Expectancy of a joint density A machine consists of two components, whose life times have the joint density function
$
f(x,y)=
\begin{cases}
1/50, & \text{for }x>0,y>0,x+y<10 \\
0, & \text{otherwise}
\end{cases}
$
The machine operates until both components fail.
Calculate the expected operational time of the machine.
With the boundaries I have found the vertices at (0,0) , (10,0) and (0,10) but I dont have the idea how to continue it seems like the answer is just 5. Hope anyone can give me a hand on this question. Thanks!
 A: For given $x,y>0$, the lifetime of the machine is $\max\{x,y\}$ because it stops when the two components fail. The life expectancy if therefore the expectancy of $\max\{x,y\}$ 
$$t=\mathbb{E}\left[\max\{x,y\}\right]=
\iint \max\{x,y\}f(x,y)\mathrm dx\mathrm dy.$$
At this moment you may want to draw a small graph:

(I have edited the end of the answer.)
In the white triangles we have $\max\{x,y\}=x$, and in the gray triangle  $\max\{x,y\}=y$. To compute the integral, it is useful to divide the domain into subdomains. 
The integral on the grey triangle is 
$$I_1=\frac1{50}\int_0^5\mathrm dx\int_x^{10-x}y\mathrm dy=\frac{1}{50}\int_0^5\frac12\left((100-x)^2-x^2\right)\mathrm dx=\frac52.$$
The integral on the first white triangle is
$$I_2=\frac1{50}\int_0^5\mathrm dx\int_0^xx\mathrm dy=\frac56.$$
The third integral is 
$$I_3=\frac{1}{50}\int_5^{10}\mathrm dx\int_0^{10-x}x\mathrm dy=\frac53.$$
So the total is $I_1+I_2+I_3=5$. Note that by symmetry you have $I_1=I_2+I_3$ which makes the computation much easier.
