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2D surface integral in MATLAB for Finite Element Calculation

  1. Mar 19, 2010 #1
    Hi everyone,

    As part of a project, I am required to numerically compute the expression

    [tex]b_{i}^{e} &=& \frac{E_{0}^{i}k_0^2(\epsilon_r-1/\mu_r)}{2\Delta^e}\left[\iint\limits_{\Omega^e}(a_i^e + b_i^e x + c_i^e y)e^{-jk_0 x} dx dy\right] \nonumber\\&&- \frac{jk_0 E_0^i r'}{2\Delta^e \mu_r}\left[\int_{\phi_{1}^{s_2}}^{\phi_{2}^{s_2}}(a_i^e \cos\phi + b_i^e r'\cos^2\phi + c_i^e r'\sin\phi\cos\phi)e^{-jk_0r'\cos\phi}d\phi\right][/tex]

    specifically, compute the integrals numerically. The problem is that [itex]\Omega^e[/itex], the domain of integration of the first integral is a triangle (whose vertex coordinates are well known).

    I am unable to figure out a way to do this integral computationally in MATLAB. That is, how does one compute an area integral in MATLAB when the x and y coordinates are coupled (and bounded to lie in a spatial region).

    If there is a documented way of doing this, or a preexisting function, I would prefer to use it and go ahead with my work, rather than reinvent the wheel. Any inputs would be greatly appreciated!

    Thanks in advance!

    Cheers
    Vivek
     
  2. jcsd
  3. Dec 29, 2015 #2
    Hi Vivek,

    One thing you could do is define a function of x and y that will return one in the domain and zero outside.

    For example, the following function returns one only in the 45-45-90 triangle near the origin given by x<3, y>0, x>y:
    domain = @(x,y) heaviside(x-y)*heaviside(3-x)*heaviside(y).

    Then you can integrate over a square including your triangular domain and multiply the integral argument by this domain function.

    I'm sure you've solved this problem by now, but I stumbled across it while searching for surface integration techniques and figured I might as well reply.
    Cheers,
    David
     
  4. Dec 29, 2015 #3

    SteamKing

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    Vivek hasn't posted in a year.

    Please resist the urge to necro post to these ancient threads.
     
  5. Dec 29, 2015 #4
    Why? This thread shows up for anyone who google searches "matlab surface integral". Wouldn't it be better if they found a reply?
     
  6. Dec 29, 2015 #5

    SteamKing

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    I'm just repeating the policy of the moderators of this forum. They prefer not to have these old posts dredged up from the past.

    However, if you wish to create a new thread of your own, dealing with this subject, you can always refer to previous threads on the same topic.

    If you seek another opinion on this point, please feel free to contact one of the moderators, say via PM, and start a discussion about necro posting to these old threads. Policies do change at PF from time to time.
     
  7. Dec 29, 2015 #6
    Ah I see. Thank you SteamKing. I'll be happy to review the policy and comply in the future.
     
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