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Div, Curl and all that jazz: Proving identities

  • Thread starter Archduke
  • Start date
58
1
1. Homework Statement
Prove:

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

2. Homework Equations
In the previous part of the question, we proved that:

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


3. The Attempt at a Solution

OK, so, here I go!

[tex] \int\left(\nabla \times \vec{F} \right)\cdot \vec{\hat{n}}dV \\

= \int \nabla \cdot \left( \vec{F} \times \vec{\hat{n}} \right) dV [/tex]

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

[tex]

\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} [/tex]

My question is...Is [tex] \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} [/tex]?

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!
 

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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What is the orientation of the surface in each integral?
 
58
1
Erm, I'm don't know what an the orientation of a surface is, but I've had an idea.

I know that:

[tex] \vec{F} \times \vec{\hat{n}} = -\vec{\hat{n}} \times \vec{F} [/tex]

and that:

[tex] \int_{a}^{b} f(x)dx = - \int_{b}^{a} f(x)dx [/tex]

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. :frown:
 

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