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Mathematically, from Stokes' theorem, it can be inferred that

$$\int_{\text{any closed surface}} (\nabla \times C) \cdot n \cdot da = 0.$$

But, physically, why is it so?

This is quoted from Feynman's Lectures:

We would like to see what happens when the loop shrinks down to a point, so that surface boundary disappears - the surface becomes closed. Now, if the vector $\mathbf{C}$ is everywhere finite, the line integral around the loop must go to zero as we shrink the loop. According to Stokes' theorem, the surface integral of $(\nabla \times \mathbf{C})_n$ must also vanish.

Mathematically, from Stokes' theorem, it can be inferred that

$$\int_{\text{any closed surface}} (\nabla \times \mathbf{C}) \cdot n \cdot da = 0.$$ But physically, what is going on that is making the circulation zero in the closed surface?

Mathematically, from Stokes' theorem  , it can be inferred that $$\int_{\text{any closed surface}} (\nabla \cross C) \cdot n \cdot da = 0$$. But, physically, why is it so?

Mathematically, from Stokes' theorem, it can be inferred that

$$\int_{\text{any closed surface}} (\nabla \times C) \cdot n \cdot da = 0.$$

But, physically, why is it so?