Find $P(X\gt Y)$ using the joint density 
$f_{X,Y}(x,y) = \frac{2}{3} (x+2y)$ for  $0 < x < 1, 0 < y < 1$; find $P(X\gt Y)$.

I got 1/9 by evaluating $$\int_0^1\int_0^{x-1} \frac{2}{3}(x+2y) dy dx$$
 A: I will use the standard notation, with (lowercase) $f$
for the density function $f(x,y)=\frac23(x+2y)$. Then
$$
P(X\gt Y) =
\int_{0}^{1}
\int_{0}^{x}
f(x,y) \,
dy \, dx \, .
$$
If you were taking summations, in the discrete case, and $x,y$ had integer values, then your upper limit for $y$ would be $x-1$. But in the continuous case, the upper limit for $y$ must be $x$ (the integral only accumulates value over an interval, so you shouldn't worry about including the endpoint $y=x$).
That this is the density and not the CDF on $R=[0,1]^2$ is evident because $0\le f\le 2$ and $\int_{R}f(x,y)\,dxdy=1$ as shown below, where we simultaneously work out the answer.
$$
\eqalign{
\int_{R}f(x,y)\,dxdy
&=
\int_{0}^{1}
\int_{0}^{1}
\frac23\left(x+2y\right)
\, dy \, dx
\\&=
\frac23
\int_{0}^{1}
\left[
xy+y^2
\right]_{0}^{1}
\, dx
\\&=
\frac23
\int_{0}^{1}
\left(x+1\right)
\, dx
\\&=
\frac23
\left[
\frac12x^2+x
\right]_{0}^{1}
\, dx
\\&=
\frac23\cdot
\frac32=1
}
\qquad
\qquad
\eqalign{
P(X\gt Y)
&=
\int_{0}^{1}
\int_{0}^{x}
\frac23\left(x+2y\right)
\, dy \, dx
\\&=
\frac23
\int_{0}^{1}
\left[xy+y^2\right]_{0}^{x}
\, dx
\\&=
\frac43
\int_{0}^{1}
x^2
\, dx
\\&=
\frac43
\left[
\frac13
x^3
\right]_{0}^{1}
\, dx
\\&=\frac49
}

$$
A: You want your bounds to represent your event, so to speak.  The event x > y could be satisfied by any value of x, and any value of y such that y < x.  Since y depends on x, I agree with the order of integration.  But why do you have y ranging from 0 to x-1?  That integral is not necessarily interpretable in the context of probability theory.  How could you repair your upper bound on y to reflect the event x > y?
