fnding the probability of a joint density If i have a joint density such as 
$f_X,Y(x,y)=6xy$ if $0<x<y<1$
how would i find a probability such as $P(X<2Y)$ or $P(XY\leq \frac{1}{2})$
For $P(X<2Y)$ would we set $y=\frac{1}{2}x$ and $y=x$ so i get something like
$$
\int_{0}^1 \int_{\frac{x}{2}}^x 6 xy \, dxdy 
$$
then solve? Not sure if this is correct though?
Could someone please explain how this is done! I have an exam on this topic later today and no solutions to my practice exam. 
Many thanks
 A: Draw axes in the usual way, and draw the region $A$ on which our joint density function "lives." In this case, it is the triangle with corners $(1,0)$, $(1,1)$, and $(0,1)$. 
For $X\lt 2Y$, draw the line $x=2y$, that is, $y=\frac{x}{2}$. The event $X\lt 2Y$ occurs if the pair $(X,Y)$ lands in the part of our triangle $A$ which is above the line $y=\frac{x}{2}$. Let $B$ be this region. Our probability is 
$$\iint_B 6xy\,dx\,dy.$$
This integral is not completely pleasant to evaluate. It is substantially easier to find the probability that $(X,Y)$ lies in the part $C$ of $A$ which is below the line $ y=\frac{x}{2}$. Then our required probability is $1$ minus this. So our probability is 
$$1-\iint_C 6xy\,dx \,dy.$$
We now evaluate the integral, by integrating first with respect to $y$, and then with respect to $x$. The line $y=\frac{x}{2}$ meets the hypotenuse $x+y=1$ of triangle $A$ at $x=\frac{2}{3}$. Thus
$$\iint_C 6xy \,dx\,dy=\int_{x=2/3}^1 \left(\int_{y=1-x}^{x/2} 6xy\,dy\right)\,dx.$$
For $XY\le \frac{1}{2}$, the procedure is similar. Draw the curve $xy=\frac{1}{2}$. Let $D$ be the part of $A$ which is below this hyperbola. We want $$\iint_D 6xy\,dx\,dy$.  The hyperbola is everywhere above the line $x+y=1$. As in the previous problem, it is easier to find first the probability that $XY\gt \frac{1}{2}$. 
We leave the rest to you, but the probability that $XY\gt \frac{1}{2}$ is
$$\int_{x=1/2}^1 \left(\int_{y=1/(2x)}^{1} 6xy\,dy\right)\,dx.$$
