Calculate $\iint_R \frac{1}{x+y}dydx$ where $R$ is the region bounded by $x=0,y=0,x+y=1,x+y=4$ by using the mapping $T(u,v)=(u-uv,uv)$.
I find this exercise super confusing. $R$ is a parallelogram, so we need to find a simpler region; since $T$ is not linear, we can't assume that it maps parallelograms into parallelograms. Besides, I don't get whether we need to find an $R^*$ such that $R^*=T(R)$ or an $R^*$ such that $T(R^*)=R$.
If the former, then by setting $x=u-uv,y=uv$, we get
$$u-uv=0 \\ uv=0 \\ u = 1 \\ u = 4$$
But I don't know how to interpret this. The first equation has the solution set $\{(u,v) : u=0 \text{ or } v=1\}$ and the second equation has both axes as a solution set. There is no single region bounded by this equations.
I tried finding parametric curves $c_1,\dots,c_2$ for $R$'s borders and then finding the composition $T \circ c_i$ but the results were ghastly (parabolas).
I'd really like to learn how to be methodical with these problems.
Consider the region $R$ as covered by the segments $I(s)$, $1\le s\le 4$, which are connecting the points $(s,0)$ and $(0,s)$ placed on the axes. For a point $(x,y)$ on $I(s)$ we have of course the sum $s=x+y$. Now we also want to parametrize the segments. A natural choice is $x = su$, $y=s(1-u)$, where $u$ runs in $[0,1]$. We are thus in the situation to consider the relations $$ \begin{aligned} x &= su\ ,\\ y &= s(1-u)\ , \end{aligned} $$ and $$ \begin{aligned} s &= x+y\ , \\ u &= x/(x+y)\ , \end{aligned} $$ which connect $(x,y)\in R$ with $(s,u)\in[1,4]\times[0,1]$.
We have formally: $$ \begin{aligned} dx\wedge dy &= d(su)\wedge d(s(1-u)) \\ &=(s\; du+u\; ds)\wedge(-s\; du+(1-u)\; ds) \\ &=(s(1-u)-u(-s))\; du\wedge ds \\ &=s\; du\wedge ds \ , \end{aligned} $$ and the integral becomes $$ \iint_R\frac 1{x+y}\; dx\ dy = \iint_{[0,1]\times[1,4]}\frac 1s\; s\; du\; ds = \iint_{[0,1]\times[1,4]} du\; ds = \operatorname{Area}([0,1]\times[1,4]) =3\ . $$