Application of change of variables in $\mathbb{R}^2$

Question

In the context of a measure theory and integration couse the following exercise came up:

Let $$Q := (0,1) \times (0,1) \subset \mathbb{R}^2$$ and consider the map $$ f : Q \to \mathbb{R},\, (x,y) \mapsto \frac{1}{1 - xy}.$$

Assume that $f$ is Lebesgue-integrable and compute the value of the integral $$\int_{Q} f(x,y)\, d\lambda^2(x,y)$$ using the substitution (i.e. linear transformation)

$$\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} u \\ v \end{pmatrix}.$$

My first question is: how do you come up with that quite specific substitution?

The second question is about the solution to the problem that I actually received but did not fully comprehend since many steps have been omitted. According to the solution, the first thing one should have done there is to come up with the set $$Q' := \{ (u,v) \in \mathbb{R}^2 \vert 0 < u < \sqrt{2}, -\min(u, \sqrt{2} - u) < v < \min(u, \sqrt{2} - u)\}$$ quite magically $\textit{by definition}$ and then perform the relatively easy change of variables under the linear transformation $\phi : Q' \to Q$ given above. Hence, my second question: How to come up with the set $Q'$ analytically? In the solution it was introduced by definition 'falling from the sky'. Can one derive it in any way? I would have never arrived there frankly just by guessing. I heard that there might be a geometrical reason which is based on the rotation of the cube $Q$ but I couldn't read much sense into that.

Answer 1

Looking at $Q^{'}$ is just changing the perspective from the original cube to the square $Q$ as a set over the diagonal of the cube (of length $\sqrt{2}$) bounded by the graphs of two functions (the ones with the $\min$ in the defintion).

The linear transformation is just a stretched rotation which moves the $x$ axis into the diagonal of the original square. (It has the additional nice property of transforming $xy$ into $(u^2-v^2)/2$)

Answer 2

A general rotation matrix in $\mathbb{R}^2$ is given by $\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}$. For $\theta = \frac{\pi}{4}$ we get exactly your matrix above. So the transformation is a rotation of $45°$ of some cube $Q'$ to the cube $Q := (0,1) \times (0,1) \subset \mathbb{R}^2$. Now we have to write down the set $Q'$. When you draw a picture like then you can define $Q'$ and it is exactly $Q' = \{ (u,v) \in \mathbb{R}^2 \vert 0 < u < \sqrt{2}, -\min(u, \sqrt{2} - u) < v < \min(u, \sqrt{2} - u)\}$ like stated.

The good thing about the above transformation is, that the matrix is orthogonal with determinant 1. So when you transform the integral there is no strange determinant term so can ignore it. on the other hand $xy=(\frac{1}{\sqrt{2}}u-\frac{1}{\sqrt{2}}v)(\frac{1}{\sqrt{2}}u+\frac{1}{\sqrt{2}}v)=\frac{1}{2}(u^2+v^2)$ so you have no product of different variables which makes the integrand more handy.