probability - Calculating the expected value of a random variable that's a function of a random variable

I am working on the following problem:

I'm having a hard time putting all of this information together:

The cost of the maintenance is $Z = X + Y$, where $X$ is the cost of the first machine and $Y$ is the cost of the second machine.

The payment $T$ by the insurer is $Z$ if $0 \leq Z \leq 6000$ and $6000$ if $Z > 6000$.

$$T(Z) = \left\{ \begin{array}{lr} Z & \text{if } 0 \leq Z \leq 6000\\ 6000 & \text{if } Z > 6000 \end{array} \right.$$

We want to find $E[T(Z)] = T(Z)\cdot Pr[T(Z)].$

From here I am uncertain about how to proceed. I know that a uniform random variable on $[0, 4000]$ has expected value $\frac{4000}{2} = 2000$. This has a probability of $\frac{1}{3}$ of occurring.

I would think $E[X] = E[Y] = \frac{4000}{6}$, and so $E[Z] = \frac{4000}{3}$.

I am just unsure of how to factor in what the insurer is actually paying.

Answer

The probability of $(0,0)$ occurring is $\frac23\cdot\frac23$; the probability of $(0,y)$ occurring, for $0 $$\frac49\cdot0 + \int_0^{4000} \frac{y\,dy}{18000} + \int_0^{4000} \frac{x\,dx}{18000} + \int_0^{4000} \int_0^{4000} \frac{\min\{x+y,6000\}\,dx\,dy}{12000^2}.$$
Solving this calculus problem (splitting the last integral at the line $x+y=6000$) gives the answer $\frac{35750}{27} \approx 1324.07$.