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I've been trying to prove that this vector field:
$$ \vec{F}=\left(\frac{y}{\left(x-1\right)^{2}+y^{2}},\frac{1-x}{\left(x-1\right)^{2}+y^{2}}\right) $$
Is conservative in:
$$ D=\left\{ \left(x,y\right)\mid1\leq\left(x+1\right)^{2}+y^{2}\leq2\right\} $$
Proof attempt #1:
To my understanding, it holds that in a simply-connected domain, a vector field $P\hat{x}+Q\hat{y}$ whose partial derivatives are continuous is conservative if and only if $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$.
I confirmed that in the circle centered at $\left(-1,0\right)$ with a radius of $2$, which is a superset of $D$, the partial derivatives are indeed continuous, and furthermore that:
$$ \frac{\partial Q}{\partial x} = \frac{\left(x-1\right)^{2}-y^{2}}{\left(\left(x-1\right)^{2}+y^{2}\right)^{2}} = \frac{\partial P}{\partial y} $$
So in that circle, $\vec{F}$ is indeed conservative.
Thus: for any closed, piecewise-smooth curve in that circle, we have that the line integral of $\vec{F}$ over that curve is $0$. Since any such curve in $D$ is also a curve in the circle (and since adding that hole-of-radius-1 in the middle of it doesn't affect any of the components of a line integral: the field in all points on the curve and any parametrization of the curve, then it also doesn't affect the line integral itself), we have that every such curve in D also has that property, and so $\vec{F}$ is conservative there, as well.

Does the above proof seem okay (if I make it a tad more rigorous, maybe)?

Proof attempt #2:
Another possible proof method that I'd considered might be using Green's Theorem to show that the line integral over the unit circle centered at $\left(-1,0\right)$ results in $0$, and then use a theorem (which I believe I can prove) by which if a field has the aforementioned property of $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$ (and probably a few other requirements) then any closed, simple curve around such a "hole" in its domain would result in the exact same value, so not exactly conservative because that value doesn't have to be $0$, but since I already showed that one such integral does evaluate to $0$, then all of them do, so the field is indeed conservative.
Does this one make sense?

AnonA
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    Both of those seem fine. I see the problem is being cute by using a field with a delta function curl at $(1,0)$, but said delta function is on the outside of $D$. – eyeballfrog Jun 24 '23 at 11:59
  • Thanks, @eyeballfrog – iirc, before trying to show that it is conservative in this $D$, I showed that it isn't conservative in a similar circle, centered at the problematic point ($(1,0)$) – I did this by computing a line integral over a circle around the center and showing it is nonzero. I'm not sure what curl/delta function are yet, but I assume this is related? – AnonA Jun 24 '23 at 12:07
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    Yeah, the vector field has zero curl at all points except $(1,0)$, where it's undefined. However, the integral around any curve that contains $(1,0)$ will have the same constant value. So it's as though the field has nonzero total curl, but it's all "concentrated" at $(1,0)$. A delta function refers to this phenomenon, where all of the value of a function seems to be concentrated at a single point. It's a kind of handwavey thing, because no actual function can have this property, but it turns out there is a way to put it on solid ground with more complicated math. – eyeballfrog Jun 24 '23 at 12:12
  • The vector field is yet another tweak of this one. We can safely say that it is conservative in any domain that does not allow closed curves around the singularity. (Your $D$ fulfils that condition.) A look at the picture in the link should make visually clear on what domains the field is conservative. – Kurt G. Jun 24 '23 at 12:28
  • Thanks a lot for this, guys – very helpful and interesting :) Looking forward to learning more about this. – AnonA Jun 24 '23 at 13:29

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