Triple Integral

  • #1
"Suppose that a smooth vector field [tex]F(x,y,z)[/tex] given on a region [tex]D[/tex] has the property that on the bounding surface [tex]S[/tex] it is perpenticular to the surface. Show that
[tex]\int\int\int_D \bigtriangledown \times F dV = 0[/tex]
in the sense that each component of [tex]\bigtriangledown \times F[/tex] has integral 0 over D."
I'm really stumped on this one.
 

Answers and Replies

  • #2
Physics Monkey
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You can proceed by examing each component of the curl of F in turn. Take the x component for instance, it looks something like
[tex] (\vec{\nabla}\times\vec{F})_x = \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} [/tex]. Since integration is linear you can consider each piece in turn. Can you find a way to represent
[tex]
\frac{\partial F_z}{\partial y}
[/tex]
as a divergence. Hint: to make progress, apply Gauss' Theorem to the result.
 
  • #3
I suppose you can express it as [tex]\bigtriangledown(0,F_3,0)[/tex] so by Gauss's theorem,

[tex]\int\int\int_D\bigtriangledown(0,F_3,0)dV=\int\int_{\partial D}(0,F_3,0)d\vec{S}[/tex].

If the field were tangent to the surface, the dot product between the normal vector and it would be zero, but since it's already perpenticular...
 
  • #4
Physics Monkey
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You aren't quite there yet, but you're close. In particular, you can't just conclude that
[tex]\int\int _{\partial D}(0,F_3,0)d\vec{S}[/tex]
equals zero. However, when you combine this term with the other term in the curl, then maybe you can say something ...
 
  • #5
It boils down to

[tex]\int\int \bigtriangledown(0,F_3,0)-\bigtriangledown(0,0,F_2)\vec{N}dS[/tex]

The integrand becomes

[tex]\bigtriangledown(0,F_3,F_2)\vec{N}[/tex]

Since the normal is parallel to the field, it becomes (F3F2-F2F3)|N|, and this is 0.

Did I get this right? If so thanks for the help!
 
Last edited:
  • #6
Physics Monkey
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I don't know why you have the gradient operator still there, but yes, you've pretty much got it.
 

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