Curl of the electrostatic field In Griffith's EM text he calculates the curl for the E field of a point charge (at the origin). 
He shows that the line integral of an arbitrary closed loop is zero:
$$
\oint E\cdot dl = 0
$$
and then immediately invokes Stoke's Theorem to conclude that the curl is 0. However, this step is not obvious to me. From Stoke's Theorem we know that
$$
\int (\nabla\times E)\cdot da=0.
$$
Why are we allowed to conclude that the integrand is 0?
 A: 
In Griffith's EM text he calculates the curl for the E field of a
  point charge (at the origin). 
He shows that the line integral of an arbitrary closed loop is zero:
  $$ \oint E\cdot dl = 0 $$
and then immediately invokes Stoke's Theorem to conclude that the curl
  is 0. However, this step is not obvious to me. From Stoke's Theorem we
  know that $$ \int (\nabla\times E)\cdot da=0. $$
Why are we allowed to conclude that the integrand is 0?

It is not zero in general.
But, if (as you state) he has shown that 
$$ \oint E\cdot dl = 0 $$
for arbitrary closed loops,
then it is true that 
$$ \int (\nabla\times E)\cdot da=0 $$
for arbitrary areas.
Because
$$ \int (\nabla\times E)\cdot da=\oint E\cdot dl. $$
And so the integrand $\nabla\times E$ is zero in this case because the integral is zero for arbitrary areas.
A: If the closed line integral of a vector field is 0, then by the fundamental theorem for line integrals it can be expressed as a gradient of some scalar function. Substituting the gradient of this scalar function for $E$ in the second equation shows that the second integral must be 0, because the curl of a gradient is always 0.
