Integration over an arbitrary triangle / Change-of-variables formula

Question

Let $a, b \in \Bbb R^2 \setminus \{0\}$, $a_1b_2 \neq a_2b_1$, and let $D \subset \Bbb R^2$ be a triangle with the edges $0, a$ and $b$. Furthermore, let $S \in \Bbb R^2$ be a simplex with the edges $0, (1, 0)$ and $(0, 1)$. Determine a linear map $T: \Bbb R^2 \rightarrow \Bbb R^2$ with $TS = D$ and calculate it's determinant.

Then, calculate

$$\int_D x d\lambda^2(x, y)$$

A simplex in $\Bbb R^2$ should just be a triangle too, so we are searching for a linear map that transforms $S$ to the position where $D$ is. Since we are working with the canonical vectors here, this should be fairly easy. The linear map is given by

\begin{pmatrix} a_1 \ a_2 \\ b_1 \ b_2 \end{pmatrix}

This yields the mapping

$$(0, 0) \rightarrow (0, 0)$$ $$(1, 0) \rightarrow (a_1, a_2)$$ $$(0, 1) \rightarrow (b_1, b_2)$$

The determinant of the matrix is $a_1b_2 - a_2b_1$, and it differs from $0$ by premise. Now in order to calculate the integral, I guess I have to change the surface I integrate over, because the edges of $D$ are arbitrary.

I was told that this should work by applying the Change-of-variables formula, but I don't see how to do it.

Answer 1

The linear map $T$ that produces your triangle $D$ with vertices $0$, $a$, and $b$ as image of the standard simplex $S$ has matrix $$\left[\matrix{a_1&b_1\cr a_2&b_2\cr}\right]$$ (different from your proposal), and acts in the form $$T:\quad (u,v)\mapsto \left\{\eqalign{x&=a_1u+b_1v\cr y&=a_2u+b_2v\cr}\right.\quad.$$ It follows that the pullback of the function $f(x,y):=x$ is given by $$\hat f(u,v)=f(a_1u+b_1v,a_2u+b_2v)=a_1 u+b_1 v\ .$$ The multivariable change of variables formula $$\int_D f(x,y)\>{\rm d}(x,y)=\int_S\hat f(u,v)\>|J_T(u,v)|\>{\rm d}(u,v)\tag{1}$$then gives $$\int_D x\>{\rm d}(x,y)=\int_S (a_1 u+b_1 v)\>|J_T(u,v)|\>{\rm d}(u,v)\ .$$ Here $|J_T(u,v)|=|{\rm det}(T)|=|a_1 b_2-a_2 b_1|$ is a constant, and $$\int_S (a_1 u+b_1 v)\>{\rm d}(u,v)=\int_0^1\int_0^{1-u}(a_1 u+b_1 v)dv\>du\ ,$$ which I may leave to you.