As noted by @YashaswiMohanty the expectation of $Y$ can sometimes
be found without explicting the probability distribution function.
Assume that all r.vs are of continuous type and $X_1$ is exponential
with rate $\lambda >0$. We can consider the survival function
$\bar{F}_Y(y) := 1 - F_Y(y)$
$$
\bar{F}_Y(y) = \text{Pr}\{ \min(Z_1,\,Z_2) > y\}
=\text{Pr}\{[Z_1 > y] \cap [Z_2
> y] \}.
$$
Then by conditioning on $X_1$ we can use the independence
\begin{align*}
\bar{F}_Y(y)
&=\int_0^\infty \text{Pr}\{[Z_1 > y] \cap [Z_2
> y] \, \vert \, X_1 = x_1\} f_{X_1}(x_1) \,\text{d}x_1\\
&= \int_0^\infty \text{Pr}\{[U_1 > y + x_1] \cap [U_2
> y + x_1] \, \vert \, X_1 = x_1\} \, f_{X_1}(x_1) \,\text{d}x_1\\
&= \int_0^\infty \bar{F}_U(y + x_1)^2 \lambda \, e^{-\lambda x_1}\, \text{d}x_1
\end{align*}
There are some cases where we can get a closed form expression. For
instance if $U_i$ are exponential with rate $\gamma$ i.e.,
$\bar{F}_U(u) = e^{-\gamma u}$ for $u >0$.
Interestingly, this is a simple and efficient way to generate a couple
of random variables with tail dependence.