The question has evolved in the comments, and now may be asking the following. Let $J$ be the minimum of $X$ and $Y$. What is the variance of $J$? We assume that the joint density function of $X$ and $Y$ is $f(x,y)$.
An answer goes as follows. Let $m(x,y)$ be the minimum of $x$ and $y$. Then
As for the variance, it is $E(J^2)-(E(J))^2$, and
In evaluating the integrals, we probably will want to use the following strategy, which we illustrate with the integral for the mean of $J$. Divide the plane into two parts, the part below $y=x$ and the part above. Then our integral is the sum of the integrals over the two parts.
In the part with $y\lt x$, we have $m(x,y)=y$. So our integral over this part is
The integral over the part where $x\lt y$ is obtained in the same way, except for some minor changes. It is
The integral for calculating $E(J^2)$ can be broken up in exactly the same way. Instead of integrating $yf(x,y)$ or $xf(x,y)$ over suitable regions, we will be integrating $y^2f(x,y)$ and $x^2f(x,y)$ over the same regions.