Vector field doesn't vanish in a simply connected region

Question

Suppose I have the following vector field $$\underline{G} = \frac{-y}{x^2+y^2}\underline{i}+\frac{x}{x^2+y^2}\underline{j}$$ this is clearly defined for $(x,y)\neq (0,0)$, i.e. the $z$ axis.

Suppose I want to find the work along the unit circle $C$ centred at the origin and lying on the plane $z=0$, parametrised as $\underline{r}(t) = \cos(t)\underline{i}+\sin(t)\underline{j}$ for $0\leq t\leq 2\pi$. Now if we calculate the line integral we see that $$\int_C\underline{G}\cdot d\underline{r} = 2\pi\neq 0$$

the reason I gave in the question was that this happens because the region inside the circle is not simply connected, indeed we cannot shrunk the circle to the origin!

Then I tried to show this even further and say, consider the unit circle centered at $(2,2)$ (so it doesn't contain the origin, nor the axes) then the work along $C_2 : \underline{r}_2(t) = (\cos(t)+2)\underline{i}+(\sin(t)+2)\underline{j}$ should be zero as the region there is simply connected! However it is not zero..

where did I go wrong? Are my assumptions wrong?

Edit

It works indeed, wrote it wrong on Mathematica

Answer 1

Yes, the integral on the unit circle around $\;(2,2)\;$ is zero, and the reason is simple: the vector field has a potencial function there:

$$\phi(x,y)=\arctan\frac yx$$

so the value of the integral around the unit circle centered at $\;(2,2)\;$ is zero...and in fact, over any closed, simple smooth curve not containing the origin.

Answer 2

The integral should come to zero. We have $$G(r_2(t)) = \frac{-\sin t-2}{4\cos t+4\sin t+9}\mathbf{i}+\frac{\cos t+2}{4\cos t+4\sin t+9}\mathbf{j}$$ $$r_2'(t) = -\sin(t)\mathbf{i}+\cos(t) \mathbf{j}$$ And so $$G(r_2(t))\cdot r_2'(t) = \frac{1+2\sin t +2\cos t}{4\cos t+4\sin t +9} = \frac{1}{2}-\frac{3.5}{4\cos t+4\sin t+9}$$ The integral of which should come out to zero.

Alternatively, define $f$ in the upper half plane, so that for $x\geq 0$ we have $f(x,y) = \tan^{-1}(y/x)$, and for $x < 0$, $f(x,y)=\pi+\tan^{-1}{y/x}$. That is, given $(x,y)$ in the upper half plane, $f$ gives us the polar angle (between $0$ and $\pi$). Then $G = \nabla f$, and so all integrals around a closed curve should be zero.

Note: The reason this doesn't work on $\Bbb{R}^2$ is because the polar angle cannot be defined in a continuous manner on the whole plane.