2
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.