Calculating $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$

The problem is "Calculating $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$, where L is a smooth, simple closed, and postively oriented curve that does not pass through the orgin".

Here is my solution:
Let $$P(x,y) = \frac{-y}{x^2 + y^2}, Q(x,y) = \frac{x}{x^2 + y^2}$$ Get $$\frac{\partial{P}}{\partial{y}} = \frac{y^2 - x^2}{(x^2 + y^2)^2} = \frac{\partial{Q}}{\partial{x}}$$ Acorrding to the Green formula: $$\oint_{L} \frac{xdy - ydx}{x^2 + y^2} = \iint (\frac{\partial{Q}}{\partial{x}} - \frac{\partial{P}}{\partial{y}})dxdy = 0$$

What's wrong with my solution?

Solution 1:

That is valid as long as $L$ is a simple closed curve and the origin is not "inside" $L$, i.e., in the bounded region determined by $L$. For example, this would work if $L$ is a unit circle centered at $(2,0)$, but not if it is the unit circle centered at the origin, and not if it is not a simple closed curve.

(As Timothy Wagner already pointed out before I finished writing, you need to check the hypotheses on the theorem you invoked to see when it works.)

If $L$ is closed, your integral should be $2\pi$ times the number of times $L$ winds around the origin.

Solution 2:

Hint: What does the hypothesis of Green's theorem say about the requirements on functions $P$ and $Q$?

Solution 3:

Just to add that I think that the value of the integral (i.e., 2Pi times the winding number) depends only on the homotopy-type of the curve, i.e., if a curve C is homotopic to the circle L that winds around the origin, then the integral over C will have the same value as the integral over L.