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Math Help - Line integral problem

  1. #1
    Junior Member
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    Line integral problem

    Am having trouble with line integrals and can't manage to get the following problem correct;

    <br />
\int\int_S y.dS<br />

    Where S is the part of the plane 2x + y + 3z = 6 for x=>0, y=>0, z=>0

    I know that ;



    And worked out the magnitude of the normal to be (14)^1/2, but i'm not sure if i'm parametrizing the plane correctly and i'm not sure of my limits of integration.
    Any help would be greatly appreciated.

    Cheers.
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  2. #2
    MHF Contributor Amer's Avatar
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    y=6-2x-3z

    \frac{\partial}{\partial x}y=-2....\frac{\partial}{\partial z}y=-3

    ds=\sqrt{\left(\frac{dy}{dx}\right)^2+\left(\frac{  dy}{dz}\right)^2+1}dxdz

    ds=\sqrt{14}dxdz

    \int_{0}^{2}\int_{0}^{3-\dfrac{3z}{2}}(6-2x-3z)\sqrt{14}dxdz



    fixed thanks for calculas26
    Last edited by Amer; June 4th 2009 at 08:31 AM.
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  3. #3
    MHF Contributor Calculus26's Avatar
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    A couple of things here:

    1. This is a surface integral not a line integral

    2. dS = 14^(1/2) not 13^(1/2)

    3. The inner integration limit should be 0 to 3 - 3z/2

    4. the integration order is dx dz not dxdy

    Monster what parameterization did you use? Hard to tell if its correct
    without knowing what it is.
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