Stokes' Theorem

  • #1
Given c(t) = [cos t, sin t, 2 + sin (t/2)] where t [tex]\epsilon[/tex] [0, 2pi] and F(x,y,z) = (2-y + x2, x + sin y, [tex]\sqrt{}[/tex]z4+1) --- Find [tex]\int[/tex]F.dS over c(0, 2pi).

I've no idea how to do this... any help would be awesome! Thanks!

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  • #2
tiny-tim
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I've no idea how to do this... any help would be awesome! Thanks!

Well, what does Stokes' Theorem say?
 
  • #3
HallsofIvy
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Actually, I see no reason to use Stoke's theorem. It looks like it should be relatively easy to integrate
[tex]\int_c F\cdot dc[/tex]
directly.

If c= [cos(t), sin(t), 2+ sin(t/2)], what is dc?
 
  • #4
Stokes' Theorem says that [tex]\int[/tex]F.dr inside a boundary C = [tex]\int[/tex]curlF.dS over a surface S... but how can I use it here?

And is dc = (-sin t, cos t, cos (t/2)/2) ??
 
  • #5
tiny-tim
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Stokes' Theorem says ... but how can I use it here?

What is curlF here? :smile:
 
  • #6
Curl F is 3 k. I'm having trouble figuring out dS... I know for [tex]\int[/tex]curlF.dS I need a surface with the given boundary curve for dS... but I'm stuck there.
 
  • #7
HallsofIvy
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The reason I suggested NOT using Stoke's theorem but integrating directly on the path is that when [itex]\theta= 0[/itex], [itex](cos(\theta), sin(\theta), 2+ sin(\theta/2))= (1, 0, 2)[/itex] and when [itex]\theta= 2\pi[/itex], [itex](cos(\theta), sin(\theta), 2+ sin(\theta/2))= (1, 0, 1)[/itex]. This is NOT a closed path and so Stoke's theorem cannot be used.

For some reason, perhaps having just waked up, I was thinking "cosine" instead of "sine". Yes, they are the same point, yes, this is a closed path.
 
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  • #8
But if the end points of the curve are the same, isn't that the definition of a closed curve?
 
  • #9
tiny-tim
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hi calculus-stud! :smile:
Curl F is 3 k. I'm having trouble figuring out dS... I know for [tex]\int[/tex]curlF.dS I need a surface with the given boundary curve for dS... but I'm stuck there.

(I make it 2k … have you copied F wrong? :confused:)

think … what is the flux of a (0,0,3) field through any closed curve? :wink:

(and yes, it looks closed to me too)
 
  • #10
Yeah it should be F = (-2y + x2....) sorry. So, curl is 3k.

The flux is zero? So, we don't really need to integrate or Stokes' theorem here?
 
  • #11
tiny-tim
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hi calculus-stud! :smile:

(thanks for the pm)
calculus-stud said:
The flux is zero?
no

what would be the flux of a (0,0,3) field through, for example, a rectangle or an ellipse in the z = 0 plane? :wink:
 
  • #12
Hey, umm, I'm kinda lost here. Can you give some more clues?
 
  • #13
Ok, does it have to do with the orientation of the curve? I'm really stuck here ... any hints?
 
  • #14
tiny-tim
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what curve? :confused:

this is the integral of a vector going through a surface
 
  • #15
Oh right, sorry... but then what is the surface? I'm so lost... I thought you were asking for flux of the vector field through any closed curve...
 
  • #16
tiny-tim
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what's the flux of (0,0,3) going through the horizontal rectangle (0,0,0) (0,2,0) (2,2,0) (2,0,0) ?

or through the horizontal circle, radius r centre (0,0,0) ?
 
  • #17
Dick
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Let me add another hint to tiny tim's hint. Imagine water flowing through a circular pipe at uniform speed v. If you take a flat disc cross-section perpendicular to the flow it's easy to compute the flux through the disc using stokes. If you take a more complicated wiggly cross section of the pipe, it could be much harder to compute the flux through that using stokes. But you don't have to. It will have the same flux as the easy cross section. Do you see why?
 

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