If $X \sim R(0,a), Y \sim R(0,b)$ both independent and $a>b$, find the distribution of $XY$ geometrically. If $X \sim R(0,a), Y \sim R(0,b)$ both independent and $a>b$, find the distribution of $XY$ geometrically.
I took $Z=XY \\$
$P(0<Z<ab)=1 \\$
Now, $F_Z(z)=P(Z \le z)=P(XY \le z)=P(Y \le \frac{z}{X})$
Now, will there be two cases?
(i)$0<z<\sqrt{a-b} \\$
(ii)$\sqrt{a-b} <z<ab$?
If a draw the sample space it would be a rectangle and the curve would be a hyperbola.
 A: In the topic question it is requested to fing distribution geometrically. Note that the pair $(X,Y)$ is the point choosen at random in the rectangle $[0,a]\times[0,b]$ in $\mathbb R^2$. Draw this rectangle and the region $y<z/x$ below the curve: 

The CDF of $Z=XY$ is $$F_Z(z)=\mathbb P(Z\leq z)=\mathbb P((X,Y)\in D)$$
where $0<z<ab$ and $D$ is the filled region. 
To find probability geometrically we should find the area of this region and divide it to total area of the rectangle:
$$
F_Z(z)=\mathbb P((X,Y)\in D) = \frac1{ab}\left(\frac{z}{b}\cdot b+\int\limits_{z/b}^a \frac{z}{x}\,dx\right)=\frac{z-z\ln\left(\frac{z}{ab}\right)}{ab}.
$$
Finally, the CDF is
$$
F_Z(z) = \begin{cases} 0, & z\leq 0 \cr \frac{z-z\ln\left(\frac{z}{ab}\right)}{ab}, & 0<z<ab \cr 1 & z \geq ab
\end{cases}
$$
and the pdf of $Z$ is 
$$
f_Z(z) = \begin{cases} 0, & z\not \in (0,ab) \cr \frac{-\ln\left(\frac{z}{ab}\right)}{ab}, & 0<z<ab 
\end{cases}
$$
A: I am not sure what you mean by finding the distribution geometrically, but the standard derivation would be to use the probability density functions.
If $X \sim \text{Unif}(0,a), \, Y \sim \text{Unif}(0,b)$ are uniformly distributed then they respectively have probability density functions
$$ f_X(x) = \mathbf{1}_{[0,a]}(x), \qquad \qquad f_Y(x) = \mathbf{1}_{[0,b]}(x),$$
where I use $\mathbf 1_{A}$ to denote the indicator function that equals one for values in the set $A$ and is $0$ elsewhere.
Given that $X,Y$ are assumed to be independent, then we can apply the formula for the pdf of the product of independent variables $Z = XY$ given to be
$$ f_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z/x) \frac1{|x|} dx.$$
In particular we have
\begin{align*}
f_Z(z) & = \int_{-\infty}^\infty \mathbf{1}_{[0,a]}(x)\mathbf{1}_{[0,b]}(z/x) \frac{1}{|x|} dx \\
& = \int_{0}^a \mathbf{1}_{[0,b]}(z/x) \frac{1}{|x|} dx \\
& = \int_{z/b}^a \frac{1}{|x|} dx \\
& = \big[ \log(x) \,\big]^{a}_{z/b} \\
& = \log(a) + \log(b) - \log(z).
\end{align*} 
Note that to get the third line we used that the pdf of $Y$ is only non zero for $0 < z/x < b$, i.e. $x > z/b$. So we have
$$f_Z(z) = \begin{cases}
- \log\left( \frac{z}{ab} \right) & \text{if $0 < z \leq ab$,} \\
0 & \text{else.}
\end{cases}
$$
So far as I'm aware this distribution does not have a standard name.
