Div, Curl and all that jazz: Proving identities

[Archduke]

Homework Statement

Prove:

$$\int\left(\nabla \times \vec{F}\right)\cdot d\vec{V} = \oint \left(\vec{\hat{n}} \times \vec{F} \right) dS$$

Homework Equations

In the previous part of the question, we proved that:

$$\nabla \cdot \left( \vec{F} \times \vec{d} \right) = \vec{d} \cdot \nabla \times \vec {F}$$
(where d is a constant vector)
And also, it looks like we'll need to use the Divergence theorem.

The Attempt at a Solution

OK, so, here I go!

$$\int\left(\nabla \times \vec{F} \right)\cdot \vec{\hat{n}}dV \\<br /> <br /> = \int \nabla \cdot \left( \vec{F} \times \vec{\hat{n}} \right) dV$$

By the relation above proved from the previous part of the question. Next, I used the divergence theorem:

$$<br /> \int \nabla \cdot \left( \vec{F} \times \vec{\hat{n}} \right) dV = \oint \left( \vec{F} \times \vec{\hat{n}} \right) \cdot d\vec{S}$$

My question is...Is $$\oint \left( \vec{F} \times \vec{\hat{n}} \right) \cdot d\vec{S} = \oint \left(\vec{\hat{n}} \times \vec{F} \right) \cdot d \vec{S}$$?

My initial thought is that it isn't, as the cross product isn't commutative. If that is thecase, where else have I gone wrong?

Cheers!

 

[HallsofIvy]

What is the orientation of the surface in each integral?

 

[Archduke]

Erm, I'm don't know what an the orientation of a surface is, but I've had an idea.

I know that:

$$\vec{F} \times \vec{\hat{n}} = -\vec{\hat{n}} \times \vec{F}$$

and that:

$$\int_{a}^{b} f(x)dx = - \int_{b}^{a} f(x)dx$$

But, since it's a closed integral, I guess if we reverse the 'order' of integration, it doesn't matter if we do: the start and end points are the same...and the minuses cancel. Seems a bit of mathematical trickery to me, though.