Does there exist $f$ such that $Df = \omega = \frac{-ydx + xdy}{x^2 + y^2}$?

Question

Does there exist $f \in \mathbb{R}^2 \setminus \{0 \}$ such that $Df = \omega = \frac{-ydx + xdy}{x^2 + y^2}$?

I know that $d\omega = 0$, and that the vector field that corresponds to $\omega$ is $F = \frac{x}{|x|^2}$, but I'm not sure how to continue from here.. Any help would be appreciated!

use http://math.stackexchange.com/questions/12022/calculating-oint-l-fracxdy-ydxx2-y2?rq=1 for @anomaly hint — , Mar 13 '16 at 18:29
Yeah, wrong vector field in the text of the question, now. The one you wrote is conservative (i.e., that $\omega$ is exact). — Ted Shifrin, Mar 14 '16 at 01:10

score 1 · Accepted Answer · answered Mar 13 '16 at 18:31

1

If such $f$ exists, then the integral of $\omega$ over any closed curve in $\Bbb R^2\setminus \{(0,0)\}$ must be zero, by the fundamental theorem of calculus. Check that $$\int_{\Bbb S^1}\omega \neq 0$$and conclude that such $f$ can't exist.

answered Mar 13 '16 at 18:31

Ivo Terek

77,665

Does there exist $f$ such that $Df = \omega = \frac{-ydx + xdy}{x^2 + y^2}$?

1 Answers1