## § Change of basis from triangle x y to barycentric

- If we have $\int_T f(x, y)dx dy$ for a triangle $T$, we would often like to change to barycentric coordinates to compute $\int_{p=0}0^1 \int_{q=0}^p f(p, q) dp dq$. But what is the relationship between these two integrals?
- Note that when we parametrize $p, q$ by as $\{ (p, q) : p \in [0, 1], q \in [0, p] \}$, we are drawing a right triangle whose base is on the $x$ axis.