Integrating a 2 - form Example
\[\int_S \omega^2\]
where \[\omega^2 =-2 dx_2 \wedge dx_1 + dx_1 \wedge dx_3\]
,, \[K=((u_1,u_2): 0 \leq u_1 \leq 1 , 0 \leq u_2 \leq \pi )\]
and \[S\]
is the image of \[K\]
under the function \[\mathbf{f} (u_1, u_2) =(u_1 cos u_2 ,u_1 sin u_2, u_2 )\]
We use this result.
\[\int_S \omega^2 = \int_K [F_1 (\mathbf{f} (u_1 , u_2)) \frac{ \partial (f_2,f_3)}{\partial (u_1,u_2)}+ F_2 (\mathbf{f} (u_1 , u_2)) \frac{ \partial (f_3,f_1)}{\partial (u_1,u_2)} + F_3 (\mathbf{f} (u_1 , u_2)) \frac{ \partial (f_1,f_2)}{\partial (u_1,u_2)}]du_1 du_2\]
\[F_1 =0, \: F_2 =-1, \: F_3 =2\]
\[ \frac{ \partial (f_2,f_3)}{\partial (u_1,u_2)} = \left| \begin{array}{cc} sin u_2 & cos u_2 \\ 0 & 1 \end{array} \right| = sin u_2\]
\[ \frac{ \partial (f_3,f_1)}{\partial (u_1,u_2)} = \left| \begin{array}{cc} 0 & 1 \\ cos u_2 & -u_1 sin u_2 \end{array} \right| = cos u_2\]
\[ \frac{ \partial (f_1,f_2)}{\partial (u_1,u_2)} = \left| \begin{array}{cc} cos u_2 & -u_1 sin u_2 \\ sin u_2 & u_1 cos u_2 \end{array} \right| = u_1\]
Then
\[\int_S \omega^2 = \int_{u_2 =0}^{/pi} \int_{u_1 =0}^1 ((-1)(-cos u_2)+2u_1)du_1 du_2 = \int_{u_2 =0}^{\pi} (cos u_2 +1) du_2 = \pi \]
