Editing Barycentric coordinate system (section)

====Integration over a triangle or tetrahedron ====

The integral of a function over the domain of the triangle can be annoying to compute in a cartesian coordinate system.  One generally has to split the triangle up into two halves, and great messiness follows.  Instead, it is often easier to make a [[Integration by substitution|change of variables]] to any two barycentric coordinates, e.g. <math>\lambda_1,\lambda_2</math>.  Under this change of variables,

<math display=block>
\int_{T} f(\mathbf{r}) \ d\mathbf{r} = 2A \int_{0}^{1} \int_{0}^{1 - \lambda_2} f(\lambda_1 \mathbf{r}_1 + \lambda_2 \mathbf{r}_2 +
(1 - \lambda_1 - \lambda_2) \mathbf{r}_3) \ d\lambda_1 \ d\lambda_2
</math>

where {{mvar|A}} is the [[Triangle#Using coordinates|area]] of the triangle.  This result follows from the fact that a rectangle in barycentric coordinates corresponds to a quadrilateral in cartesian coordinates, and the ratio of the areas of the corresponding shapes in the corresponding coordinate systems is given by <math>2A</math>. Similarly, for integration over a tetrahedron, instead of breaking up the integral into two or three separate pieces, one could switch to 3D tetrahedral coordinates under the change of variables

<math display="block">
\int\int_{T} f(\mathbf{r}) \ d\mathbf{r} 
= 6V \int_{0}^{1} \int_{0}^{1 - \lambda_3} \int_ {0}^{1-\lambda_2-\lambda_3} 
f(\lambda_1\mathbf{r}_1 + \lambda_2\mathbf{r}_2 +
\lambda_3\mathbf{r}_3 + (1-\lambda_1-\lambda_2-\lambda_3)\mathbf{r}_4)
\ d\lambda_1 \ d\lambda_2 \ d\lambda_3
</math>where {{mvar|V}} is the volume of the tetrahedron.