# Why does $\oint\mathbf{E}\cdot{d}\boldsymbol\ell=0$ imply $\nabla\times\mathbf{E}=\mathbf{0}$?

I am looking at Griffith's Electrodynamics textbook and on page 76 he is discussing the curl of electric field in electrostatics. He claims that since $$\oint_C\mathbf{E}\cdot{d}\boldsymbol\ell=0$$ then $$\nabla\times\mathbf{E}=\mathbf{0}$$ I don't follow this logic. Although I know that curl of $\mathbf{E}$ in statics is $\mathbf{0}$, I can't see how you can simply apply Stokes' theorem to equate the two statements.

If we take Stokes' original theorem, we have $\oint\mathbf{E}\cdot{d}\boldsymbol\ell=\int\nabla\times\mathbf{E}\cdot{d}\mathbf{a}=0$. How does this imply $\nabla\times\mathbf{E}=\mathbf{0}$? Griffiths seem to imply that this step is pretty easy, but I can't see it!

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What he is saying is: if the loop integral of every loop is zero then the curl must be zero everywhere. Convince yourself this is true. – BebopButUnsteady Jul 1 '13 at 21:56

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