Finding the joint distribution of two dependent random variables given conditions 
Suppose $X \sim U(0,1)$ and $Y|X=x \sim U(0,1-x)$. Find the joint
  distribution of $(X,Y)$ given that $X\le \frac{1}{2}$, $Y\le
 \frac{1}{2}.$

My attempt:
The pdf of $X$ is $f_{X}(x) = 1$ for $x \in (0,1)$ and the pdf of $Y|X=x$ is $f_{Y|X}(y|x) = \frac{1}{1-x}$ for some $x\in (0,1)$ and $y\in (0,1-x)$.
Hence the (unconditional) joint density function should be $f_{XY}(x,y) = f_{Y|X}(y|x)f_X(x) = \frac{1}{1-x} $ where $x \in (0,1)$ and $y\in (0,1-x)$.
To find the distribution of $(X,Y)$ given the constraint let $A = \{(X,Y): \quad X\le \frac{1}{2} \cap Y\le \frac{1}{2} \}$.
Then, what I am trying to find is (using the CDF method) $$ \begin{align}\mathbb P (X\le x,Y\le y\vert (X,Y)\in A) &= \frac {\mathbb P\biggl(X\le x, Y\le y, X\le \frac 12, Y\le \frac {1}{2}\mathstrut \mathstrut \biggr) }{\mathbb P\Bigl(X\le \frac {1}{2},Y\le \frac 12\mathstrut \mathstrut \Bigr) }\mathstrut \\ &=   \frac {\mathbb P\Bigl(X\le \text{min}\Bigl(x, \frac {1}{2}\mathstrut \Bigr), Y\le \text{min}(y, \frac {1}{2})\Bigr) }{\mathbb P(X\le \frac 12, Y\le \frac 12\mathstrut \mathstrut ) }\mathstrut  \end{align} $$
at this point I'm not sure how to continue. For $(x,y) \in [0, \frac{1}{2}]^2$ the numerator becomes the (unconditional) CDF of just $(X,Y)$ but for e.g. $x > \frac{1}{2}$ and $y < \frac{1}{2}$ I'm not sure how to identify the distribution/simplify the expression.
 A: \begin{align}
P(Y\le y, X\le x)&= \int_{-\infty }^x P\left(Y\le y\ |\,X=t\right)f_X(t)dt\\
&=\cases{\displaystyle  0& if $\ x\le0\ $or$\ y\le0$\\
\displaystyle\int_0^x\frac{y}{1-t}dt&if $\ 0<x\le1-y,y< 1$\\
\displaystyle\int_0^{1-y} \frac{y}{1-t}dt+\int_{1-y}^xdt&if $\ 0<y<1, 1-y<x<1$\\
\displaystyle\int_0^{1-y} \frac{y}{1-t}dt+\int_{1-y}^1dt&if $\ 1\le x, 0<y<1 $\\
\displaystyle\int_0^xdt&if $\ 0<x<1, 1\le y$\\
\displaystyle 1&if $\ 1\le x,y$
}\\
&=\cases{
\displaystyle  0&if $\ x\le0\ $or$\ y\le0$\\
-y\ln(1-x)& if $\ 0<x\le1-y,y< 1$\\
-y\ln(y)+ x+y-1 &if  $\ 0<y<1, 1-y<x<1$\\
-y\ln(y)+y&if $\ 1\le x, 0\le y<1$\\
x&if $\ 0<x<1, 1\le y$\\
1&if $\ 1\le x,y$}
\end{align}
Therefore $\ P\left(Y\le\frac{1}{2}, X\le\frac{1}{2}\right)=\frac{\ln(2)}{2}\ $, and
\begin{align}
P\left(X\le x, Y\le y\left|\ Y\le\frac{1}{2}, X\le\frac{1}{2}\right.\right)&=\frac{P\left(X\le\min\left(x,\frac{1}{2}\right), Y\le\min\left(y,\frac{1}{2}\right)\right)}{P \left(Y\le\frac{1}{2}, X\le\frac{1}{2}\right)}\\
&=\cases{
0&if $\ x\le0\ $ or $\ y\le 0\ $\\
\displaystyle-\frac{2y\ln(1-x)}{\ln(2)}&if $\ 0<x,y\le\frac{1}{2}, $\\
2y&if $\ \frac{1}{2}<x, 0<y\le\frac{1}{2} $\\
\displaystyle-\frac{\ln(1-x)}{\ln(2)}&if $\ 0<x\le \frac{1}{2}, \frac{1}{2}<y $\\
1&if $\ \frac{1}{2}<x,y$
}
\end{align}
A: Does this figure describe the situation?

