Integrating Joint Random Variable Distributions - Defining Integration Intervals I'm currently studying for a stats quiz using the course text book, and I'm discovering that I'm a little rusty on setting up regions for double integrals. The current question I'm hung up on is as follows: 
If the joint probability density of $X$ and $Y$ is given by: 
$$f(x,y) = \begin{cases} 2 & \text{for } x>0, y>0,x+y<1 \\ 0 & \text{elsewhere}\end{cases}$$
find:
a. $P(X \leq \frac{1}{2}, Y \leq \frac{1}{2})$
b. $P(X+Y > \frac{2}{3})$
c. $P(X > 2Y)$
I got part a correct, with the region of integration being $A = \{(x,y) \mid 0 < x < \frac{1}{2}, 0 < y < \frac{1}{2}\}$, but b and c I'm finding a bit more difficult. I have yet to attempt c, but I've set the following regions up for b:
$$A = \{(x,y) \mid 0 < x < \frac{2}{3}, 0 < y < 1-x\}$$ and $$A = \{(x,y) \mid \frac{2}{3} < x < 1, 0 < y < 1-x\}$$
Both regions yielded incorrect answers $\frac{8}{9}$ and $\frac{1}{9}$ respectively, with the correct answer being $\frac{5}{9}$. Generalized, how does one go about setting up the region of integration when given bounds as described above?
Thank you.
 A: These things are best sketched out first.
Region (a):  $\Pr[X \le 1/2 \cap Y \le 1/2]:$

It becomes obvious that we should set it up as $$\int_{x=0}^{1/2} \int_{y=0}^{1/2} f(x,y) \, dy \, dx.$$
Region (b):  $\Pr[X+Y > 2/3]:$

This is trickier.  You could use two double integrals, by visualizing a vertical line at $x = 2/3$ splitting the red region into a parallelogram and a triangle:  $$\int_{x=0}^{2/3} \int_{y=2/3-x}^{1-x} f(x,y) \, dy \, dx + \int_{x=2/3}^1 \int_{y=0}^{1-x} f(x,y) \, dy \, dx.$$  Or, you could calculate the complement, which is $1$ minus the integral over the blue triangle:  $$\Pr[X+Y > 2/3] = 1 - \Pr[X+Y < 2/3] = 1 - \int_{x=0}^{2/3} \int_{y=0}^{2/3-x} f(x,y) \, dy \, dx.$$
Region (c):  $\Pr[X > 2Y]:$

This one suggests integrating in horizontal strips rather than vertical ones:  $$\int_{y=0}^{1/3} \int_{x=2y}^{1-y} f(x,y) \, dx \, dy.$$  The $1/3$ comes from the fact that the intersection of the two lines $x = 2y$ and $x + y = 1$ occurs at the point $(x,y) = (2/3, 1/3)$.  Note that the limits of the integral with respect to $x$ are simply the equations of these two lines, where $x$ is solved in terms of $y$.
A: In this case, we are integrating $2$ over a suitable region, so the integration amounts to finding an area, amd multiplying by $2$.
For Problem b), the easiest approach is to first find $\Pr(X+Y\le \frac{2}{3})$. This is twice the area of the first-quadrant region below the line $x+y=\frac{2}{3}$. That area is $\frac{2}{9}$, so  this probability is $\frac{4}{9}$. It follows that $\Pr(X+Y\gt \frac{2}{3})=\frac{5}{9}$.
For Problem c), draw the line $y=\frac{x}{2}$. We want the probability that $(X,Y)$ lands below the line $y=\frac{x}{2}$.
Consider the triangle $T$ whose vertices are $(0,0)$, $(1,0)$, and $((0,1)$. This is where our joint density function "lives." Let $A$ be the part of this triangle which is below the line $y=\frac{x}{2}$. We want to integrate $2$ over $A$. Equivalently, we want to find the area of $A$, and then multiply the result by $2$. This is a geometry problem.
For the line $y=\frac{x}{2}$ meets the line $x+y=1$ at $(\frac{2}{3}, \frac{1}{3})$. So $A$ is the triangle with corners $(0,0)$, $(1,0)$, and $(\frac{2}{3},\frac{1}{3})$.  Its area is not hard to find.  
Remark: If you really want to integrate, you can do it as follows. We can either integrate first with respect to $y$, or first with respect to $x$. We can read off the limits of integration from the picture.
If we do $y$ first, the double integral has to be split into two parts, $x=0$ to $x=\frac{2}{3}$, and $x=\frac{2}{3}$ to $x=1$.  For the first integral, $y$ goes from $0$ to \frac{x}{2}$.
If we integrate first with respect to $x$, things are a bit more pleasant, $x$ goes from $2y$ to $1-y$, and then $y$ goes from $0$ to $\frac{1}{3}$. 
