Integrating a Triangle

For a specified Meshes.Triangle surface with area $A$, let $u$ and $v$ be Barycentric coordinates that span the surface.

\[\int_\triangle f(\bar{r}) \, \text{d}A = \iint_\triangle f\left( \bar{r}(u,v) \right) \, \left( \text{d}u \wedge \text{d}v \right)\]

Since the geometric transformation from the originally-arbitrary domain to a Barycentric domain is linear, the magnitude of the surface element $\text{d}u \wedge \text{d}v$ is constant throughout the integration domain. This constant will be equal to twice the magnitude of $A$.

\[\int_\triangle f(\bar{r}) \, \text{d}A = 2A \int_0^1 \int_0^{1-v} f\left( \bar{r}(u,v) \right) \, \text{d}u \, \text{d}v\]

This non-rectangular Barycentric domain prevents a direct application of most numerical integration methods. It can be directly integrated, albeit inefficiently, using nested Gauss-Kronrod quadrature rules. Alternatively, additional transformation could be applied to map this domain onto a rectangular domain.

WORK IN PROGRESS: continued derivation to detail this barycentric-rectangular domain transformation