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The exercise consists of calculating the integral $\int_{\gamma} \mathbf{F} \cdot d \mathbf{r}$ where

$\mathbf{F}(x,y)=\left( - \frac{y-1}{(x-1)^2 + (y-1)^2}, \frac{x-1}{(x-1)^2 + (y-1)^2} \right) + \left(\frac{y}{x^2 + y^2}, \frac{-x}{x^2 + y^2} \right)$

and $\gamma$ is the circle $x^2 + y^2 = 4$ oriented anticlockwise, and to determine if the integral is independent of its path on the set $D=\{(x,y): (x,y) \neq (0,0) \text{ and } (x,y) \neq (1,1) \}$

For the first part, I concluded that using Green's theorem is not an option, since the disc $x^2 + y^2 \leq 4$ contains singular points. Since it is appears to be a bit of a tricky function, I also chose not to parametrise it directly.

Instead I opted to determine the potential function to the integral. I show that $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, and, solving it like a system of equations, found that the potential function is $U(x,y) = \arctan{(\frac{y-1}{x-1})} - \arctan{(\frac{y}{x})}$. If I $U$ actually is a conservative vector field, I can then evaluate the integral simply by using the starting points and end points of $U$, by letting the starting point and end point to be, for instance, $(\sqrt{2}, \sqrt{2})$. Since $\gamma$ is a closed curve, the integral is zero.

I then argue that, on the set $D$, the integral is independent of its path, since $\mathbf{F}$ is conservative, and the potential function $U$ does not have any singular points on $D$.

I am a bit unsure whether this is the right approach, but I cannot see how else I would solve this exercise. Any input would be much appreciated.

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  • $\begingroup$ $\mathbf F$ is not in fact path-independent over the entire region. Try integrating along a convenient closed curve that encloses one of the singular points. It is, however path-independent over any simply-connected open subset of $D$. $\endgroup$ – amd May 2 '18 at 21:41
  • $\begingroup$ Thank you, this clears things up. To follow up, does this invalidate my answer to the value of the integral? If so, any suggestions on how to go about the problem? $\endgroup$ – chrstnsn May 3 '18 at 7:25
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If $\mathbf F$ is path-independent, then its integral along any simple closed curve must be zero. It suffices, then, to demonstrate a closed path $\Gamma$ along which the integral doesn’t vanish.

This particular vector field is the sum of two copies of one of the standard examples of an irrotational field that is not conservative. Let $$\mathbf G(x,y) = \left(\frac{y}{x^2 + y^2}, \frac{-x}{x^2 + y^2} \right)$$ so that $\mathbf F(x,y) = \mathbf G(x,y)-\mathbf G(x-1,y-1)$. Choose for our $\Gamma$ the unit circle with counterclockwise orientation. By linearity of the integral, we can consider each copy of $\mathbf G$ separately. For the first term, use the obvious parameterization to get $$\int_\Gamma \mathbf G\cdot d\mathbf r = \int_0^{2\pi}(\sin t,-\cos t)\cdot(-\sin t,\cos t)\,dt = -\int_0^{2\pi}dt = -2\pi.$$ For the other term, you can use Green’s theorem since $\Gamma$ doesn’t enclose a singular point of $\mathbf G(x-1,y-1)$. That integral vanishes on $\Gamma$, therefore $$\int_\Gamma \mathbf F\cdot d\mathbf r = -2\pi \ne 0$$ and so $\mathbf F$ is not conservative on $D$.

Finally, as far as integrating $\mathbf F$ over the larger circle goes, you could switch to the complex domain and use the residue theorem, but otherwise no particular trick for simplifying the computation comes to mind. Parameterizing $\gamma$ in the obvious way eventually leads to the integral $$-\int_0^{2\pi} {\cos t+\sin t-1 \over 2\cos t+2\sin t-3} dt.$$ This happens to be the equal to $-\int_\Gamma \mathbf G(x-1,y-1)\cdot d\mathbf r = 0$ ($\Gamma$ is the unit circle), but the only reason I know this is that I had expanded the latter integral while exploring the path-independence of $\mathbf F$ earlier.

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Your field ${\bf F}$ has singularities at the points ${\bf 0}$ and ${\bf p}=(1,1)$. It can be written as $${\bf F}={\bf A}_{\bf p}-{\bf A}_{\bf 0}\ ,$$ whereby $${\bf A}_{\bf 0}(x,y)=\nabla{\rm arg}(x,y)=\left({-y\over x^2+y^2},{x\over x^2+y^2}\right)\ ,$$ and ${\bf A}_{\bf p}$ is the field ${\bf A}_{\bf 0}$ translated to ${\bf p}$. When $x>0$ then $${\bf A}_{\bf 0}(x,y)=\nabla\arctan{y\over x}\ .$$ For any curve $\gamma$ the line integral $$\int_\gamma{\bf A}_{\bf 0}\cdot d{\bf z}$$ is equal to the sum of the increments $\Delta\phi$ of the polar angle $\phi$ along $\gamma$. If $\gamma$ is a loop going $n_{\bf 0}$ times counterclockwise around the origin ${\bf 0}$ then the sum of these increments is $2\pi n_{\bf 0}$. Similarly for ${\bf A}_{\bf p}$, where we now have to count the number $n_{\bf p}$ of times the loop $\gamma$ encircles the point ${\bf p}$.

Taken together we obtain that for any loop $\gamma$ avoiding ${\bf 0}$ and ${\bf p}$ we have $$\int_\gamma{\bf F}\cdot d{\bf z}=2\pi(n_{\bf p}-n_{\bf 0})\ .$$

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