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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.

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    $$\begin{pmatrix}1&1\\2&-3\end{pmatrix}^{-1} =\frac15\begin{pmatrix}3&1\\2&-1\end{pmatrix}$$ – 2017-02-10
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    Could you elaborate a little as to how to use the inverse here? What does it say with regard to u and v? – 2017-02-10
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    Well, the thought was $M' = \phi^{-1}(M)$, but in fact, you have that backwards. $\phi : M \to M'$, so it really isn't helpful at all. – 2017-02-11

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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 $$\frac15\int_0^4\int_0^1\sqrt{x}dxdy$$

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    There are several things I do not get. First, how do you come up with $\sqrt{x}$ as integrand? Second, according to the classical change of variables in $\mathbb{R}^n$, isn't the substitution backwards? Initially, we have to consider the integral $\int_M ... $. Since we are given a linear transformation $T: M' \to M$ we now want to calculate $\int_{M'} ... $ instead. So we have to determine $M'$ to do so. Third, you mention how helpful it is to actually change variables but then there are still $x$ and $y$ under the transformed integral sign? – 2017-02-11
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    1st: the function being integrated is $f(u,v) = \sqrt{u+v}$. Now $x = u + v$. Therefore in terms of $x$ and $y$, the function is $\sqrt{x}$. 2nd: No, you are given a transformation $T: M \to M'$. It starts with the $u, v$ that $M$ is defined in terms of, and carries that to a new set of coordinates $x, y$. The target $M'$ is the space for $x, y$. 3rd; "Still"? We started with $u, v$, not $x, y$. $x, y$ are the new variables, while $u, v$ are the originals. – 2017-02-12
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    Thank you! I looked at it in a completely different way. Now I will go over it again and in case no more questions pop up I will mark it as the accepted answer. Once more, thank you. – 2017-02-12
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    What I noticed: Shouldn't it be $\frac{1}{5}$ in front of the integral since you have $dxdy = 5dudv \Leftrightarrow \frac{1}{5}dxdy = dudv$? – 2017-02-12
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    Glad you caught that!. I've corrected it. – 2017-02-12