I have looked at several elementary proofs (i.e., using only basic calculus, and not using differential form or manifold) of Stokes' theorem in books and Wikipedia, and all seem to use the fact that the mixed second partials of the parametrization of the surface are equal ($\psi_{uv}=\psi_{vu}$). But the statement of Stokes' theorem assumes that the surface is smooth, which only requires $C^1$ parametrization (and nonvanishing Jacobian). Is it possible that in Stokes' theorem, we need that the surface is $C^2$, not $C^1$ ?
2026-03-27 20:31:39.1774643499
On elementary proof of Stokes' theorem
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