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Math Help - Stoke's Theorem

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    Stoke's Theorem

    Q:let S be the surface x^2+y^2+z^2=1,z>=0.Use stokes theorem to evaluate
    line integral of (2x-y)dx-ydy-zdz
    where line integral is around circle x^2+y^2=1,z=0,oriented anti clockwise.
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    MHF Contributor Calculus26's Avatar
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    See attachment
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    Quote Originally Posted by Calculus26 View Post
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    every thing is clear to me except,how you calculate normal N
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    Quote Originally Posted by Calculus26 View Post
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    This answer of \pi can't be right. If you calculate directly

    \int_c (2x-y)\,dx - y \,dy - z \,dz the vector field {\bf F} = <2x-y, -y, -z> is conservative and hence the line integral is zero!
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    MHF Contributor Calculus26's Avatar
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    No the curlF is not 0 therefore the vector field is not conservative

    If it were what would the potential function be
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    Quote Originally Posted by danny arrigo View Post
    This answer of \pi can't be right. If you calculate directly

    \int_c (2x-y)\,dx - y \,dy - z \,dz the vector field {\bf F} = <2x-y, -y, -z> is conservative and hence the line integral is zero!
    sorry F is not conservative?????
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    MHF Contributor Calculus26's Avatar
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    No F is not -- the condition is If F is conservative curl F = 0

    If you calculate the line integral directly

    z= 0 so F = (2x-y)i -yj

    note this fails the test for conservative as well since d(2x-y)/dy does not equal d(-y)/dx

    If you parameterize C by x =cos(t) y = sin(t)

    you will also get pi
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  8. #8
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    Quote Originally Posted by Mathventure View Post
    sorry F is not conservative?????
    Quote Originally Posted by Calculus26 View Post
    No F is not -- the condition is If F is conservative curl F = 0

    If you calculate the line integral directly

    z= 0 so F = (2x-y)i -yj

    note this fails the test for conservative as well since d(2x-y)/dy does not equal d(-y)/dx

    If you parameterize C by x =cos(t) y = sin(t)

    you will also get pi
    Your both right - how silly of me.
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