Use Stokes Theorem to show a relationship

[princessme]

Homework Statement

Use the Stokes' Theorem to show that

$\int$f(∇ X A) dS = $\int$(A X ∇f) dS + $\oint$f A dl

Homework Equations

Use vector calculus identities. Hint given : Start with the last integral in the above relation.

The Attempt at a Solution

To be honest, I really don't know how to start doing this. I could understand Stokes theorem when needed to evaluate the circulation, vector and surface, but I don't have any idea on how to start to prove this relation. Please help?

 

[ArcanaNoir]

Do you know any identities? What does stokes theorem state?

 

[princessme]

Do you know any identities? What does stokes theorem state?

The ones that I know are like the ones states in this webpage : http://hyperphysics.phy-astr.gsu.edu/hbase/vecal2.html

Stokes theorem is $\oint$F.dr = $\int$∇ X F . n dS

 

[I like Serena]

Did you know that by convention $\mathbf n dS$ is the same as $d\mathbf S$?

So what do you get if you set $\mathbf F = f \mathbf A$ and $\mathbf r = \mathbf l$?

Is there another identity on your web page that you can apply then?

 

[princessme]

Did you know that by convention $\mathbf n dS$ is the same as $d\mathbf S$?

So what do you get if you set $\mathbf F = f \mathbf A$ and $\mathbf r = \mathbf l$?

Is there another identity on your web page that you can apply then?

Sorry but I still don't get it..quite confused. How should I actually begin proving it?

 

[I like Serena]

Sorry but I still don't get it..quite confused. How should I actually begin my proving it?

Start with $\oint$F.dr = $\int$∇ X F . dS

What do you get if you substitute $\mathbf F = f \mathbf A$ and $\mathbf r = \mathbf l$?

 

[princessme]

Start with $\oint$F.dr = $\int$∇ X F . dS

What do you get if you substitute $\mathbf F = f \mathbf A$ and $\mathbf r = \mathbf l$?

I'll get $\oint$fA.dl = $\int$∇ X F . dS

 

[I like Serena]

I'll get $\oint$fA.dl = $\int$∇ X F . dS

Good!
But there's still an $\mathbf F$ in there.
Can you replace that too?

 

[princessme]

Good!
But there's still an $\mathbf F$ in there.
Can you replace that too?

So it will be $\oint$fA.dl = $\int$∇ X fA . dS

 

[I like Serena]

So it will be $\oint$fA.dl = $\int$∇ X fA . dS

Right!
Actually, let's make that $\oint$(fA).dl = $\int$∇ X (fA) . dS

Do you have another identity on your web page for something like that?
Perhaps something like $\nabla \times (u\mathbf A)$?

 

[princessme]

Right!
Actually, let's make that $\oint$(fA).dl = $\int$∇ X (fA) . dS

Do you have another identity on your web page for something like that?
Perhaps something like $\nabla \times (u\mathbf A)$?

Okay!

Yeah ∇ x (uA) = u∇ x A - A x ∇u

 

[I like Serena]

Okay!

Yeah ∇ x (uA) = u∇ x A - A x ∇u

Good!

Substitute?

 

[princessme]

Good!

Substitute?

$\oint$(fA).dl = $\int$ (u∇ x A - A x ∇u ) dS ?

I can see where is this going now..But the original form they wanted is f(∇ x A) dS . so u∇ x A dS is equivalent to that form?

 

[I like Serena]

$\oint$(fA).dl = $\int$ (u∇ x A - A x ∇u ) dS ?

I can see where is this going now..But the original form they wanted is f(∇ x A) dS . so u∇ x A dS is equivalent to that form?

Yes.
You were supposed to replace u by f as part of the substitution.

 

[princessme]

Yes.
You were supposed to replace u by f as part of the substitution.

alright. So the it doesn't matter that the f in the original form is outside the bracket and not just stuck with ∇?

 

[I like Serena]

alright. So the it doesn't matter that the f in the original form is outside the bracket and not just stuck with ∇?

In this case it doesn't matter. Both forms are the same.
But as a rule, you should read $f\nabla \times \mathbf A$ as $f(\nabla \times \mathbf A)$.

 

[princessme]

In this case it doesn't matter. Both forms are the same.
But as a rule, you should read $f\nabla \times \mathbf A$ as $f(\nabla \times \mathbf A)$.

Okay thank you for the great help!

 

[princessme]

In this case it doesn't matter. Both forms are the same.
But as a rule, you should read $f\nabla \times \mathbf A$ as $f(\nabla \times \mathbf A)$.

Hi there. I have another question. I realized that I made a mistake on my Stokes' Theorem equation at first. The surface integral should be a double integral. But for the one that I need to prove, it is just a single integral. Will that make any difference to my answer?

 

[I like Serena]

Hi there. I have another question. I realized that I made a mistake on my Stokes' Theorem equation at first. The surface integral should be a double integral. But for the one that I need to prove, it is just a single integral. Will that make any difference to my answer?

That's not really a mistake.
You are integrating with $d\mathbf S$ which is an infinitesimal surface element.
You can either use one or two integral symbols - it means the same thing.

Formally, it should be a single integral symbol, since there is only one infinitesimal specified.
It should only be a double integral if your integrating with the infinitesimals $dx dy$ or some such.