How can Stoke's theorem be applied to vector fields?

[yoghurt54]

Homework Statement

$$\nabla \times f \vec{v} = f (\nabla \times \vec{v}) + ( \nabla f) \times \vec{v}$$

Use with Stoke's theorem

$$\oint _C \vec{A} . \vec{dr} = \int \int _S (\nabla \times \vec{A}) . \vec{dS}$$

to show that

$$\oint _c f \vec{dr} = \int \int _S \vec{dS} \times \nabla f$$

Homework Equations

I think the only that you really need to know is the the scalar triple product.

The Attempt at a Solution

Ok, I allowed

$$\vec{A} = f \vec{v}$$

And subsituted into Stokes' theorem

$$\oint _C f\vec{v} . \vec{dr} = \int \int _S (\nabla \times f \vec{v}) . \vec{dS}$$

which gives

$$\oint _C f\vec{v} . \vec{dr} = \int \int _S f ((\nabla \times \vec{v}) + ( \nabla f) \times \vec{v} ). \vec{dS}$$

My problem is that I got to this stage, and thought that this would work really great if

$$\nabla \times \vec{v} = 0$$.

So I got stuck on this for a while, looked at my textbook, and they said that we let

$$\vec{A} = f \vec{v}$$ where $$\vec{v}$$ is a CONSTANT VECTOR.

Hence, the curl of a constant vector is zero, and the RHS becomes

$$\int \int _S f ( \nabla f) \times \vec{v} ). \vec{dS}$$

which we rearrange with the triple scalar product identity to give

$$\int \int _S f ( \vec{dS} \times \nabla f) ) .\vec{v}$$

So now the thing looks like this:

$$\oint _C f\vec{v} . \vec{dr} = \int \int _S f ( \vec{dS} \times \nabla f) ) .\vec{v}$$

and cancelling $$\vec{v}$$ from both sides as it's constant, gives us:

$$\oint _c f \vec{dr} = \int \int _S \vec{dS} \times \nabla f$$

The problem

My problem is that this restricts the kind of vector field $$\vec{A}$$ can be, as it has to be a product of some function times a constant vector.

Does this make the derived identity non-universal, or is this a necessary step in order to arrive at the identity?

 

[gabbagabbahey]

My problem is that this restricts the kind of vector field $$\vec{A}$$ can be, as it has to be a product of some function times a constant vector.

Does this make the derived identity non-universal, or is this a necessary step in order to arrive at the identity?

The vector identity you proved involves only $f$. So, who cares what you restricted $\textbf{A}$ to be? The identity is valid for any scalar function $f$, and that's all you were asked to prove.

 

[yoghurt54]

Yeah you're right. I don't know why I got hung up on this, I guess I was reading too much into it.