Example problem: Compute $\int _A 3x^2+y \, \mathrm{d}A$, where $A$ is a triangle with vertices $(0,0),(2,0),(1,1)$.
Can this be computed using barycentric coordinates? I found a Wikipedia article on that matter, but I don't really understand it. Could you work through this example? I'd really appreciate that!
Your integral is $\displaystyle \int_0^{1} \int_y^{2-y} (3x^2 + y) \, dx \, dy = \frac{23}{6}$
Now we can always use change of variable to any two barycentric coordinates such that limits of our integral over a triangular region becomes
$0 \leq \lambda_1 \leq 1 - \lambda_2, 0 \leq \lambda_2 \leq 1$
In barycentric coordinates,
$\lambda_1 + \lambda_2 + \lambda_3 = 1 \,$ so $\lambda_3 = 1 - \lambda_1 - \lambda_2$
Given the vertices of the triangle $(0, 0), (2, 0), (1, 1)$, we have
$x = \lambda_1 x_1 + \lambda_2 x_2 + \lambda_3 x_3 = 1 - \lambda_1 + \lambda_2$
Similarly, $y = 1 - \lambda_1 - \lambda_2$
So our integral becomes
$\displaystyle I = 2 A \int_0^1 \int_0^{1-\lambda_2} [3 (1 - \lambda_1 + \lambda_2)^2 + 1 - \lambda_1 - \lambda_2] \, d \lambda_1 \, d\lambda_2 = \frac{23}{6}$
$A = 1$ is the area of the triangle (which is easy to find as it is isosceles triangle with base $2$ and altitude $1$).
While in this case finding the integral is easier with the regular method, it is not always the case and we have to split our integral into two. Barycentric coordinates can be easier in such cases.