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Flow in a circular duct: Simsons Rule

  1. Oct 9, 2009 #1
    I want to integrate log(ri)log(rj) over the unit circle where i,j=1,2,....,N and ri=distance between a point (x,y) and a singularity outside the domain. The answers will form the aij of a matrix.

    The method I am using is to pattern the singuarities so that they form a circle around the domain. Therefore every singularity is a distance 2 from the centre of the domain. So does every log(ri/j)=2?

    Also what are the limits of intergration? Are they -1 and 1 for dy and -sqrt(1-y2) and sqrt(1-y2) for dx?

    I understand this is not specifically a Simpsons Rule question but once these confusions have been sorted I will be going on to solve the integral using Simpsons on Matlab.

    Thank you
     
  2. jcsd
  3. Oct 9, 2009 #2

    LCKurtz

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    Just trying to understand your question. Here's my take on it for you to clear up:

    You apparently have N singularity points:

    P1(x1,y1),P2(x2,y2)...PN(xN,yN)

    For each aij you want to do the following integral, where P(x,y) represents a point in the unit circle and || || is the Euclidean norm?

    [tex]a_{ij} = \int \int_C \log(||P-P_i||\cdot \log||P-P_j||\ dA[/tex]

    [tex]= \int \int_C \log(\sqrt{(x-x_i)^2+(y-y_i)^2})\cdot \log(\sqrt{(x-x_j)^2+(y-y_j)^2})\ dA[/tex]

    To answer your question about the limits, yes, you have them correct for a dx dy integral
     
  4. Oct 9, 2009 #3
    Correct. I am unsure whether I have to integrate for different points (x,y) in the unit circle (domain) or just choose one, in which case the easiest point to choose would be (0,0) and log(ri) would be 2 for every i=1,...n.

    Hopefully I have managed to attach the report that I am basing my problem on. Mine is a more simple problem of the circular pipe but the calculations are similar. The question I have asked is regarding the matrix A at the top of page 3. I am trying ot find the unknowns cj.

    Thanks
     

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