scratch

§ Change of Basis from Triangle X Y to Barycentric

created 2023-05-19
  • If we have ∫Tf(x,y)dxdy\int_T f(x, y)dx dy∫T​f(x,y)dxdy for a triangle TTT, we would often like to change to barycentric coordinates to compute ∫p=001∫q=0pf(p,q)dpdq\int_{p=0}0^1 \int_{q=0}^p f(p, q) dp dq∫p=0​01∫q=0p​f(p,q)dpdq. But what is the relationship between these two integrals?
  • Note that when we parametrize p,qp, qp,q by as {(p,q):p∈[0,1],q∈[0,p]}\{ (p, q) : p \in [0, 1], q \in [0, p] \}{(p,q):p∈[0,1],q∈[0,p]}, we are drawing a right triangle whose base is on the xxx axis.
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