Show that $\oint f\nabla g \cdot d\alpha = -\oint g \nabla f \cdot d\alpha $ for every piecewise smooth Jordan curve $C$ in $S$.

Question

If $f$ and $g$ are continuously differntiable in an open connected set $S$ in the plane, show that $\oint f\nabla g \cdot d\alpha = -\oint g \nabla f \cdot d\alpha $ for every piecewise smooth Jordan curve $C$ in $S$.

Attempt:$\oint f\nabla g \cdot d\alpha + \oint g \nabla g \cdot d\alpha = f ~d(g(\alpha)) + g~d(f(\alpha)) = \iint_R ( \dfrac {\partial g}{\partial f}- \dfrac {\partial f}{\partial g} ) df~dg $ .

( By Green's Theorem)

How do I proceed now? We need to prove that $\dfrac {\partial g(\alpha)}{\partial f(\alpha)}= \dfrac {\partial f(\alpha)}{\partial g(\alpha)}$ ?

I just have a connecting idea that in an open connected space, a gradient follows the same behavior what we are looking to prove, but then, nothing is mentioned about the presence of a gradient function either.

Thank you very much for the help.

Answer 1

Since $f,g$ are in the pane, there are existing the following total differentials: $df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy := A_f dx + B_f dy$ and

$dg = \frac{\partial g}{\partial x} dx + \frac{\partial g}{\partial y} dy := A_g dx + B_g dy$. It arises the requirement $\frac{\partial B_f}{\partial x} = \frac{\partial A_f}{\partial y}$ etc. due to the theorem of Schwarz. To prove the Statement, expand $f,g$ in Terms of total derivatives by $x$ and $y$ and Show that $fdg+gdf = dF$ for a function $F$ because a closed path integral over a closed path vanishes by using the requirement for a total differential.

Answer 2

Hint: By the Gradient Theorem, $\displaystyle\oint_C \nabla h\cdot d\alpha =0$. Set $h=fg$, and expand.