# Stokes Theorem

## 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

Last edited:

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

I made an error. It is a squared term.

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

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.

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 dont understand this?

I dont 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 $\vec{e}_x$.