Distribution of $Z=\min(X,Y)$ for given couple of random variables 
Let X,Y be continuous random variables with distribution function $f(x,y)=\frac {(x+2y)e^{-x-y}} 3 (x,y>0)$.
a)Calculate expected value of XY
b) Assume we define $Z=min(x,y)$ find its distribution. Can its variance and expected value can be calculated without calculating the distribution?

About A: Can I calculate expected value by simply calculating $\iint_{\mathbb R^2} xyf(x,y)dxdy$?
About b:Assume define Z to be $Z(x,y)=\begin{cases} x &x<y \\y&y<x\end{cases}$ I don't know how to calculate its distribution stright. In the lecture we mentioned we can do so by using the formula $\mathbb E[g(x,y)]=\displaystyle\iint_{\mathbb R^2}g(x,y)f(x,y)dxdx$. Except using the  aforementioned formula how can I get the distribution of Z?
 A: 
Can its variance and expected value can be calculated without calculating the distribution?

Consider some independent random variables $U$ and $V$ with densities $u\mathrm e^{-u}\mathbf 1_{u\geqslant0}$ and $\mathrm e^{-v}\mathbf 1_{v\geqslant0}$ and define $(X,Y)$ by $(X,Y)=(U,V)$ with probability $\frac13$ and $(X,Y)=(V,U)$ with probability $\frac23$, independently of $(U,V)$. Then $(X,Y)$ has density $f$, hence $Z$ is distributed like 
$$
T=\min(U,V).
$$
Since $[T\gt z]=[U\gt z]\cap[V\gt z]$ for every $z\geqslant0$, by independence of $(U,V)$, 
$$
P[Z\gt z]=P[T\gt z]=P[U\gt z]\cdot P[V\gt z].
$$
This yields the value of $\bar F(z)=P[Z\gt z]$ since one knows that $P[U\gt z]=(z+1)\mathrm e^{-z}$ and $P[V\gt z]=\mathrm e^{-z}$. Now, the identities
$$
E[Z]=\int_0^\infty\bar F(z)\mathrm dz,\qquad E[Z^2]=\int_0^\infty2z\bar F(z)\mathrm dz,
$$
valid for every nonnegative random variable $Z$, yield the mean and variance of $Z$.

Calculate expected value of $XY$.

The construction above shows that $E[XY]=E[UV]=E[U]\cdot E[V]=2\cdot1$. This allows to check the result of the more direct computation
$$
E[XY]=\iint xyf(x,y)\mathrm dx\mathrm dy=\int_0^\infty\int_0^\infty xy\tfrac13(x+2y)\mathrm e^{-x-y}\mathrm dx\mathrm dy.
$$ 
Edit: More generally, choose two PDFs $f_1$ and $f_2$ and assume that $(X,Y)$ has density $f$ where, for some $p$ in $[0,1]$,
$$
f(x,y)=pf_1(x)f_2(y)+(1-p)f_1(y)f_2(x).
$$ 
Then $Z=\min(X,Y)$ is distributed like $\min(U,V)$ where $U$ and $V$ are independent with respective PDFs $f_1$ and $f_2$. In particular the complementary CDF of $Z$ is $\bar F=\bar F_1\cdot\bar F_2$.
A: You are right about how to find $E(XY)$. The integration procedure is integration by parts, not very pleasant.
To find the density function of $Z$, we find the cumulative distribution function of $Z$, and then differentiate. Note that $Z\gt z$ precisely if $X\gt z$ and $Y\gt z$.
The cdf of $Z$ is $1-\Pr(Z\gt z)$.
To find the probability that $X\gt z$ and $Y\gt z$,  integrate the joint density function of $X$ and $Y$ over the region $x\gt z, y\gt z$.  The integration is quite doable. 
Finally, differentiate the cdf to find the density function of $Z$. If one has enough experience with differentiation under the integral sign, some of the integration can be avoided.
As to finding the mean of $Z$ without the distribution, undoubtedly the answer is yes. I  have not written it out.
