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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 ...


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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 ...



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