Calculating the joint probability density $f(x,y)=\frac{1}{y}$ If the joint probability density of $X$ and $Y$ is given by:
$$f(x,y) = \frac{1}{y}$$
for $0<x<y, 0<y<1$
Find the probability that the sum of the values of $X$ and $Y$ will exceed 1/2.
What I have tried:
I have tried sketching out the region and got this:

However, I was not sure on which areas to shade. The corrected areas have been provided in the image as I looked at the solution. I got the framework on the graph correct, just not which areas to shade.
What should I look out for when trying to shade regions of my graph when given a pdf as described? I sort of understand why the area above 1/2 is shaded, but I do not understand why the triangle below 1/2 is also shaded.
Secondly here's my working on the integrals in the y direction:
$$\int_0^{\frac{1}{2}}dx\int^{\frac{1}{2}-x}_0 \frac{1}{y}dy + \int_{\frac{1}{2}}^1\int_0^{1-x}\frac{1}{y}dy$$
 A: According to nejimban's comment you have to integrate on the set
$$
\begin{align}
A&=\{(x,y):0<x<y<1,\ \frac{1}{2}< x+y\}\\
\end{align}
$$
Given this information you find
$$y>\frac{1}{2}-x>\frac{1}{2}-y$$
such that
$$1>y>\frac{1}{4}$$
Furthermore you find, that
$$y>x>\frac{1}{2}-y$$
However, note that for $y>\frac{1}{2}$
$$0 > \frac{1}{2}-y$$
Because of this, you have to add another boundary at $y=\frac{1}{2}$. Integrating your density function with these information in mind, gives the asked probability
$$
\begin{align}
\int_\frac{1}{4}^\frac{1}{2}\int_{\frac{1}{2}-y}^y\frac{1}{y}dx\ dy + \int_\frac{1}{2}^1\int_{0}^y\frac{1}{y}dx\ dy&=\int_\frac{1}{4}^\frac{1}{2}\frac{1}{y}\Big(\int_{\frac{1}{2}-y}^ydx\Big)\ dy + \int_\frac{1}{2}^1\frac{1}{y}\Big(\int_{0}^ydx\Big)\ dy\\
&=\int_\frac{1}{4}^\frac{1}{2}\frac{1}{y}(2y-\frac{1}{2})\ dy + \int_\frac{1}{2}^1\ dy\\
&=\int_\frac{1}{4}^\frac{1}{2}(2-\frac{1}{2y})\ dy + \int_\frac{1}{2}^1\ dy\\
&=\frac{1}{2}-\frac{1}{2}\ln(2x)\Big|_{x=\frac{1}{4}}^{x=\frac{1}{2}}+\frac{1}{2}\\
&=1-\frac{1}{2}\ln(2)
\end{align}
$$
