$$\mathbf{v}=\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0\right)$$ is a vector field.
If we need to find $$\iint_{S}^{}(\nabla\times\mathbf{v})\cdot d\mathbf{a}$$ over a hemispherical surface placed on the $x$-$y$ plane centred at the origin, is it zero?
However if we use Stoke's theorem and calculate the line integral it gives a value $2\pi$. What's going on here?
Even though $\nabla\times\mathbf{v}=\mathbf{0}$, the domain of the field $\mathbf{v}$ is $D = \mathbb{R}^3 \setminus \left\{(0,0,z) ~|~ z \in \mathbb{R}\right\}$, i.e. $\mathbb{R}^3$ minus the $z$-axis. $D$ is not a simply connected region: for example, the boundary for your hemisphere surrounds the $z$-axis and cannot be continuously shrunk to a point while remaining in $D$. For any closed curve $X$ in $D$ which can be so shrunk, we have $\displaystyle\oint_X\mathbf{v}\cdot d\mathbf{r} = 0$