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Integrate on a triangular domain

  1. Jun 17, 2008 #1

    I have to integrate on a triangular domain

    \int_T f(x,y,z)dxdydz

    so I use simplex coordinates, i.e.
    [tex]x=(1-\alpha-\beta)x_1+\alpha x_2+\beta x_3[/tex]
    [tex]y=(1-\alpha-\beta)y_1+\alpha y_2+\beta y_3[/tex]
    [tex]z=(1-\alpha-\beta)z_1+\alpha z_2+\beta z_3[/tex]

    where [tex](x_i,y_i,z_i)[/tex] are the vertices of the triangle and [tex] 0\le\alpha\le1,\ 0\le\beta\le1-\alpha[/tex]

    So what is the Jacobian? When I try to calculate it, the matrix is not square because I have two variables and three equations!
  2. jcsd
  3. Jun 18, 2008 #2


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    Hi daudaudaudau! :smile:

    By "triangular domain", do you mean a two-dimensional surface?

    If so, you can't have a triple integral, can you (not enough independent variables)? :rolleyes:

    Which is why the matrix isn't square! :smile:
  4. Jun 22, 2008 #3
    Can you tell us what "f" is? Or did you already figure your problem out?

    Tiny Tim's comment is "right on" given the way you stated your problem, but, maybe you mistated it.
  5. Jun 22, 2008 #4
    "f" is just some scalar function, I don't think it matters.

    Here is what I am doing: I have tessellated a sphere into a mesh of triangles, and I have to integrate over these individual triangles. So obviously the triangles are plane but they also have both x, y, and z coordinates because they make up the 3D sphere.

    I hope you understand me now.
  6. Jun 22, 2008 #5
    Oh, yeah, I got you now. You have a differential 3-form f(x,y,z)dxdydz on R3 and you need to cut it down so that it can be integrated over a two-dimensional surface in R3. I remember Apostle's calculus book had an explanation of how to do that that I understood when I was a first year graduate student. I never actually had to do it. If anything occurs to me, I'll let you know. At least I understand what you are trying to do.
    Deacon John
  7. Jun 22, 2008 #6
    I can see my library has one(don't know if it's the right one, it just says "Apostol - Calculus"), so I will go and have a look at it. Thank you.
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