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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.

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    The vector field $v$ has a singularity along the $z$-axis, so there is no domain with boundary $\Gamma$ on which $v$ is smooth. You could of course consider a domain with a boundary containing $\Gamma$ (for example, an annulus) but this would only switch the path of integration (to another curve around the $z$-axis). – 2017-01-16

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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").

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    What do you think about the cylinder $D=\{(x,y,z)\mid x^2+y^2=1, z\geq 0\}$ ? Then $D$ is simply connected and $\partial D=\Gamma$. – 2017-01-16
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    That cylinder contains the z axis which is the problem – 2017-01-16
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    @Triatticus: It's indeed inside, but not on the surface, and since I integrate on the surface, it should be a problem, no ? – 2017-01-19
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    The cylinder will not contain the z axis, the surface you create will be an annulus as v isn't defined at the origin and so D isn't simply connected in a sense that any path circling the origin cannot be squeezed to a ppint – 2017-01-19