- #1

yoghurt54

- 19

- 0

## Homework Statement

[tex] \nabla \times f \vec{v} = f (\nabla \times \vec{v}) + ( \nabla f) \times \vec{v} [/tex]

Use with Stoke's theorem

[tex] \oint _C \vec{A} . \vec{dr} = \int \int _S (\nabla \times \vec{A}) . \vec{dS} [/tex]

to show that

[tex] \oint _c f \vec{dr} = \int \int _S \vec{dS} \times \nabla f[/tex]

## 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

[tex] \vec{A} = f \vec{v} [/tex]

And subsituted into Stokes' theorem

[tex] \oint _C f\vec{v} . \vec{dr} = \int \int _S (\nabla \times f \vec{v}) . \vec{dS} [/tex]

which gives

[tex] \oint _C f\vec{v} . \vec{dr} = \int \int _S f ((\nabla \times \vec{v}) + ( \nabla f) \times \vec{v} ). \vec{dS} [/tex]

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

[tex] \nabla \times \vec{v} = 0 [/tex].

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

[tex] \vec{A} = f \vec{v} [/tex] where [tex] \vec{v} [/tex] is a

**CONSTANT VECTOR**.

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

[tex] \int \int _S f ( \nabla f) \times \vec{v} ). \vec{dS} [/tex]

which we rearrange with the triple scalar product identity to give

[tex] \int \int _S f ( \vec{dS} \times \nabla f) ) .\vec{v} [/tex]

So now the thing looks like this:

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

[/tex]

and cancelling [tex] \vec{v} [/tex] from both sides as it's constant, gives us:

[tex] \oint _c f \vec{dr} = \int \int _S \vec{dS} \times \nabla f[/tex]

**The problem**

My problem is that this restricts the kind of vector field [tex] \vec{A} [/tex] 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?

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