> Let $X$, $Y$, and $Z$ be three independent uniform random variables on $[0, 1]$. Compute the probability $P(XY < Z^2)$. 
Let $X$, $Y$, and $Z$ be three independent uniform random variables on $[0, 1]$. Compute the probability $P(XY < Z^2)$.

Here is what I've done:
\begin{align}
 P(XY<Z^2) &= \int_{-\infty}^{\infty} P(XY<Z^2|Z=z)f_z(z)dz \\
 &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} P(XY<z^2|Y=y)f_y(y)f_z(z)dzdy \\
 &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} P(X<\frac{z^2}{y})f_y(y)f_z(z)dzdy\\
 &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} F_X(\frac{z^2}{y})dzdy
\end{align}
since $f_y(y)=f_z(z)=1$ and $F_X(x)=x$, we get
$$\int_{0}^{1}\int_{0}^{1}\frac{z^2}{y}dzdy$$
There must be something wrong...
Can anyone help me with this?
 A: Nice plan, but try it while leaving  $Z$ till last, since that is where the square is.  This also avoids having any inverses in the probability statement.
$$\begin{split}\mathsf P(XY\leq Z^2) &=\int_0^1\int_0^1\mathsf P(Z^2\geq xy)~\mathsf d x~\mathsf d y\\ & = \int_0^1\int_0^1 (1-\sqrt{xy})~\mathsf d x~\mathsf dy\\&~~\vdots\end{split}$$
It works.

PS: Your problem was that you should have used $\mathsf P(X\leq z^2/y) = \min(z^2/y, 1)$ not just $z^2/y$.  You cannot have a proability excceeding 1 which our formula would have given when $y<z^2$.
A: Your missing the point, that $\forall y \notin [0,1] f_y(y)=0$ and for $x$ too.
I get \begin{align}
P(XY<Z^2)
&=\int_0^1 \int_0^1 \int_0^1 1_{xy<z^2}dx dy dz\\
&=\int_0^1 \int_0^1 \int_0^1 1_{x<z^2/y}dx dy dz\\
&=\int_0^1 \int_0^1 \int_0^{\min(z^2/y,1)}dx dy dz\\
&=\int_0^1 \int_0^1 \min(z^2/y,1) dy dz\\
&=\int_0^1 \int_0^{z^2} \min(z^2/y,1) dy + \int_{z^2}^1 \min(z^2/y,1) dy dz\\
&=\int_0^1 \int_0^{z^2} 1 dy + \int_{z^2}^1 z^2/y dy dz\\
&=\int_0^1 z^2 + z^2 (-\ln(z^2)) dz\\
&=\int_0^1 z^2(1-\ln(z^2)) dz\\
&=[(5 z^3)/9 - 1/3 z^3 \ln(z^2)]_0^1\\&=5/9
\end{align}
I did the last integral with Wolframalpha
