During a course I encountered the following problem for which the exact integral has to be computed.After days of thinking about it, I can't seem to find out how to proceed. It is like this:
Let $M := \{ (u,v) \in \mathbb{R}^2 \vert 0 < u + v < 1, 0 < 2u - 3v < 4 \}$ be a set.
Now define $f: M \to \mathbb{R}_+, (u,v) \mapsto \sqrt{u + v}$ and the substitution $\begin{pmatrix} u \\ v \end{pmatrix} \mapsto \begin{pmatrix} 1 & 1 \\ 2 & -3 \end{pmatrix} \begin{pmatrix} u \\ v \end{pmatrix}. $
The question now is to evaluate $\int_{M} f(u,v)\, d\lambda^2(u,v)$ by the above change of variables. So, obviously we are looking for a set $M'$ s.t. the linear map $\phi: M' \to M, \begin{pmatrix} u \\ v \end{pmatrix} \mapsto \begin{pmatrix} 1 & 1 \\ 2 & -3 \end{pmatrix} \begin{pmatrix} u \\ v \end{pmatrix}$ is bijective in order to apply change of variables. The problem now is that I do not know how to construct or find the appropriate set $M'$. According to $M$ it is $$0 < u + v < 1 \\ 0 < 2u - 3v < 4,$$ but how do I find the inequalities for $u$ and $v$? After I have found them, I can simply substitute $u + v$ and $2u - 3v$ in the integrand and integrate over $M'$. But how do I find $M'$? I tried to scale the first one with $2$ and then attempt to eliminate $u$ in the second one but there is nothing sensible in the end.
Maybe part of the problem you are finding this so hard is notational. I'm not even sure what $d\lambda^2(u,v)$ is supposed to mean, so I am going to assume it is someone's convoluted concept of how the area differential should be denoted. When using a change of variables, it is helpful to actually change the variables. Let $\begin{pmatrix}x\\y\end{pmatrix} = M\begin{pmatrix}u\\v\end{pmatrix}$. Or more simply, let $x = u + v$ and $y = 2u-3v$. Then by straight substitution, the domain of integration becomes $M' = \{(x, y) \mid 0 < x < 1, 0 < y < 4\}$. Now $$dxdy = \begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{vmatrix}dudv = |M|dudv = 5dudv$$ and so $$dudv = \frac 15 dxdy$$ So your integral just reduces to $$\frac15\int_0^4\int_0^1\sqrt{x}dxdy$$