Finding area via double integrals, transformation and change of variables. Find the area of the closed region D formed by the lines $x + y = c$ ,
$x + y = d$, $y = ax$, and $y = bx$ where $(0 < c < d, 0 < a < b)$ by first transforming it into a rectangle with area D' with sides d - c and b - a in the u-v plane and then via double integration.
Attempt at a solution: I found the variable relationship to be $x=\frac{u}{1+v}$ and $y=\frac{\left(u\cdot v\right)}{1+v}$ but am unable to form the double integral as the closed region D seems to be neither horizontally nor vertically simple.

 A: If the region is neither horizontally nor vertically simple, you have not succeeded in transforming it into a rectangle.
Start with $(x,y)$

Transform $(x,u=y+x)$

We'd like to map the red and green lines to vertical lines, say to $v=c$ and $v=d$ respectively, so transform $(v=\frac{(d-c) \left(x-\frac{u}{a+1}\right)}{\frac{u}{b+1}-\frac{u}{a+1}}+c , u)$

A: Put
$$u:=x+y,\quad v:={y\over x}\ ,$$
and solve for $x$ and $y$:
$$x(u,v)={u\over1+v}, \quad y(u,v)={uv\over 1+v}\ .$$Inspection of the figure then shows that
$$\psi:\quad (u,v)\mapsto \bigl(x(u,v),y(u,v)\bigr)$$
maps the rectangle $R:=[c,d]\times[a,b]$ in the $(u,v)$-plane bijectively onto your domain $D$ in the $(x,y)$-plane. One computes
$$[d\psi(u,v)]=\left[\matrix{{1\over 1+v}&-{u\over(1+v)^2}\cr{v\over1+v}&{u\over(1+v)^2}\cr}\right]\ ,$$
so that we obtain the Jacobian
$$J_\psi(u,v)={\rm det}[d\psi(u,v)]={u\over(1+v)^2}\ .$$
This leads to
$${\rm area}(D)=\int_D1\>{\rm d}(x,y)=\int_R1\> J_\psi(u,v)\>{\rm d}(u,v)=\int_c^d\int_a^b {u\over(1+v)^2}\>dv\>du\ ,$$
which I may leave to you.
