Can someone help me? Any sugestion will be helpful. According to Courant and John in Introduction to Calculus and Analysis, Vol. 2 chapter 5 section 5.1, if we take:
\begin{equation} L = f(x,y)dy - g(x,y)dx \end{equation}
the differential form of L will be:
\begin{equation} dL = dfdy - dgdx \end{equation}
\begin{equation} dL = (f_xdx +f_ydy)dy - (g_xdx + g_ydy)dx \end{equation}
\begin{equation} dL = f_xdxdy + f_ydydy - g_xdxdx - g_ydydx \end{equation}
By alternating differential forms we know that $dxdy = -dydx$ and $ dxdx = dydy = 0$.
So, the answer will be:
\begin{equation} dL = (f_x + g_y)dxdy \end{equation}
I know that alternating functions have an important role in the calculus of determinants, as well in vectorial product. But, I do not understand the relation here, I tried to reach the answer performing a determinant, and the best I did was this:
\begin{equation} \begin{vmatrix} f_xdx & dx \\ g_xdx & dy \end{vmatrix} + \begin{vmatrix} f_ydy & dx \\ g_ydy & dy \end{vmatrix} \end{equation}
But it makes no sense to me. So, why to use differential alternating form ? Why it is applicable in this situation ? This doubt has been tormenting me for a long time.
What is related to the determinant is the operation $dxdy$ which should be written as $dx\wedge dy$ and is defined as a bilinear map $\mathbb R^2\times \mathbb R^2\to \mathbb R$ by $$dx\wedge dy\ (v,w)=dx(v)dy(w)-dx(w)dy(v),$$ $$\qquad =v^1w^2-v^2w^1.$$ This number is the area of the parallelogram spawned by the vectors $v,w$.
So, your $dL$ is proportional to $dx\wedge dy$, with the scalar $f_x+g_y$ as proportion.