I know that a surface integral is used to calculate the flux of a vector field across a surface. I know that Stokes's Theorem is used to calculate the flux of the curl across a surface in the direction of the normal vector.
So I'm guessing that the flux of a vector field across a surface is not the same thing as the flux of the curl across a surface?
Can we use Stokes's Theorem to calculate the flux of a vector field across a surface?
Can we use a surface integral to calculate the flux of the curl across a surface in the direction of the normal vector?
What's the difference between the flux of a vector field across a surface and the flux of the curl across a surface in the direction of the normal vector?
What's the difference between calculating the two-form used in Stokes's Theorem:
$$ \iint \nabla x F \cdot \vec{n} d\sigma$$
and the two form used in the vector Surface Integral:
$$ \iint F \cdot \vec{n} d\sigma$$?
Let $ F$ be a vector field, $ \vec{n}$ be the normal vector
Is the two-form used in Stokes's Theorem a Surface Integral?
For the first integral you can use Stokes' Theorem directly and compute the surface integral over a surface $M$ as a line integral over the boundary $\partial M$ (properly oriented):
$$\iint_M (\nabla \times F) \cdot \hat{n} d\sigma = \int_{\partial M} F\cdot \hat{T} ds$$
For the second, you have to find a vector potential for $F$ - that is, to express $F$ as $\nabla \times G$ for some to-be-determined-by-you vector field $G$: $$\iint_M F\cdot \hat{n} d\sigma = \iint_M (\nabla \times G) \cdot \hat{n} d\sigma = \int_{\partial M} G\cdot \hat{T} ds$$
So:
"Can we use Stokes's Theorem to calculate the flux of a vector field across a surface?" Yes, if you find a vector potential for the given vector field. Since the divergence of a curl is zero, that would not be possible if the divergence of $F$ were not zero.
"Can we use a surface integral to calculate the flux of the curl across a surface in the direction of the normal vector?" Yes, but the computation would likely be simplified by using Stokes' Theorem - hence computing a line integral instead of a surface integral.