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Scalar field

  • Thread starter JaysFan31
  • Start date
  • #1
JaysFan31

Homework Statement


Let S be a simple parametrically defined surface with boundary C as in Stokes' Theorem. Let f and g be two continuously differentiable scalar fields defined on S. Let n be a choice of unit normal on S. If grad(f) is perpendicular to grad(g) x n everywhere on S, show that integral of fgrad(g)*dr=0.
Note: x is cross product and * is dot product


Homework Equations


Stokes' Theorem:
integral of F*dr=integral of curlF*ndS


The Attempt at a Solution


I'm pretty confused on this one.
First off, what does the perpendicular part mean? What does it tell me? Doesn't it just tell me that the cross product exists everywhere? How does this information help me solve the problem and show that the integration equals zero?
 

Answers and Replies

  • #2
HallsofIvy
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You do not know what "perpendicular" means? It means here exactly what it means in elementary geometry- that the two vectors are at right angles to one another. What you are really asking is "What does it tell me". No, it does not tell you that the cross product exists everywhere! As long as the two vectors exists, their cross product exists. What it tells you is much simpler than that: that the dot product of the two vectors is 0.
 
  • #3
Dick
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To solve this (and the last post) all you need are some vector identities. Here they are at no extra charge. f is a scalar, a,b,c are vectors. (. is dot, * is scalar multiply).

curl(grad(f))=0
a.(bxc)=b.(cxa)=c.(axb)
curl(f*a)=grad(f)xa+f*curl(a)

I'm not charging for this service because I'm sure you already had them.
 
  • #4
JaysFan31
Ok. So I get that
grad(f) . (grad(g) x n)=0 where x is cross product and . is dot product since the vectors are perpendicular.

Using the second identity shown,

n . (grad(f) x grad(g))=0.
This equals (grad(f) x grad(g)) . n=0.

How do I show however that (grad(f) x grad(g))=fgrad(g) to make it work?
 
  • #5
Dick
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They AREN'T equal. Use one of your free identities to compute curl(f grad(g)). The point to Stokes is integrating curl(F) over the interior and integrating F over the boundary. curl(F) and F are TWO DIFFERENT FIELDS.
 

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