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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?

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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.

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  • $\begingroup$ Can we use stokes’s theorem to find surface integrals? $\endgroup$ – user515045 Jan 5 '18 at 2:41
  • $\begingroup$ Would you use stokes’s theorem to find the flux across a surface not a solid by finding some F that equals $\nabla $ $\times G$? $\endgroup$ – user515045 Jan 5 '18 at 2:41
  • $\begingroup$ We know that $$\iint_M (\nabla \times F) \cdot \hat{n} d\sigma = \int_{\partial M} F\cdot \hat{T} ds$$. Does $$\iint_M (\nabla \times F) \cdot \hat{n} d\sigma = \int_{\partial M} F\cdot \hat{N} ds$$ also hold? $ \hat N$ is the unit normal vector to a closed curve. where $ \hat N$ is normal to $ \hat T$ $\endgroup$ – user515045 Jan 5 '18 at 2:58
  • $\begingroup$ @Lasuiqw: No. There are three unit vectors involved here: $\hat{n}$ is normal to $M$, hence to $\partial M$ as well; $\hat{N}$ is tangent to $M$ but normal to $\partial M$ and pointing away from $M$, and $\hat{T}$ is tangent to $\partial M$. Try representing these three vectors when $M$ is the Northern hemisphere and $\partial M$ is the Equator, and compare to the same drawing when $M$ is the Southern hemisphere and $\partial M$ is stilll the Equator. $\endgroup$ – Catalin Zara Jan 5 '18 at 3:04

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