Let $A=\{(x,y)\in\mathbb R^2|x\lt y\lt 2x, 1-x\lt y\lt 3-x\}$ be a set and evaluate $\int_A \frac{y}{x} d\lambda_2(x,y)$.
I calculate that $x\in(\frac{1}{3},\frac{3}{2})$ and $y$ depends on x with respect to the two inequalities. The problem is to set the bounds of $y$. I draw the set A,but it doesn't helped.
In $\mathbb{R}^2$ a region D is classified as type-I if D can be expressed as $$D=\{(x,y):a\le x\le b, f(x)\le y\le h(x)\}$$ As I want to avoid integrating log, I'll integrate w.r.t y first and to do this I'll break up the region of integration in type-I regions. Had I wanted to integrate w.r.t x first, I would have broken up A into type-II regions (whose definition can be easily obtained by interchanging x and y in the logical predicate part of the set-builder notation in the definition of type-I).
The color-coded regions B,C,D in the figure above are type-I as $$\begin{align} &B=\{(x,y):1\le x\le{3\over2}, x\le y\le3-x\}\\ &C=\{(x,y):{1\over2}\le x\le 1, x\le y\le 2x\}\\ &D=\{(x,y):{1\over3}\le x\le {1\over2}, 1-x\le y\le 2x\} \end{align}$$ So the contribution of B to the integral is $$ \int_{1}^{3\over2}\int_{x}^{3-x}{y\over x}dydx=\int_{1}^{3\over2}{9-6x\over 2x}dx={9\over2}\ln\left({3\over2}\right)-{3\over2} $$ Similarly the rest of the integral can be computed.