Finding the p.d.f. of T=XY using the C.D.F. Technique (Functions of Random Variable) Suppose X and Y are continuous r.v. with join p.d.f. $$f(x,y)=3y, 0 \le x \le y \le 1$$
Find the p.d.f. of $T=xy$
So I know that we can consider for $ t < 0, P(T \le t)=0$ and if $t \ge 1, P(T \le t)=1$ Therefore we consider $t \in (0,1)$. But I don't know how to set up the limits of integration. Please help! 
 A: Note that the joint density of $X$ and $Y$ "lives" in the triangle with corners $(0,0)$, $(1,1)$, and $(0,1)$. Draw that triangle, and for some fixed $t$ between $0$ and $1$, draw the first quadrant part of the hyperbola $xy=t$. It is essential that the picture be drawn if the rest of the post is to be understood.
Then $\Pr(X\le t)$ is the probability that $(X,Y)$ lands in the part of the triangle that is below the hyperbola. Call that region $K$.
If we take a look at $K$, we notice that whatever we integrate with respect to first, we will have to split the region of integration. 
Note that the hyperbola meets the top line $y=1$ of the triangle at $x=t$. Draw the line $x=t$, and let $A$ be the part of $K$ to the left of $x=t$, and $B$ the part of $K$ to the right of $x=t$. We want
$$\iint_A 3y\,dy\,dc+\iint_B 3y\,dy\,dx.$$
Now express each of the two integrals above as iterated integrals.
For the first, $y$ goes from $x$ to $1$, and then $x$ goes from $0$ to $t$. We get 
$$\int_{x=0}^t \left(\int_{y=x}^1 3y\,dy\right)\,dx.$$
For the second integral, $y$ goes from $x$ to $t/x$, and then $x$ goes from $t$ to $\sqrt{t}$. The $\sqrt{t}$ is because the line $y=x$ of our diagram meets the hyperbola at $x=y=\sqrt{t}$.
Remark: There is a nicer way. Look at the part of the triangle that is above the hyperbola. Integrate over that. That gives the probability that $XY\gt t$, from which we can easily find $\Pr(XY\le t)$.
If we do things right, we don't have to break up the integral. Integrate first with respect to $x$. The picture tells us that $x$ goes from $t/y$ to $y$. And then $y$ goes from $\sqrt{t}$ to $1$.
As a small bonus, the integration is marginally easier. An antiderivative is $3xy$, so the inner integral is $3y^2-3t$. Now it remains to find
$\int_{\sqrt{t}}^1 (3y^2-3t)\,dy$.
