partial f_1}{\partial 2}) dx_2 \wedge dx_ = \int_{\partial D} f_1 dx_1 +f_2 dx_2 +f_3 dx_3 \]
.
which is equivalent to
 
\[\int_D \mathbf{\nabla} \times \mathbf{F} dD = \int_{\partial D} \mathbf{F} \cdot d \mathbf{r} \]
  where  
\[\mathbf{F} =(f_1 ,f_2 , f_3)^T\]
.

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