I want to integrate a polynomial expression over a 4-node element in 3D. Several books on FEA cover the case where integrating is performed over an arbitrary flat 4-noned element. The usual procedure in this case is to find Jacobi matrix and use it's determinant to change the integration basis to the normalized one in which I have the simpler integration limits [-1;1] and the Gauss-Legendre quadrature technique is used easily.

In other words $\displaystyle\int_S f(x,y)\ \mathrm{d}x\,\mathrm{d}y\,$ is reduced to the form of $\displaystyle\int^{-1}_{1}\int^{-1}_{1} \tilde{f}(e,n)\ \left|\det(J)\right|\,\mathrm{d}e\,\mathrm{d}n$

But in 2D case I change the flat arbitrary element to the flat one but well-shaped square 2 by 2.

3D 4-noded element isn't flat in general but I suppose it still can be mapped with 2D coordinate system which is somehow related to cartesian coordinate system. I can't figure out how to express {x,y,z} in terms of {e,n} and what would be the size of the Jacobi matrix in this case (it's supposed to be square).

2D and 3D domains


1 Answer 1


You're integrating a function on a 2-dimensional manifold embedded in $\mathbb{R}^{3}$; books in analysis on manifolds (like Munkres' accessible book, or Lee's books on manifolds) are helpful in discussing the theory defining this type of integral.

Let's suppose that $f$ is a real-valued function defined on the manifold $\mathcal{M}$, which is your 4-node 3-D element.

You want to calculate:

\begin{align} \int_{\mathcal{M}} f \,\mathrm{d}S. \end{align}

Suppose that $\varphi$ is a function that maps $[-1,1]^{2}$ to $\mathcal{M}$. Then

\begin{align} \int_{\mathcal{M}} f \,\mathrm{d}S = \int_{[-1,1]^{2}} f(\varphi(x,y))\Big(\det\big(\mathrm{D}\varphi^{\mathrm{T}}(x,y)\mathrm{D}\varphi(x,y)\big)\Big)^{1/2}\, \mathrm{d}x \,\mathrm{d}y \end{align}

(I used this set of notes to refresh my memory.) Above, $\mathrm{D}\varphi$ is the Jacobian matrix of $\varphi$, and $\mathrm{D}\varphi^{\mathrm{T}}$ is its transpose.

Once you can write the integral over $[-1,1]^{2}$, then you can use numerical methods to evaluate it.

Some comments:

  • I'm pretty sure your 4-node 3-D element is a manifold. If it is, the function $\varphi$ exists (by definition), is piecewise continuous (for topological manifolds), and is invertible. It's up to you to find a function with those properties.
  • The argument above assumes that $\mathcal{M}$ is a smooth manifold, which implies that there exists a $\varphi$ that is continuously differentiable. In your case, the element you describe may not be continuously differentiable. If that is true, you could probably still partition your manifold into two smooth manifolds, and then the argument above still holds. Again, you must find $\varphi$ satisfying the properties of invertibility and continuous differentiability.
  • $\begingroup$ Thanks a lot. The book I'm reading covers only the case where a square (2 by 2) Jacobi matrix is involved to keep things simple. The expression above if I got it right makes it possible to use arbitrary sized (2 by 3) Jacobi matrices. Unfortunately I'm still getting $\det(\mathrm{D}\varphi^{\mathrm{T}}(x,y)\mathrm{D}\varphi(x,y))=0$ at the moment but it's much better than I had earlier. I'll create another thread on the proper choice of mapping function. Thanks again. $\endgroup$
    – danny_23
    Oct 7, 2012 at 17:45
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    $\begingroup$ Your Jacobian matrix $\mathrm{D}\varphi$ should be 3 by 2, so $\mathrm{D}\varphi^{\mathrm{T}}\mathrm{D}\varphi$ should be a 2 by 2 matrix. $\endgroup$ Oct 7, 2012 at 19:35
  • 2
    $\begingroup$ Geoff, that's correct. I put a simple general formula plus a worked out example here: theoretical-physics.net/dev/src/math/integration.html $\endgroup$ Oct 8, 2012 at 4:04

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