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We can write the Divergence Theorem in terms of forms. In fact, if  
\[\omega^2 = f_1 dx_2 \wedge dx_3 + f_2 dx_3 \wedge dx_1 +f_3 dx_1 \wedge dx_2\]
  is defined on a region  
\[D\]
  of  
\[\mathbb{R}^3\]
  with surface  
\[S\]
  then the statement  
\[\int_D d \omega^2 = \int_S \omega^2 \]
  (1) is equivalent to the Divergence Theorem.
To see this not that with  
\[\omega^2\]
  as above,
\[d \omega^2 = (\frac{\partial f_1}{\partial x_1} + \frac{\partial f_2}{\partial x_2} + \frac{\partial f_3}{\partial x_3})dx_1 \wedge dx_2 \wedge dx_3\]

\[\frac{\partial f_1}{\partial x_1} + \frac{\partial f_2}{\partial x_2} + \frac{\partial f_3}{\partial x_3}\]
  is the divergence of the vector field  
\[\mathbf{F} = (f_1,f_2,f_3)^T\]

Substituting for  
\[\omega^2 , \: d \omega^2\]
  in (12) gives
\[\int_D (\frac{\partial f_1}{\partial x_1} + \frac{\partial f_2}{\partial x_2} + \frac{\partial f_3}{\partial x_3})dx_1 \wedge dx_2 \wedge dx_3 = \int_S f_1 dx_2 \wedge dx_3 + f_2 dx_3 \wedge dx_1 +f_3 dx_1 \wedge dx_2 \]

This is equivalent to the Divergence Theorem.