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Math Help - flux: surface integral

  1. #1
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    flux: surface integral

    Hi,

    I need to evaluate the flux integral \iint  F \cdot \hat{n}dS

    for portion of a plane x +2y+3z = 6 in the first octant and vector field F= < xy, -x^2, (x+z) >

    How do i go about doing this?

    I started of by doing it through parametrising the surface with r(s,t) = (3s,3t,(2-2t-s))

    With normal vector equalling cross product of tangents N= (3,-6,9)

    Then my bounds for integration were:

      0\leqslant s \leqslant 6-2t ,       0\leqslant  t \leqslant 6

    then i proceeded to integrate but it became really messy and did not seem right!

    so firstly, do i need to do it by parametrising the surface? secondly, are my bounds for integration correct? and lastly, It doesnt look like my normal vector is point in the right direction.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by olski1 View Post
    ... and did not seem right!

    Well, you have chosen a correct and standard method. Perhaps they are asking for the flux integral on the four faces of the tetrahedron. In such case,


    \iint_S\vec{F}\;\vec{n}\;dS=\iiint_T\textrm{div}\;  (\vec{F})\;dV
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  3. #3
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    Quote Originally Posted by olski1 View Post
    Hi,

    I need to evaluate the flux integral \iint  F \cdot \hat{n}dS

    for portion of a plane x +2y+3z = 6 in the first octant and vector field F= < xy, -x^2, (x+z) >

    How do i go about doing this?

    I started of by doing it through parametrising the surface with r(s,t) = (3s,3t,(2-2t-s))

    With normal vector equalling cross product of tangents N= (3,-6,9)
    You may have forgotten to multiply the "j" term by -1. I get N=(1, 2, 3) which is a multiple of (3, 6, 9).

    You don't say how this is to be oriented. (1, 2, 3) gives the normal vector of the plane "oriented upward" (or "to the right" or "to the back") while (-1, -2, -3) would give the normal for the plane "oriented downward" (or "to the left" or "to the front").

    Then my bounds for integration were:

      0\leqslant s \leqslant 6-2t ,       0\leqslant  t \leqslant 6

    then i proceeded to integrate but it became really messy and did not seem right!

    so firstly, do i need to do it by parametrising the surface? secondly, are my bounds for integration correct? and lastly, It doesnt look like my normal vector is point in the right direction.
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