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Math Help - Stokes Theorum

  1. #1
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    Stokes Theorum

    I'm not sure if this is the right place to put this?

    Show that: curl(fu) = fcurlu + (gradf) x u

    By using this in relation to Stoke's theorum, show that for a simple closed curve C,

    (Line integral) fdS = -(surface integral) (gradf) x dS

    Firstly, is this a typo? I've never heard of a line integral of a function wrt to surface element?


    I think Stoke's Theorum is:
    (Line integral)(A.dl) = (surface integral)((curlA).dS)

    I did the first part fine. I then tried to do

    (surface integral) (gradf)xdS = (surface integral) (curl(fdS) - fcurldS)

    I then got stuck. Please help!
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  2. #2
    Oli
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    I think it is a typo.

    I put fu=fdS.
    Stick it into the identity you made up to find a value to curl(fdS).

    Re-arrange to find an identity for (gradf)xdS.

    Then stick (gradf)xdS into an integral sign, use your second identity and Stokes theorem, and note that curl(dS) = 0 (I think).

    Then the solution should pop out.
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  3. #3
    Oli
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    Sorry, I was wrong, it is harder than that.
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  4. #4
    Member Danshader's Avatar
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    curl(uF) = u(curlF) + grad(u)xF

    let F (x,y,z) = f(x,y,z)i + g(x,y,z)j + h(x,y,z)k
    then uF= u fi + ugi + uhk


    curl(uF) = {d/dy(uh) - d/dz(ug)}i + {d/dz(uf) - d/dy(uh)}j + {d/dx(ug) - d/dy(uf)}k
    =
    [ u{dh/dy-dg/dz}i + u{df/dz-dh/dy}j + u(dy/dx - df/dy}k ] + [ {(du/dy)(h) - (du/dz)(g)}i + {(du/dz)(f) - (du/dy)(h)}j + {(du/dx)(y) - (du/dy)(f)}k ]
    =u(curlF) + grad(u)xF
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