Stokes Theorem and flux

Question

Recently I came across Stokes Theorem,
$$\oint_C\vec F\cdot d\vec r = \iint_S(\vec\nabla\times \vec F)\cdot \hat{n}\;dS$$ I was confused about why the RHS had $\hat{n}\;dS$. This is usually associated with finding flux through certain areas. Like in the divergence theorem. But in the case of Stokes theorem, since it's just a line integral of a closed curve in 3d space, why is there an expression that is usually associated with flux? I tried looking at the proof of the theorem but it didn't helps me much.

Answer 1

Stokes' theorem can be derived informally. I'm sure by now you've seen plenty of proofs of Stokes' theorem for graphs. Like the Divergence theorem, Stokes' theorem is a statement which relates the global property of a vector field with its local property. The global property would be the closed line integral side. It measures the tendency of $\vec F$ to circulate around the curve $C$. The flux integral side says to sum of all the local circulations (or curl) through the surface $S$ whose boundary is $C$. These two expression are equivalent due to internal cancelation. In other words, only the boundary matters.

You may find more conceptual information here: https://mathinsight.org/stokes_theorem_idea