Conservative Vector Field Problem

[triangles]

Say we have the vector field

\boldsymbol{F}(x,y)=\frac{1}{x^2 + y^2}\begin{bmatrix}
-y\\
x
\end{bmatrix}

and we have a circle of any radius r

\varphi (t)=\begin{bmatrix}
r\ cos (t)\\
r\ sin (t)
\end{bmatrix}, 0\leq t\leq 2\pi

Then I get that the line integral of F over phi is always two pi.

\int_{\varphi }^{ }F\cdot ds = 2\pi

I also checked the line line integral of F over a positively oriented square connecting (1, -1), (1, 1), (-1, 1), (-1,-1) and it was also 2pi.
Then I checked a bunch of random triangles (containing the origin) and they all worked out to be 2pi.
But F=\bigtriangledown \psi where \psi(x,y)=-\arctan (x/y) (right? unless I've made some error).
So I know that the line integral of F over any closed path should be zero except when the region contained inside the path contains the origin because that region is not simply connected (since F is undefined at the origin) so green's theorem doesn't work. (Because if we applied green's theorem we would get zero right? since the integrand would be zero by equality of mixed partials?) I haven't learned why the region needs to be simply connected, only that it does.
When I computed a not closed path, say a line segment connecting (1, -1) to (1, 1) by saying that

\int_{\varphi }^{ }F\cdot ds=\int_{a}^{b}F(\varphi (t))\cdot \varphi '(t)dt=\int_{a}^{b}\bigtriangledown \psi (\varphi (t))\cdot \varphi '(t)dt=\int_{a}^{b}\frac{d}{dt}(\psi (\varphi (t))))dt=\psi (\varphi (b))-\psi (\varphi (a))

I get -pi/2, whereas direct computation gives me pi/2. So F is not the gradient here or something? What is causing this weirdness? Is there any easy way to see why the line integral is always 2pi? This has me very confused.

[skeptical scientist]

To explain why you're getting the wrong answer computing your integral:

Your F=-arctan(x/y) doesn't just have a singularity at the origin, but along the entire x-axis. It is true that psi is the gradient of a function on any simply connected region not containing the origin, and so the integral along any closed curve not containing the origin will be zero. However, psi is the gradient of F on the upper half plane, and the lower half plane, but not on any region overlapping the x-axis. So when you try to compute the integral along a curve starting in the lower half plane and ending in the upper half plane using F, you get the wrong answer. In fact, as you cross the x-axis, F has a jump discontinuity of size π, so that's what your answer is off by.

Check what happens when you use F=arctan(y/x) instead, which has a singularity along the x-axis rather than the y-axis.

I'm not sure if there's an easy way to see that the integral is 2π along any circle. However, using the fact that the integral is zero along any loop not containing the origin, and is 2π along the unit circle (which is easy enough to check, since you just get ∫₀^2π 1), one can see that for any closed loop whatsoever, the integral will just be 2π times the number of times the loop "winds around" the origin (counting clockwise loops as negative).

God it's been a long time since I've done multivariable calc.

[Ben-oni]

You're actually integrating

\oint_{C}\frac{-i}{z}dz

where C is a simply connected loop in the complex plane.

\frac{1}{2i\pi}\oint_{C}\frac{1}{z}dz

is the winding number of C, the number of times it loops around the origin. This is a trivial consequence of the Residue Theorem. As you've already noticed, any loop that does not surround the center integrates to zero, so you can add and subtract these zero loops to create a tighter bounding loop around the origin. In fact, so long as two loops are homotopic (one can be deformed in a continuous manner into the other), they must have the same integral. Moreover, this is an if and only if condition: if two loops have the same integral, they must be homotopic.

[triangles]

I did other examples and noticed the differences between the correct integral and the integral calculated using the gradient were always pi or 2pi. Now I see that it was pi when I crossed the x axis once, and 2pi when I crossed it twice. Thanks. I have convinced myself that the line integral around any circle is 2pi because the radius cancels and you're left with \int_{0}^{2\pi}dt like with the unit circle, but I want to know why it is 2pi for every loop.
Ben-oni, your answer seems interesting, but I'm not well-versed enough to really understand it. Though, I understand the idea that since I know the unit circle has an integral of 2pi, I can morph any closed curve containing the origin into the unit circle by adding or subtracting other paths that do not contain the origin, and thus do not change the value of the integral. Is that right? Is there some simple way to give me more intuition on that, like some exercises I could do that's not heavy on complex analysis?

EDIT: nevermind. After thinking about this for a while, I'm satisfied with my understanding of this problem for now. Thank you both.