Stokes Theorem

[Gregg]

Homework Statement

Prove that

$\oint_{\partial S} ||\vec{F}||^2 d\vec{F} = -\int\int_S 2 \vec{F}\times d\vec{A}$

Homework Equations

Identities:

$\nabla \times (||\vec{F}||^2 \vec{k}) = 2\vec{F} \times \vec{k}$

For $\vec{k}$ constant i.e. $\nabla \times \vec{k} = 0$

Stokes Theorem

$\oint_{\partial S} \vec{B} \cdot d\vec{x} = \int\int_S (\nabla \times \vec{B})\cdot d \vec{A}$

The Attempt at a Solution

So I need to use that identity $\nabla \times (||\vec{F}||^2 \vec{k}) = 2\vec{F} \times \vec{k}$

The problem is that Stokes theorem is in a different form. The constant vector here I think is the k=dA.

I really can't think of what to do

 

[DryRun]

Well, $\vec F$ is the vector field, but I'm not sure what $||\vec F||$ represents in the original equation.

 

[Gregg]

I made an error. It is a squared term.

 

[DryRun]

Assuming that there are no more mistakes in your first post, $\vec k$ represents the outward unit normal vector from the surface, S.

 

[clamtrox]

Maybe you should use Stokes' theorem for each component of the original vector-valued integral
$$\vec{I}= \oint ||\vec{F}||^2 d\vec{F}$$
$$I_x = \oint (||\vec{F}||^2 \vec{e}_x) \cdot d\vec{F}$$
etc. Now it's in the correct form.

 

[Gregg]

Maybe you should use Stokes' theorem for each component of the original vector-valued integral
$$\vec{I}= \oint ||\vec{F}||^2 d\vec{F}$$
$$I_x = \oint (||\vec{F}||^2 \vec{e}_x) \cdot d\vec{F}$$
etc. Now it's in the correct form.

I don't understand this?

 

[clamtrox]

I don't understand this?

$\vec{e}_x$ is the unit vector in x-direction. The lower integral is just the x-component of your full integral. You can calculate this by taking the dot product with $\vec{e}_x$.