## Integrating a Two Form

Let
$\omega^x =F_1 (\mathbf{x}) dx_2 \wedge dx_3+ F_2 (\mathbf{x}) dx_3 \wedge dx_1 + F_3 (\mathbf{x}) dx_1 \wedge dx_2$
where
$F_1 , \: F_2 , , \: F_3$
are continuous in
$\mathbb{R}^3$
.
Let
$\mathbf{f} : \mathbb{R}^2 \rightarrow \mathbb{R}^3$
be continuous and differentiable on a closed bounded rectangle
$K \subset \mathbb{R}^2$
.
We want to find an expression for
$\int_S \omega^2$
where
$\mathbf{f}(K)=S$

$dx_i \wedge dx_j (\mathbf{a}, \mathbf{b}) = det \left( \begin{array}{cc} dx_1 (\mathbf{a}) & dx_1 (\mathbf{b}) & \\ dx_2 (\mathbf{a}) & dx_2 (\mathbf{b})\end{array} \right)$

Then
\begin{aligned} \omega^2_{\mathbf{f} (\mathbf{u})} (\frac{\partial \mathbf{f}}{\partial u_1} , \frac{\partial \mathbf{f}}{\partial u_2}) &=F_1 (\mathbf{f} (\mathbf{u})) det \left( \begin{array}{cc} dx_2 (\frac{\partial \mathbf{f}}{\partial u_1}) & dx_2 (\frac{\partial \mathbf{f}}{\partial u_2}) & \\ dx_3 (\frac{\partial \mathbf{f}}{\partial u_1}) & dx_3 (\frac{\partial \mathbf{f}}{\partial u_2}) )\end{array} \right) \\ &+ F_2 (\mathbf{f} (\mathbf{u})) det \left( \begin{array}{cc} dx_3 (\frac{\partial \mathbf{f}}{\partial u_1}) & dx_3 (\frac{\partial \mathbf{f}}{\partial u_2}) & \\ dx_1 (\frac{\partial \mathbf{f}}{\partial u_1}) & dx_1 (\frac{\partial \mathbf{f}}{\partial u_2}) )\end{array} \right) \\ &+ F_3 (\mathbf{f} (\mathbf{u})) det \left( \begin{array}{cc} dx_1 (\frac{\partial \mathbf{f}}{\partial u_1}) & dx_1 (\frac{\partial \mathbf{f}}{\partial u_2}) & \\ dx_2 (\frac{\partial \mathbf{f}}{\partial u_1}) & dx_2 (\frac{\partial \mathbf{f}}{\partial u_2}) )\end{array} \right) \\ &= F_1 (\mathbf{f} (u_1 , u_2)) \frac{ \partial (f_2,f_3)}{\partial (u_1,u_2)}+ F_2 (\mathbf{f} (u_1 , u_2)) \frac{ \partial (f_3,f_1)}{\partial (u_1,u_2)} + F_3 (\mathbf{f} (u_1 , u_2)) \frac{ \partial (f_1,f_2)}{\partial (u_1,u_2)} \end{aligned}

Then
$\int_S \omega^2 = \int_K [F_1 (\mathbf{f} (u_1 , u_2)) \frac{ \partial (f_2,f_3)}{\partial (u_1,u_2)}+ F_2 (\mathbf{f} (u_1 , u_2)) \frac{ \partial (f_3,f_1)}{\partial (u_1,u_2)} + F_3 (\mathbf{f} (u_1 , u_2)) \frac{ \partial (f_1,f_2)}{\partial (u_1,u_2)}]du_1 du_2$