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Math Help - Flux

  1. #1
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    Flux

    F(x,y,z) = xze^y i - xze^y j + z k
    S is part of the plan x+y+z=1 in first octante with orientation to down
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  2. #2
    MHF Contributor Amer's Avatar
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    by divergence theorem

    F(x,y,z) = xze^y i - xze^y j + z k

    \int\int F.n ds=\int\int\int divF dv

    div(F)=\frac{\partial}{\partial x}(xze^y)+\frac{\partial}{\partial y}(- xze^y)+\frac{\partial}{\partial z}z

    div(F)=ze^y-xze^y+1

    z=1-x-y

    \int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y}(ze^y-xze^y+1) dzdydx

    now just evaluate the integral
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  3. #3
    Super Member Random Variable's Avatar
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    Here's how to do it directly.

    Flux =  \int \int_{S} F \cdot N \ \mathrm{d} S = \int \int_{D} F \cdot (T_{x} \times T_{y}) \ \mathrm{d}x \ \mathrm{d}y

    Let  g(x,y) = z = 1-x-y

    The surface can be parametrized by  \Phi(x,y) = \Big (x,y,g(x,y)\Big) = (x,y,1-x-y)

     T_{x} = \frac {\partial \Phi}{\partial x} = (1,0,-1)

     T_{y} = \frac {\partial \Phi}{\partial y} = (0,1,-1)

    EDIT :  T_{x} \times T_{y} = (1,1,1)

    The problem wants the normal vector pointing in the other direction, or  (-1,-1,-1)


    then  F \cdot (T_{x} \times T{y}) = -z = -1+x+y


    D is the region in the xy-plane bounded by the positive x and y axes and the line y = 1-x

    The final final integral is then  \int_{0}^{1} \int_{0}^{1-x} (-1+x+y) \ \mathrm{d}y \ \mathrm{d}x
    Last edited by Random Variable; June 3rd 2009 at 02:38 PM. Reason: fixing error
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Random Variable View Post
    Here's how to do it directly.

    Flux =  \int \int_{S} F \cdot N \ \mathrm{d} S = \int \int_{D} F \cdot (T_{x} \times T_{y}) \ \mathrm{d}x \ \mathrm{d}y

    D is the region in the xy-plane bounded by the positive x and y axes and the line y = 1-x

    Let  g(x,y) = z = 1-x-y
    """""????
    The surface can be parametrized by  \Phi(x,y) = \Big (x,y,g(x,y)\Big) = (x,y,1-x-y)

     T_{x} = \frac {\partial \Phi}{\partial x} = (1,0,-1)

     T_{y} = \frac {\partial \Phi}{\partial y} = (0,1,-1)

     T_{x} \times T_{y} = (1,1,0)
    """"""????
    The problem wants the normal vector pointing in the other direction, or  (-1,-1,0)


    but then  F \cdot (T_{x} \times T{y}) = 0 ? Is that right? Anybody?


    D is the region in the xy-plane bounded by the positive x and y axes and the line y = 1-x

    The final final integral is then  \int_{0}^{1} \int_{0}^{1-x} 0 \ \mathrm{d}y \ \mathrm{d}x which of course is just zero.

    can you please explain that for me in details how you parametrize like that
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  5. #5
    Super Member Random Variable's Avatar
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    [
    can you please explain that for me in details how you parametrize like that
    The surface can be written as z = 1 - x - y. So an obvious parametrization (at least to me) is (x, y, 1-x-y). It I had w = 1-x-y-z, then the parametrization would be (x, y, z, 1-x-y-z).

    The stuff of mine you quoted has an error in it. The answer is not zero. I fixed it above.
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  6. #6
    Super Member Random Variable's Avatar
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    I might have made a mistake somewhere because I'm not getting the same answer as your approach.
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  7. #7
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    Quote Originally Posted by Amer View Post
    by divergence theorem

    F(x,y,z) = xze^y i - xze^y j + z k

    \int\int F.n ds=\int\int\int divF dv

    div(F)=\frac{\partial}{\partial x}(xze^y)+\frac{\partial}{\partial y}(- xze^y)+\frac{\partial}{\partial z}z

    div(F)=ze^y-xze^y+1

    z=1-x-y

    \int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y}(ze^y-xze^y+1) dzdydx

    now just evaluate the integral
    Ok. Thank you
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