Double integral over the triangle 0I got stuck doing this exercise from my analysis lecture
Let $f:\mathbb{R^2}\rightarrow{\mathbb{R}}$ st
$$ f(x)=\begin{cases}
       \frac{\sqrt{1-y}}{\sqrt{x-y}} &\quad\text{if } 0<y<x<1 \\
       0 &\quad\text{otherwise} \\
     \end{cases}$$
Then $f$ is integrable over $\Omega=\{(x,y)\in{\mathbb{R^2}} \ | \text{ }  0<y<x<1 \}$. I got so far with the Tonelli-Fubini's theorem, since $f \mathbf{1}_{\Omega}:\mathbb{R^2}\rightarrow{[0,\infty)}$ is measurable, then
$$ \int_{R^2}f \mathbf{1}_{\Omega}(x,y)d(x,y)=\int_{\mathbb{R}}\int_{\mathbb{R}}f \mathbf{1}_{\Omega}(x,y)d(x,y)=\int_{\mathbb{R}}\int_{\mathbb{R}} \frac{\sqrt{1-y}}{\sqrt{x-y}} \mathbf{1}_{\Omega}(x,y)dxdy$$
Using the invariant over translation of measurable functions
$$ \int_{\mathbb{R}}\int_{\mathbb{R}} \frac{\sqrt{1-y}}{\sqrt{x-y}} \mathbf{1}_{\Omega}(x,y)dxdy = \int_{\mathbb{R}}\sqrt{1-y}\int_{\mathbb{R}} \frac{1}{\sqrt{x}}\mathbf{1}_{\Omega}(x+y,y)dxdy$$
And I don't know what to do next. Any hint?.

We have that: $\{(x,y)\in\Bbb R^2:0 < y< x < 1\}=\{(x+y, y)\in\Bbb R^2: 0<y<1, 0<x<1-y\}$
So your integral is $$\int_0^1 \sqrt{1-y~}\int_0^{1-y}\dfrac 1{\surd x}\,\mathrm d x\,\mathrm d y$$
And you may apply a change of variables to $y$ as well.
$$\int_0^1\surd y\int_0^y\dfrac 1{\surd x}\,\mathrm d x\,\mathrm d y$$

Although this is quite valid , to avoid confusion we often change symbols when apply these transformations:
$$\begin{align}&\quad\iint_{0<y<x<1} \dfrac{\sqrt{1-y}}{\sqrt{1-x}}\,\mathrm d (x,y) \\[1ex]&=\iint_{0<1-v<u+1-v<1}\dfrac{\sqrt v}{\sqrt u}\,\mathrm d (u,v)&&\text{via }{u=x-y\\v=1-y}\\&=\iint_{0<v<1,0<u<v}\dfrac{\sqrt v}{\sqrt u}\,\mathrm d (u,v)\\&=\int_0^1 \sqrt v\int_0^v\dfrac{1}{\sqrt u}\,\mathrm d u\,\mathrm d v\end{align}$$
Of course, the ultimate result is the same.
