Given that $X,Y,Z$ are independent $U(0, 1)$ distribution, find $\Pr(XY/Z \le t)$.
For $t \ge 1$,
$$\Pr(XY/Z \le t) = \int_V f(x,y,z) dzdxdy$$
Where $V$ is the volume enclosed by $XY/Z \le t$ and the unit cube in the first quadrant.
Since $X, Y$ and $Z$ are independent, $f(x,y,z) = 1$ for $(x,y,z) \in[0,1]^3$.
I can write the expression as $xy \le zt$, so the limits for $z$ in the integral is $xy/t \le z \le 1$. By looking at the graph of $xy\le zt$ for $z=1$, gives $0\le x,y\le 1$.
$$\Pr(XY/Z \le t) = \int_0^1\int_0^1\int_{xy/t}^1 1 dzdxdy = 1- \dfrac1{4t}$$
For $t \in (0, 1)$,
Integration limits for $z$ are still $xy/t \le z \le 1$ and by looking at the graph of $xy \le t$, shows $0 \le x \le 1$ for $0 \le y \le t$ and $0 \le x \le t/y$ for $t \le y \le 1$
$$\Pr(XY/Z \le t) = \int_0^t\int_0^1\int_{xy/t}^1 1 dzdxdy + \int_t^1\int_0^{t/y}\int_{xy/t}^1 1 dzdxdy = \dfrac{3t}{4} - \dfrac{t\ln(t)}{2}$$
The solution given is,
$$\Pr(XY/Z \le t) =\begin{cases} 1 - \dfrac{t}{4}& t \in (0, 1] \\ t - \dfrac{1}{4t} + \dfrac{1}{2t} \ln(t) &t \gt 1\end{cases}$$
Are my integration limits wrong? or my expression for $f(x,y,z)$ is wrong ?
Let $X,Y,Z \sim \mathcal U((0,1))$. Let $V = \frac{XY}{Z}$. We want to compute $F_V$ - CDF function of r.v $V$. Note that $V > 0$ so $F_V(t) = 0$ for $t \le 0$. Now take any $t > 1$, we have: $F_V(t) = \mathbb P(XY \le tZ) = \mathbb P((X,Y,Z) \in A_t )$, where $A_t = \{ (x,y,z) \in (0,1)^3 : xy \le zt \}$. Then $$F_V(t) = \int_{A_t} \chi_{(0,1)^3}(x,y,z) d\lambda_3(x,y,z)= \int_0^1 \int_0^{1} \int_{\frac{xy}{t}}^1 dzdydx = \int_0^1 \int_0^{1}(1-\frac{xy}{t})dydx = \int_0^1 (1 - \frac{x}{2t})dx = 1 -\frac{1}{4t}$$
And for $t \in (0,1)$ we get: $$F_V(t) = \int_0^t \int_0^1 \int_\frac{yx}{t}^1dzdydx + \int_{t}^1 \int_0^{\frac{t}{x}} \int_{\frac{xy}{t}}^1 dzdydx = \int_0^t (1 - \frac{x}{2t})dx + \int_t^1 (\frac{t}{x} - \frac{t}{2x})dx = t - \frac{1}{4}t + \frac{t}{2}\int_t^1 \frac{1}{x}dx = \frac{3}{4}t - \frac{t}{2}\ln(t) = \frac{t}{4}(3 - 2\ln(t))$$
So I'd say your computation is correct. The solution cannot be correct, since it isn't even bounded by $1$.