# Numerical Integration over 3D mesh.

I have a 3D mesh composed of tessellating triangles. I would like to perform an integral over the surface of this mesh of the form:

$$\int f(\Omega)R(\Omega)d\Omega$$

Where $R(\Omega)$ is the radius at a given angle and $f(\Omega)$ is an arbitary function. I have a method implemented but it is giving large errors.

My method is as follows, is there anything obviously wrong about it?

For each triangular face:

1. Calculate the value of $fR$ at each vertex of the face.
2. Calculate the average of these three values to find the average value of $fR$ across the face.
3. Multiply this value by the solid angle covered by the face.
4. Sum up this result for all of the faces in the mesh.

I implemented this from scratch and I thought it should work. Has anyone done anything similar and if so am I overlooking something?

Thanks,

• If it is 3D, then each element is a tetrahedron, and your integration is with respect to $d\Omega$ (volume). – Shuhao Cao Apr 27 '13 at 22:53
• Sorry, I think I should clarify, I would like to integrate over the surface of this mesh. I'll edit the main question. – user668074 Apr 27 '13 at 23:10
• For numerical quadrature, the sample points should be chosen on the surface. After you done the average, you have to multiply the area of each face on each integral. Also I am sure about what "solid angle covered by the face", can you make your statement more mathematical? For example using Cartesian coordinates. – Shuhao Cao Apr 27 '13 at 23:31
• My sample points are at the corners of triangles which stitch together to make a closed surface. Those should be appropriate right? Regarding the solid angle part, I am approximating the different $d\Omega$ with the solid angle covered by the triangle face. Mathematically and in the program, I am using the veritices to create a spherical triangle. The solid angle covered by that spherical triangle is $(A+B+C) - \pi$. Where $A$, $B$ and $C$ are the angles between the vertices as measured on the surface. link – user668074 Apr 27 '13 at 23:37
• I don't know if you're still looking for answers to this, but if so, what is $\Omega$? – Eric Kightley Jul 12 '13 at 19:14