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I have a question relating an identity using Stokes theorem.

$$\iint_A \left( \nabla \otimes \vec{F}\right) \cdot \vec{dA} = \oint_c \vec{F} \cdot \vec{dr}$$

But my professor mentioned something about the divergence theorem

$$\int_V \left( \nabla \cdot \vec{F} \right)dV = \oint_{\partial V} \vec{F} \cdot \vec{dA}$$

so does this mean I can say:

$$\iint_A \left( \nabla \otimes \vec{F}\right) \cdot \vec{dA} = \oint_c \vec{F} \cdot \vec{dr} = \iint_A \left( \nabla \cdot \vec{F} \right) dA$$

Because this would imply:

$$\left( \nabla \otimes \vec{F}\right) \cdot \vec{dA}= \left( \nabla \cdot \vec{F} \right) dA$$

And I don't believe this to be correct. Thank you for your help in advance!

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1 Answer 1

up vote 2 down vote accepted

It's wrong because you are equating very different things, used in very different contexts. In Stokes' Theorem, you are expressing the value of a line integral of the tangential component of a vector field about a closed circuit in terms of the normal component of the curl of that vector field over a "capping" surface of that circuit, which is an open area. On the other hand, the divergence theorem states that the net flux of the normal component of a vector field through a closed surface is equal to the amount of the divergence of that vector field in the volume defined by that surface. So the "areas" you are equating cannot be equated: one is open, and one is closed.

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Thank you very much! –  user62239 Feb 14 '13 at 2:12

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