# Probability of rectangular area formed by random point $Q$ and the $x$, $y$ axis

Given a point $$Q = (X, Y)$$ which is uniformly randomly chosen within the range of $$0 \leq X, Y \leq 1$$, it forms a rectangle with both the $$x, y$$ axis whose area is $$A = XY$$. For some $$r$$ such that $$0 \leq r \leq 1$$, find the probability that $$A \leq r$$.

My thoughts: Since the point is randomly chosen, $$\mathrm{P}(A \leq r)$$ is simply the area under $$xy = r$$, or $$y = \frac{r}{x}$$ within the range, divided by the area of the whole range, which is $$1$$.

To calculate the area, I tried to divide the area into two pieces: the rectangle whose upper-right corner is point $$M$$, and the area at the right.

The left-hand side is easy: since point $$M$$ should be at where equations $$\begin{cases}y = 1 \\ xy = r\end{cases}$$ both is satisfied, the area should be $$r$$.

The right-hand side, i think, should be the result of $$\displaystyle\int^{1}_{r}\frac{r}{x}dx = -r \ln{(r)}$$.

So $$\mathrm{P}(A \leq r) = \frac{r -r \ln{(r)}}{1} = r -r \ln{(r)}$$.

I think I got it right, but without an answer, I couldn't confirm. Also, I wonder is there any better approach. Anything helps, and thanks in advance.