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Math Help - Using Stoke's Theorem..

  1. #1
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    Using Stoke's Theorem..

    I'm having difficulty to find the surface integral using Stoke's Theorem, can anyone please help? Help will be much appreciated!
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  2. #2
    MHF Contributor arbolis's Avatar
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    I'm learning exactly the same subject so I might be wrong.
    I believe the following : the surface of the sphere is x^2+y^2+z^2=4 with the restriction of the first octant.
    So z=\sqrt {4-x^2-y^2}. You're integration over the circle x^2+y^2=4 so the boundaries of the surface integral would be \int _0^{\frac{\pi}{4}} \int _0^{2} F dA. Notice that I used polar coordinates.
    I'm not sure though, so I'll wait any further help.
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    I'm thinking you'd somehow have to use the integral of region S of Curl F multiply by dS..not sure if i'm on the right track..if anyone else can provide more help, please do!
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  4. #4
    MHF Contributor Calculus26's Avatar
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    The integration limits are correct, however in using Stokes theorem

    the integrand is( curlF*N dA) where N = -dz/dx i -dz/dy j +k where

    d/dx,d/dy are the partial derivatives
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  5. #5
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by calciihelp View Post
    I'm thinking you'd somehow have to use the integral of region S of Curl F multiply by dS..not sure if i'm on the right track..if anyone else can provide more help, please do!
    Yes, sorry.
    The boundaries are good I think. \int _0^{\frac{\pi}{4}} \int _0^{2} F dA should read \int _0^{\frac{\pi}{4}} \int _0^{2} curl F \hat n dS I believe.
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  6. #6
    MHF Contributor Calculus26's Avatar
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    Yes, sorry.
    The boundaries are good I think. should read I believe.
    Not quite int (curlF*nds) = int (curlF*NdA) once you have the region R in the plane and integration limits dS is replaced with dA and n with N
    recall n is a unit vector n = N/|N| dS = |N|dA
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  7. #7
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Calculus26 View Post
    Not quite int (curlF*nds) = int (curlF*NdA) once you have the region R in the plane and integration limits dS is replaced with dA and n with N
    recall n is a unit vector n = N/|N| dS = |N|dA
    I get it, thanks a lot.
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