Why I can't use stokes to compute $\int_{\Gamma}v\cdot dr$?

Question

Let $v:E\backslash Oz\to \mathbb R^3$ defined by $$v(x,y,z)=\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0\right).$$ Let $\Gamma=\{(x,y,z)\mid x^2+y^2=1, z=0\}$. They ask me to compute $$\int_{\Gamma}v\cdot dr.$$

I have that $Curl(v)=0$, so by Stokes (an a domain $D$ such that $\Gamma=\partial D$), I have that $$\int_{\Gamma}v\cdot dr=0.$$

But it's wrong and I don't understand why.

Answer 1

The vector field $v(x,y,z)$ is defined on a Domain that has a "hole" at the point $x=0,y=0$, i.e. the Domain is not simply connected. In this case, the LOCAL value of $curl(v)$ is equal to Zero, BUT GLOBALLY, i.e. if you will compute a line integral over $\Gamma$ you will get a different value!

Note that Stoke's Theorem can be applied only if $D$ is simply connected (no "holes").