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.


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.

  • $\begingroup$ Thanks for your answer, but your matrix maps, for example, $(1,0)$ onto $(a_1, b_1)$, and this is not the desired result, is it? Edit: Ah, I was wrong. $\endgroup$
    – Borol
    Feb 9 '17 at 17:41
  • $\begingroup$ As I said, I was told that it would be able to solve this with the "Change-of-variables formula", which differs from your own formula. Do you have an idea how to do it in this case? $\endgroup$
    – Borol
    Feb 10 '17 at 8:33
  • $\begingroup$ The formula $(1)$ above is the change-of-variables formula. By the way: The transform matrix in your question is still wrong. – That's my last word in this matter. $\endgroup$ Feb 10 '17 at 8:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.