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One-dimensional integration - flux

  1. Aug 6, 2017 #1
    <Moderator's note: Moved from a technical forum and thus no template.>

    Calculate flux of the vector field $$F=(-y, x, z^2)$$ through the tetraeder $$T(ABCD)$$ with the corner points $$A= (\frac{3}{2}, 0, 0), B= (0, \frac{\sqrt 3}{2},0), C = (0, -\frac{\sqrt 3}{2},0), D = (\frac{1}{2},0 , \sqrt 2)$$ by one dimensional integrtion.

    I calculated div, where $$divF = 2z$$

    I can see from the points distribution that $$z\in [0,\sqrt 2] $$ And according to theorem of Gauss I should apply this formula.

    $$\iiint_TdivFdv$$

    How can I seperate the integrals to get one dimensional?
     
    Last edited by a moderator: Aug 7, 2017
  2. jcsd
  3. Aug 7, 2017 #2

    Charles Link

    User Avatar
    Homework Helper

    The tetrahedron they give you has its base it the x-y plane. What is the area of the base? Can you write an expression for the cross-sectional area ## A (z) ## as a function of z, so that ## dv= d^3x=A(z) dz ##? All you need then is to proceed with the integral. ## \\ ## Additional comment: I believe this post might belong in the homework section. If it is homework/schoolwork, please use the Homework section with the homework template on future posts. Thanks. :)
     
  4. Aug 7, 2017 #3

    fresh_42

    Staff: Mentor

    Right.
     
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