Q:

Calculating the probability of $ (Z-1)^2 \leq XY$

We randomly choose three numbers $X, Y, Z$ $\in [0,1]$. Calculate the probability that $ (Z-1)^2 \leq XY$.
I have tried to observe just the "edge" i.e. $ (Z-1)^2  = XY$ but I am pretty much stuck on calculating the boundaries for double integral. Any hint helps!

A:

Since $(Z-1)^2\le XY$ is equivalent to $1-Z\le \sqrt{XY}$ or $1-\sqrt{XY}\le Z$,
it follows that
$$\eqalign{\mathbb{P}((Z-1)^2\le XY)&=\mathbb{P}(1-\sqrt{XY}\le Z)\cr
&=\int_{x=0}^1\int_{y=0}^1\int_{z=1-\sqrt{xy}}^1dzdydx\cr
&=\int_{x=0}^1\int_{y=0}^1\sqrt{xy}dydx\cr
&=\int_{x=0}^1\sqrt{x}dx\int_{y=0}^1\sqrt{y}dy=\frac{4}{9}.
}$$

