Joint density function of $T_1,T_2$ and expectation of $E[T_1 ^2 +T_2 ^2 ]$ 
Given that $T_1,T_2$ are random variables representing the useful life (in
  hours) of two electrical appliance.

The joint probability function of two variables distributed uniformly
in the domain of:
$$0\leq t_1\leq t_2 \leq L$$ ($L$ a positive  constant )
I need to prove that the Joint density function of $T_1,T_2$ is equal to  $\frac{2}{L^2}$.
And to calculate the expectation of $$E\left[T_1 ^2 +T_2 ^2 \right]$$
I'd like to know how to begin with this question?
Thank you.
 A: $$
\operatorname{E}(T_1^2+T_2^2) = \iint\limits_\text{domain} (t_1^2 + t_2^2) f(t_1,t_2)\, d(t_1,t_2)
$$
where $f$ is the joint density function.  To say that the pair $(T_1,T_2)$ is uniformly distirbuted in the domain means that $f$ is constant in the domain.  Thus the expected value is
$$
\operatorname{E}(T_1^2+T_2^2) = \iint\limits_\text{domain} (t_1^2 + t_2^2)  c \, d(t_1,t_2)
$$
where the constant $c$ must be so chosen that
$$
\iint\limits_\text{domain} c\,d(t_1,t_2) = 1.
$$
The domain is defined by $$0 \le t_1 \le t_2 \le L.$$  Either of the double integrals above can be written as an iterated integral in either of two ways.


*

*First look at $$ \int_0^L \cdots \, dt_1 $$ i.e. $t_1$ goes from $0$ to $L$.  Then for any particualr value of $t_1$ between $0$ and $L$, the other variable $t_2$ goes from $t_1$ to $L$, so we have $$ \int_0^L \left( \int_{t_1}^L \cdots \, dt_2  \right) \, dt_1. $$

*Now look at $$ \int_0^L \cdots \, dt_2 $$ since $t_2$ goes from $0$ to $L$, but then for any fixed value of $t_2$, the other variable $t_1$ goes from $0$ to $t_2$, and we have $$ \int_0^L \left( \int_0^{t_2} \cdots \, dt_1 \right) \, dt_2 $$


Either approach will give the same answer.
