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3d: Stoke's Theorem

  1. Jul 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Use Stokes' Theorem to evaluate ∫∫curl F dS, where F(x,y,z) = xyzi + xyj + x^2yzk, and S consists of the top and the four sides (but not the bottom) of the cube with vertices (±1,±1,±1), oriented outward.


    2. Relevant equations
    Stokes' Theorem:
    ∫∫curl F dS = ∫F dr

    a hint they gave: "What is the boundary of S? can you find another surface S1 with the same boundary?"

    3. The attempt at a solution
    im not sure im lost :(. im not sure how to make it so it stays within the limits of that cube. idk how to find S1
     
  2. jcsd
  3. Jul 18, 2008 #2
    It's been a few years, so I am not really sure on this. Since it's oriented outwards, and there's 4 sides, the sides opposite each other would cancel each other out... (I think, but hopefully I get corrected if not) so the only surface that contributes is the top.

    Again, I would wait for someone else's reply because I haven't looked at this for a few years, and I'm still a student too.
     
  4. Jul 18, 2008 #3
    im still not sure what to do :( how would i set it up as F dr ? and idk what the limits would be either
     
  5. Jul 18, 2008 #4

    tiny-tim

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    Hi jaredmt! :smile:

    Well, they've given you a cube with one side missing …

    so the obvious surface with the same boundary is … ? :smile:
     
  6. Jul 18, 2008 #5
    the side they left open is z = -1 and it is within (±1,±1,-1)
    so i guess z = -1 , -1<x<1 , -1<y<1
    idk im confused with this entire section of the chapter. any more hints? haha. i might need a walk-through on how to find what im supposed to integrate. the book's example is completely different and doesnt help. if i could just figure out what to do for this problem i will probably be able to do the other 1s from this section
     
  7. Jul 18, 2008 #6

    Defennder

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    Since you are told to find the flux of curl F through S, by Stokes theorem that equates to finding the closed line integral of the closed path which bounds S. You've already noted that the closed path would be the perimeter of the square of the cube's bottom face at z=-1. So all you need to do is the evaluate the line integral along that path. However to do this you need to break up the square perimeter into 4 lines and perform a line integral along each of them and then add up the result, which would be a lot more tedious. So what was suggested by Tiny-tim is for you apply Stokes' theorem again and calculate the flux through the other surface which the closed path bounds. You would have to parametrize that surface first though before doing the surface integral.

    Bear in mind that in 3D space a closed path bounds an infinite number of (non-closed) surfaces and Stokes theorem guarantees that no matter which surface you choose your answer will be the same. Hence the trick when applying Stokes theorem is to choose a surface which is easy to parametrise and integrate over.

    Of course you have to take note of the orientation Stokes theorem works for.
     
  8. Jul 19, 2008 #7
    ok well i finally did this problem after carefully rereading a lot of stuff. i got the book's answer which is 0 but that was the same answer on the problem before this so i just want someone to confirm if i did this right.

    edit: F(x,y,z) = xyzi + xyj + x^2yzk

    when i parametrized i got:
    x=x y=y z=-1
    r(x,y) = <x,y,-1>
    rx = <1,0,0>
    ry = <0,1,0>
    rx X ry = k

    f(r(x,y)) = <-xy,xy,-yx^2>

    then i did: ∫∫curlF·ds = ∫∫F·dr = ∫∫F(r(x,y))·(rx X ry)dA = ∫∫F(r(x,y))·kdA
    = ∫∫<-xy,xy,-yx^2>·<0,0,1>dA = ∫∫(-yx^2)dA

    and i set the limits:
    -1<x<1
    -1<y<1

    and ∫∫(-yx^2)dydx = 0

    k thanks for all the help so far, just let me know if u think this looks right.
     
  9. Jul 19, 2008 #8

    Defennder

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    It's odd you managed to get the answer despite the fact that your calculations involve the flux of F through the surface and not curl F which is what you were told to find. You should find find curl F first then calculate the surface integral.
     
  10. Jul 20, 2008 #9

    HallsofIvy

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    No, that's not odd at all. Stoke's theorem says precisely that the integral of curl F over the surface is the same as the integral of F around the boundary.
     
  11. Jul 20, 2008 #10

    Defennder

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    Yes but he evaluated the integral of F over the surface and not curl F which was what he was supposed to find.
     
  12. Jul 20, 2008 #11
    well it said to use Stoke's theorem to evaluate it. i thought that the theorem meant that i had to change it to the other form. idk maybe i should do it the other way too. if i did curlF dS, dS = nds = (rx X ry) = k right?

    it seems like sometimes dS = dr unless im doing something wrong
     
    Last edited: Jul 20, 2008
  13. Jul 20, 2008 #12

    HallsofIvy

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    Yes, but Defennnder's point, that I missed, is that you integrated F, not curl F, around the boundary!
     
  14. Jul 20, 2008 #13

    tiny-tim

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    Stoke's Theorem twice

    hi jaredmt! :smile:
    ah, but there's two interpretations of that …

    either that you use it once, and integrate around the boundary,

    or that you use it twice, and integrate over the bottom face (because that's equal to the boundary integration, which is equal to the integration over the original five faces).

    Of course, both methods work.

    Since the original hint in the question was:
    I think they intended you to use the second method. :smile:
     
  15. Jul 20, 2008 #14
    o ok. but for curlF dS, does dS = (rx X ry) ? im getting confused cus it seems like dS = dr
     
  16. Jul 21, 2008 #15

    Defennder

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    dS is a differential element of a surface while dr appears to be a differential line segment. They're not the same. dS, given your parametrisation is dxdy. rx X ry is the normal vector associated the differential surface element.
     
  17. Jul 21, 2008 #16
    ok so ur saying that i can only find dr when i have r(t). i cant find dr in the form of r(x,y).
    crap. haha it makes sense though. this means that previously i didnt find Fdr i just found FdS.

    anyways. i redid it with ∫∫curlF·dS = ∫∫curlF·(rx X ry)dA and got 0 again thankfully.

    but one last question. when i did curlF i just did the curl of the original problem [curl F(x,y,z)] then took the answer and replaced Z with -1. in this case it didnt matter (i dont think) cus the k term was all that i needed since (rx X ry) = k. but should i replace Z with -1 before or after i find curlF? cus im not sure if maybe i should have found curl F(r(x,y)). i think the way i did it was right but i just wanna make sure
     
  18. Jul 21, 2008 #17

    Defennder

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    curl is applied to a vector function. Substituting [tex]F_z(x,y,z)[/tex] with -1 in accordance with the surface parametrisation is something that is done before calculating curl F. Just look at the formula for evaluating surface integral over a parametrised surface.

    EDIT: Here's the formula for reference:
    [tex]\iint \vec{F}(x,y,z) \cdot d\vec{S} = \iint \vec{F}(u,v) \cdot (\vec{r_u} \times \vec{r_v}) \ dudv [/tex] where this time the surface is given in terms of parameters u,v.

    There's supposed to be an S under the double integral on the left and a D under the double integral on the right. I can't figure out how to do that in latex.
     
    Last edited: Jul 21, 2008
  19. Jul 21, 2008 #18
    ok i think i got it now. this is the most difficult section in this entire course IMO and they squeezed it in 4 pages... and skipped some steps in the examples too so that didnt help lol. anyways thanks for the help :)
     
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