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Stock's theorem?

  1. Sep 16, 2004 #1

    ori

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    c is a curve of the cutting of the to surfaces:
    x^2+y^2=1
    z=xy
    at the point (1,0,0) the tangent to C is toward j^
    so what is
    S(xz^2-y)dx+(3x-yz^2)dy+(zx^2-zy^2)dz
    C
    ?

    hmm..
    i calculated this vectorian field rotor: it's (0,0,4)
    i know i should use stocks and make it
    SS(0,0,4)*n^ ds
    but how can i build a surface (so i could know the integration borders and what is the normal)?

    thanks
     
  2. jcsd
  3. Sep 16, 2004 #2

    mathwonk

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    come again?
     
  4. Sep 17, 2004 #3

    HallsofIvy

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    (Ori, I hope you are not offended by our confusion. I assume that English is not your native language and I assure you that your English is far better than my command of whatever language is your native language!) :smile:

    I'm not at all sure what "stocks" are here and I THINK that "vector rotor" is the curl. I'm pretty sure this person is using "S" to indicate integral and the problem is to integrate the vector function (xz^2-y)dx+(3x-yz^2)dy+(zx^2-zy^2)dz
    around the intersection of the surfaces given by x^2+y^2=1 and z=xy.

    Aha! Stoke's theorem! x2+ y2= 1 is the cylinder along the z-axis. We can write that in parametric equations as x= cos(θ), y= sin(θ), z= z (with θ and z as parameters) and the intersection of that with z= xy is x= cos(θ), y= sin(θ), z= sin(θ)cos(θ).
    Integrating the vector function around that curve would be tricky but doable. To use Stoke's theorem you need to find the curl of the given vector function.

    You don't really need to find an expression for a surface inside that boundary. The nice thing about Stoke's theorem is that it applies to ANY surface having that boundary.

    Take as your surface the surface z= xy itself. Here is how I would do it:
    f(x,y,z)= xy- z= 0 has z= xy as a "level surface": div f= yi+ xj- zk is perpendicular to that surface and we can "normalize" to the projection onto the xy-plane by dividing by the k component: Integrate the dot product
    ((xz^2-y)i+(3x-yz^2)j+(zx^2-zy^2)k).(-y/z, -x/z,1)dx dy.

    You will, of course, need to use z= xy to reduce the integral to x,y only. The integration is over the unit disk so you may want to convert to polar coordinates.
     
    Last edited: Sep 17, 2004
  5. Sep 17, 2004 #4

    ori

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    10x
    my native lang is hebrew
     
  6. Sep 17, 2004 #5

    HallsofIvy

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    Your English is fine. It is "Stoke's theorem", not "Stock's" and "vector rotor" for "curl" makes sense to me! I've also noticed that in another thread you referred to Green's "sentence". The English word is "formula" (if not "theorem").
     
  7. Sep 17, 2004 #6
    Oh right Stokes' Theorem--- I was wondering for a second hehe.
     
  8. Sep 18, 2004 #7

    ori

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    look guys
    if i was writing stoke's theorem it was borring
    if im writing stocks theorem, it's attracts ppl to read - maybe there's a new mathmatical theorem about stocks etc.. plus it sound like stoke's..
    (kiddin :rolleyes: , it just a spelling mistake)
     
  9. Sep 18, 2004 #8

    mathwonk

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    by the way, the original term for curl, in maxwell's book, was "rotation". and "stokes" theorem is apparently due originally to lord kelvin.
     
    Last edited: Sep 18, 2004
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