$X$ and $Y$ are $\text{unif}(0,1)$. Find the Joint Distribution of $S$ and $U$, where $S=X+Y$, $U=\min(X,Y)$.
I know the density of $S$ is $f(s)=s$ for $0<s<1$, $f(s)=2-s$ for $1<s<2$, and $0$ otherwise. The density of $U$ is $f(u)=2-2u$, for $0<u<1$. How to find the joint distribution $f(s,u)$?
As I stated in my comment, without assuming independence between $X$ and $Y$ the problem cannot be solved.
Assuming independence and using the jacobian method, easy find that
$$f_{US}(u,s)=2\cdot \mathbb{1}_{(0;1)}(u)\cdot \mathbb{1}_{(2u;1+u)}(s)$$
that is defined in the following purple area
Just for a double check, you can derive the marginals
$$f_U(u)=2\int_{2u}^{u+1}ds=2(1-u)$$
$$f_S(s)=2\int_{0}^{\frac{s}{2}}du=s$$
for $0<s<1$
and
$$f_S(s)=2\int_{s-1}^{\frac{s}{2}}du=2-s$$
for $1<s<2$
as already known
The easiest way to derive the joint distribution of $(S,U)$ is the following.
Let's avoid to assume independence between $X,Y$ but state that $X<Y$ now $f(x,y)=2$, the jacobian is 1 and the $f(s,u)$ is again uniform in the support I showed. Note that the support area is 0.5, so perfectly coherent with the joint density =2.