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

I used the following approach :
Step 1 : Calculated the Probability distribution for $XY$. It turns out to be $P(XY \leq K) = \frac 1K$.  
Step 2 : Calculated the Probability distribution for $Z^2$. It turns out to be $P(Z^2 \leq L) = P(-\root \of{L} \leq Z \leq \root \of{L}) = \root \of{L}$.  
Step 3: Calculate the joint density function by multiplying the above functions and differentiating. I get $$f_{Q_1Q_2} (q_1,q_2) = \frac {-1}{2q_1^2\root\of{q_2}}$$. where $Q_1 = XY$ , $Q_2 =Z^2$.  
Step 4: Calculate the probability using the integral below. 
$$\int_{0}^{1}\int_{q_1}^{1}\frac{-1}{2q_1^2\root\of{q_2}}dq_2dq_1$$
There is definitely something wrong with this procedure. Any help will be appreciated.
 A: Alternative (avoiding calculation of CDF's and PDF's):
$$\begin{aligned}P\left(XY\leq Z^{2}\right) & =\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\left[xy\leq z^{2}\right]dzdxdy\\
 & =\int_{0}^{1}\int_{0}^{1}1-x^{\frac{1}{2}}y^{\frac{1}{2}}dxdy\\
 & =1-\int_{0}^{1}\int_{0}^{1}x^{\frac{1}{2}}y^{\frac{1}{2}}dxdy\\
 & =1-\int_{0}^{1}y^{\frac{1}{2}}\int_{0}^{1}x^{\frac{1}{2}}dxdy\\
 & =1-\int_{0}^{1}x^{\frac{1}{2}}dx\times\int_{0}^{1}y^{\frac{1}{2}}xdy\\
 & =1-\left(\int_{0}^{1}x^{\frac{1}{2}}dx\right)^{2}\\
 & =1-\left(\left[\frac{2}{3}x^{\frac{3}{2}}\right]_{0}^{1}\right)^{2}\\
 & =1-\left(\frac{2}{3}\right)^{2}\\
 & =\frac{5}{9}
\end{aligned}
$$
Here $\left[xy\leq z^{2}\right]$ denotes the function $[0,1]^3\to\mathbb R$ that takes value $1$ if $xy\leq z^{2}$ and takes value $0$ otherwise.
A: I think the CDF of $XY$ is not correct.
\begin{align}
P(XY \le K) &= 1-P(XY >K)\\
&= 1- \int_{K}^1\int_{\frac{K}{x}}^1 \, dy\, dx \\
&= 1-\int_K^1 (1-\frac{K}{x}) \, dx\\
&= 1-[x-K\ln x]_K^1 \\
&= 1-(1-K+K\ln K) \\
&= K - K \ln K
\end{align}
Note that density can't be negative in step $3$.
$$f_{Q_1,Q_2}(q_1, q_2)=\frac{-\ln q_1}{2\sqrt{q_2}}$$
\begin{align}&\int_0^1\int_{q_1}^1 -\frac{\ln q_1}{2\sqrt{q_2}}\, dq_2\, dq_1 = \frac59 \end{align}
Edit to check your integral:
\begin{align}
P(XY \le K) &= \int_K^1 \int_0^{\frac{K}{x}}\, dy \, dx + \int_0^K\int_0^1 \, dy \, dx\\
&=\int_K^1 \frac{K}{x}\, dx + K\\
&=K\ln x|_K^1+K\\
&=K-K\ln K
\end{align}
A: from Does the square of uniform distribution have density function?
we know that $Z\sim U[0,1] => f_{Z^2}(t) = \frac{1}{2\sqrt{t}}$
and from here product distribution of two uniform distribution, what about 3 or more
we know that $X\sim U[0,1],Y\sim U[0,1]\,iid => F_{XY}(t) = t-t\log t$
from Finding probability $P(X<Y)$
we know that $P(X<Y) = \int F_X(t) f_Y(t)dt$
$$P(XY<Z^2) = \int_0^1P(XY<t)f_{Z^2}(t)dt 
= \int_0^1(t - t\log(t))\frac{1}{2\sqrt{t}}dt = \frac{1}{2}\int_0^1 \sqrt{t} - \sqrt{t}\log t\,dt = (\frac{5}{9} t^{3/2} - t^{3/2} \log t) |_0^1 = \frac{5}{9}$$
