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Homework Help: Flux integral

  1. Mar 5, 2010 #1
    1. The problem statement, all variables and given/known data

    Directly compute the flux integral [tex]\int\int(G*n dS)[/tex] where D is the part of the plane x+y+z=1 in the first octant, oriented upwards.

    G = -y-z-x

    My attempt at solution:

    normal vector, normalized, is [tex]1/sqrt(3) * (1,1,1)[/tex] and since z=1-y-x, the integral simplifies to

    [tex]1/sqrt(3)*\int\int(-1 dS)[/tex]

    How do I evaluate this, without parametrization of the plane?
     
  2. jcsd
  3. Mar 5, 2010 #2

    tiny-tim

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    Hi mathman44! :smile:

    (have a square-root: √ and an integral: ∫ :wink:)

    ∫∫ dS is the area, so you need the area of an equilateral triangle of side … ? :smile:
     
  4. Mar 5, 2010 #3
    Doh!

    Of side √2. Meaning the area is 1, and the flux is -1/√3... is that okay?
     
  5. Mar 6, 2010 #4

    tiny-tim

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    sin60º ≠ 1 :redface:
     
  6. Mar 6, 2010 #5
    Ha, oops... so area is √3/2 and the flux is then -1/2.
     
  7. Mar 6, 2010 #6

    tiny-tim

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    Yup! :biggrin:
    ah … now I didn't actually understand the original question :redface:

    G is presumably a vector, but what are its coordinates? :confused:
     
  8. Mar 6, 2010 #7
    G is -y(i) - z(j) - x(k)

    so G dotted with n

    = 1/√(3)*(1,1,1) * G = 1/√(3) * (-y-z-x)

    But because x+y+z=1, z=1-x-y and so -y-z-x = -1, becoming:
    1/√(3) * (-y-z-x) = 1/√(3) * (-1)

    So then flux should be -1/√3 * dS, which is -1/√3 * √3/2 = -1/2
     
  9. Mar 6, 2010 #8

    tiny-tim

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    ah! all is now clear! :smile:

    Yes, -1/2 looks good. :wink:
     
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