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Stokes Theorem = 0?

  1. Sep 9, 2007 #1
    (1)

    using stokes theorem and cutting the surface into 2 parts how can we prove that

    [tex]\int [/tex] curl A.n dS = 0

    assume the surface "S" to be smooth and closed, and "n" is the unit outward normal as usual.

    (2)
    How can you prove
    [tex]\int [/tex] curl A.n dS = 0
    using the divergence theorem?
     
  2. jcsd
  3. Sep 9, 2007 #2

    matt grime

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    What have you attempted? I mean if, in (1), you just follow the hints the answer seems like it is immediate, so how have you tried to use the hints?
     
  4. Sep 9, 2007 #3
    im confused because A hasn't been defined and neither has r(t)??
    what shall i assume those to be?
     
  5. Sep 9, 2007 #4

    HallsofIvy

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    You were the one who asked the question! What exactly is the statement? It is certainly not true that
    [tex]\int \nabla \cross \vec{A}\cdot\vec{n}dS= 0[/itex]
    for arbitrary [itex]\vec{A}[/itex]!

    What exactly are you asking? If this is homework problem, state exactly what it says. Is [itex]\vec{A}[/itex] a constant vector? It would be easy in that case.

    As far as r(t) is concerned, there was no "r(t)" in your original post!
     
  6. Sep 9, 2007 #5

    matt grime

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    Whatever A is supposed to be, or the surface, it is clear that (1) wants you to split the surface into two regions (presumably they'll need to be simply connected), then replace the integral over those two surfaces with the line integrals along the boundary - of course we may have to make some leap like taking the closure of the two subsets, but that is immaterial, and then noticing that you reverse the orientation on one, and cancel.
     
  7. Sep 9, 2007 #6
    for part (1)
    the only information provided was that :

    "assume the surface "S" to be smooth and closed, and "n" is the unit outward normal as usual."

    now what do i assume the parameters to be.. i understand that

    [itex]\int curl A.ndS = \int A.dr [/itex]

    so i assume i have to show : [itex] \int A.dr [/itex] = 0??

    but how do i start this off since i havent been given any parameters etc
     
  8. Sep 9, 2007 #7

    it doesnt state what A is.. all it says is:

    S is a smooth closed surface.. n i a unit outward normal to S... thats it..

    i havent been given A... or r(t) or the radius of the curve.. or anythin like that,... so what would i assume the parameters to be
     
  9. Sep 9, 2007 #8
    i think i have managed to work out part 2:

    if F = curl A
    the div theorem says:
    [itex]\int_v div(curlA)dV = \oint Curl A.ndS[/itex]

    div(curl A) = 0
    So: [itex]\oint CurlA.ndS = 0 [/itex]

    i understand that the identity div(curl A)=0 but how is this proved please?
     
  10. Sep 9, 2007 #9

    matt grime

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    Tediously. It is just an exercise in manipulating partial derivatives.
     
  11. Sep 9, 2007 #10

    matt grime

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    And I still fail to see what is troubling about 1, and I don't get Halls' objections at all. You have a smooth closed surface, like a sphere. Split it into two hemispheres and apply Stokes' theorem, and you're done - you have two line integrals, integrating the same function over the same boundary but with opposite orientations.
     
  12. Sep 9, 2007 #11
    to work out the question i need to work out [itex]\int A.dr = \int A(r(t)). r(t)dt [/itex]
    but i dont know the values of r. and im not given the parameter of r(t)???

    so how would i start the question when im not even given anythin to work with? please start me off here and ill attempt the rest
     
  13. Sep 9, 2007 #12

    matt grime

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    Why would you want to actually work out the integral? Change a surface integral into two surface integrals, then turn these into two line integrals. This completes the proof as far as I can tell.
     
  14. Sep 9, 2007 #13

    mathwonk

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    stokes theorem says that for any smooth oriented surface, that surface integral can be computed as an integral over the boundary of the surface.

    if the surface is closed, i.e. the boundary is empty, then the boundary integral is necessarily zero.
     
  15. Sep 9, 2007 #14
    use 2 spheres. simple as. orient them in opposite directions. they cancel out.
     
    Last edited: Sep 9, 2007
  16. Sep 9, 2007 #15
    i understand what is being told to me.. but what i need help on is showing this mathematically..

    i know the 2 spheres cancel each other out since they are oriented in different directions...

    i know i need to show that the 2 line integrals cancel each other out..

    but my problem is::::: What Are The Values That I Use For The Line Integrals? since i am not given this in the question.

    i was told to prove that [itex]\int curl A.ndS = 0 [/itex] so surely i need to show this mathematically instead of just using words...

    so please help me on how to show this mathematically!!! greatly appreciated..
    thnx.. no one seems to hav made that clear to me so far.
     
  17. Sep 9, 2007 #16

    mathwonk

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    did you understand my post? i have completely answered your question, which has nothing to do with spheres, or cutting anything open.
     
  18. Sep 9, 2007 #17
    kind of.. but the question says that you HAVE to cut the sphere into 2 pieces..

    here is what i worked out.. please verify if it is correct..

    (1)

    so is this proof correct:

    [itex]\int curl A.n dS = \int A.dr [/itex]

    cutting the sphere horizontally into 2 hemispheres we have...

    [itex]\int A.dr - \int A.dr = 0.. [/itex]

    therefore, [itex]\int curlA.n ds = 0...[/itex]

    is this proof correct??

    (2)

    for 2 i managed to show

    if F = curl A
    the div theorem says:
    [itex]\int_v div(curlA)dV = \oint Curl A.ndS[/itex]

    div(curl A) = 0
    So: [itex]\oint CurlA.ndS = 0 [/itex]

    i understand that the identity div(curl A)=0 but how is this proved please?
     
  19. Sep 9, 2007 #18
    [itex]curl A= \hat i(\frac{\partial A}{\partial y} - \frac{\partial A}{\partial z}) -\hat j(\frac{\partial A}{\partial x} - \frac{\partial A}{\partial z}) + \hat k(\frac{\partial A}{\partial x} - \frac{\partial A}{\partial y}) [/itex]
    then
    [itex]div(curl A)= \frac{\partial}{\partial x}(\frac{\partial A}{\partial y} - \frac{\partial A}{\partial z}) + \frac{\partial}{\partial y}(-\frac{\partial A}{\partial x} + \frac{\partial A}{\partial z}) + \frac{\partial}{\partial z}(\frac{\partial A}{\partial x} - \frac{\partial A}{\partial y}) = 0[/itex]
    i mean, this happens when A is a C^2 function, so you can use Schwartz theorem: mix the derivatives, and it's zero =)
     
    Last edited: Sep 9, 2007
  20. Sep 9, 2007 #19

    mathwonk

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    do you know how to prove stokes theorem? you paste together the greens theorem in a bunch of rectangles.

    i.e. green says the path integral of a one - form over the boundary of a rectangle, equals the integral of the curl of that one form over the interior of the rectangle. then if you have a more extensive region you just cut it up into rectangles and apply greens theorem to each rectangle and add.

    the path integrals over the boundaries of those rectangles that share a boundary cancel, while the area integrals over the adjacent areas add, and it leaves you with the desired statement.

    i.e. after all the canceling of boundaries traversed in opposite directions, the sum of all the path integrals equals the path integral over those portions of the boundaries only traversed once, i.e. the boundary of the original surface.

    hence if it hapens that all the boundaries cancel, i.e. that the original surface has empty boundary, then all the path integrals cancel out leaving zero.

    if they had stated the stokes theorem correctly in the first place, i.e. as saying the integral of the curl over the surface equals the path integral of the vector field over the boundary, this is simply zero when that boundary is empty.

    so i suggest you try to understand the topic as well as get the answer to the question, since the questioner is perhaps not firing on all cylinders here. i.e. pleasing the teacher is not life's ultimate goal.

    or perhaps your teacher is trying to illustrate the methid of proof of the stokes theorem? but then they should have said to deduce it from greens theorem, or maybe they gave you a very preliminary version of stokes theorem?


    as we say here so often, at some point one wants to get on the freeway, and off the exit ramps, and start making some progress.


    good luck.
     
    Last edited: Sep 10, 2007
  21. Sep 9, 2007 #20
    nice explanation mathwonk... at least, it was useful for me :D

    *anyway, i thought that green theorem was a particular case of the stokes theorem... :/, i guess i have to study this again xDD
     
    Last edited: Sep 9, 2007
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