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Math Help - Circulation

  1. #1
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    Circulation

    Determine the circulation of F(x,y,z) = (z-x)i +(x-y)j + (e^{xyz})k around the curve \alpha traveled in a counter-clockwise direction when viewed from the axis OZ, knowing that this curve is the edge of the surface: S1 x^2+y^2+z^2 = 4; z > 0
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Disregard this attachment

    See correct version on later post
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    Last edited by Calculus26; June 8th 2009 at 11:36 AM.
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  3. #3
    Super Member Random Variable's Avatar
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    EDIT: fixed errors


     \int \int_{S} curl F \cdot dS = \int_{C} F \cdot dr

    I think they mean that the boundary curve is  x^{2}+y^{2} = 4

    which can be parametrized by  r(\theta) = 2cos(\theta) \hat{i} + 2sin(\theta) \hat{j} + 0 \hat{k}

    where  0 \le \theta \le 2 \pi

    then  \int_{C} F \cdot dr = \int_{0}^{2 \pi} F(r(\theta)) \cdot r'(\theta) \ d \theta

    where  F(r(t)) = -2cos(\theta) \hat{i} + 2\big(cos(\theta) -sin (\theta)\big) \hat{j}+ \hat{k}

    and  r'(t) = -2sin(\theta) \hat{i} + 2cos(\theta) \hat{j} + 0 \hat{k}
    Last edited by Random Variable; June 8th 2009 at 11:23 AM. Reason: errors
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  4. #4
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    Quote Originally Posted by Random Variable View Post
    EDIT: fixed errors


     \int \int_{S} curl F \cdot dS = \int_{C} F \cdot dr

    I think they mean that the boundary curve is  x^{2}+y^{2} = 4

    which can be parametrized by  r(t) = 2cos(\theta) \hat{i} + 2sin(\theta) \hat{j} + 0 \hat{k}

    where  0 \le \theta \le 2 \pi

    then  \int_{C} F \cdot dr = \int_{0}^{2 \pi} F(r(\theta)) \cdot r'(\theta) \ d \theta

    where  F(r(t)) = -2cos(\theta) \hat{i} + 2\big(cos(\theta) -sin (\theta)\big) \hat{j}+ \hat{k}

    and  r'(t) = -2sin(\theta) \hat{i} + 2cos(\theta) \hat{j} + 0 \hat{k}
    How you found F(r(t)) ?
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    How you found F(r(t)) ?
    r(\theta) can be written as

     x = 2cos(\theta)
     y = 2sin(\theta)
     z = 0

    Do you know the final answer? I get  4 \pi
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  6. #6
    MHF Contributor Calculus26's Avatar
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    Verification with Stokes Thm

    previous post corrected

    See attachment
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  7. #7
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    4 \pi thanks
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