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Why? The expression given by: $$\int_{S} \ dS \ \hat{\mathbf{n}} \cdot \left( \nabla \times \mathbf{A} \right) = \oint_{C} \ \mathbf{A} \cdot d\mathbf{l}$$ is just a vector calculus rule called Stokes' theorem and is valid regardless of the type of flow. Because the $S$ is not simply connected? The $S$ in the expression represents the closed ...
Exactly because the region is not simply connected. The stokes (or Green in 2d) theorem no longer holds. Consider for example the two dimensional vector field $$\vec F(x,y)=\frac{-y\hat i+x\hat j}{x^2+y^2},$$ which has vanishing curl and circulation $2\pi$ around a unit circle centerd at the origin. If this vector field is meant to be a flow velocity field ...