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^{2}, x + sin y, [tex]\sqrt{}[/tex]z

^{4}+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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- Thread starter calculus-stud
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I've no idea how to do this... any help would be awesome! Thanks!

- #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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[tex]\int_c F\cdot dc[/tex]

directly.

If c= [cos(t), sin(t), 2+ sin(t/2)], what is dc?

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And is dc = (-sin t, cos t, cos (t/2)/2) ??

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tiny-tim

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Stokes' Theorem says ... but how can I use it here?

What is curlF here?

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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.

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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But if the end points of the curve are the same, isn't that the definition of a closed curve?

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tiny-tim

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

(and yes, it looks closed to me too)

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The flux is zero? So, we don't really need to integrate or Stokes' theorem here?

- #11

tiny-tim

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(thanks for the pm)

nocalculus-stud said:The flux is zero?

what would be the flux of a (0,0,3) field through, for example, a rectangle or an ellipse in the z = 0 plane?

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Hey, umm, I'm kinda lost here. Can you give some more clues?

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Ok, does it have to do with the orientation of the curve? I'm really stuck here ... any hints?

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tiny-tim

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this is the integral of a

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tiny-tim

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or through the horizontal circle, radius r centre (0,0,0) ?

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Dick

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