You seem to be a little confused about the likelihood function of the $U[0,\theta]$ model.
Let $f(x\mid\theta)=1/\theta$, for $0\leq x\leq\theta$, and $f(x\mid\theta)=0$, otherwise, for $\theta>0$.
For some fixed $x$, what is the graph of $f(x\mid\theta)$ as a function of $\theta$?
To draw the graph, notice -- and this is the key point -- that $x\in[0,\theta]$ if and only if $\theta\in[x,\infty)$.
So, using indicator functions, we have
$$f(x\mid\theta)=\frac{1}{\theta}I_{[0,\theta]}(x)=\frac{1}{\theta}I_{[x,\infty)}(\theta)\, .$$
After you understand this, just do the integration:
$$f(x)=\int_0^\infty f(x\mid\theta)\pi(\theta)\,d\theta = \int_x^\infty \frac{1}{\theta}\pi(\theta)\,d\theta\, .$$
From Sklar's Theorem, it follows that you can construct the joint distribution using a copula:
$$H(x,y) = C(F(x),G(y)).$$
So, you need two ingredients: the marginal distributions $(F,G)$, and the copula $C$. You mentioned that you know the marginals, so this ingredient is done. Now, you need information to construct the copula. So, if you cannot come up with enough information to select/estimate/guess/divine the copula, then you cannot construct the joint distribution.
Best Answer
As rightly pointed out by Dilip Sarwate, computing the expectation of one component as a two dimensional integral requires integrating out the other element of the vector: $$\mathbb{E}[X] = \iint x\,f_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx,$$ (which is a special case of the so-called law of the uncounscious statistician). The only simplifications I can think of is