Random variables with joint density function 
Let R be the rectangle $\ \{(x, y); 0 <= x <= 2, 0 <= y<=  1\} $, and let $\ f(x, y) =
>k(x^2+ y^2)$ on R and zero elsewhere.
(a) Find the value of k which makes f a joint density function.
(b) If X and Y are random variables with joint density function
  f(x, y), find
i. the marginal distributions of X and Y ;
ii. the expectations and variances of X and Y ;
iii. the covariance and correlation of X and Y .

For a) I set the double integral equal to 1 and got k = $\ 3\over10 $
for b)I) I got:
$\ f_X(x) = $$\ 3\over 10$$\ (x^2+ $ $1 \over 3$$)$ and 
$\ f_Y(y) = $$\ 3\over 10$$\ ($$8 \over 3$$+ 2y^2)$
I'm not sure about the rest. For b) II) I'm thinking it might be:
$E(X) = \int xf_X(x)dx$, $\int E(Y) = yf_Y(y)dy$
and
$Var(X) = \int (x-E(X))^2f_X(x)dx$, $Var(Y) = \int (y-E(Y))^2f_Y(y)dy$
I've no idea about covariance or correlation yet. I'll cross that bridge when I come to it.
Am I on the right track?
Thanks
 A: It´s all good. However for calculating the variance I do suggest using the following identity, which is easily derived($\mu_x$ is $E(X)$, in case it is not clear):
$$\begin{align}
var(X) &= E\left((X-\mu_x)^2\right)\\
&= E\left(X^2 - 2X\mu_x + \mu_x^2\right)\\
&= E(X^2) - E(2X\mu_x) + E(\mu_x^2)\\
&= E(X^2) - 2\mu_xE(X) + \mu_x^2\\
&= E(X^2)-2\mu_x^2+\mu_x^2\\
&=E(X^2)-\mu_x^2 \\
\end{align}
$$
You already have $\mu_x$, but you still have to solve an integral here for finding $E(X^2)$. However, it is in general easier than directly finding $E((X-\mu_x)^2)$. In any case, it's better if you find the variance both ways a few times, so you convince yourself the identity makes life easier most times.
For covariance and correlation there is not much of a bridge to cross. Covariance is just $E((X-E(X)(Y-E(Y))$ Again in this case, avoid solving the integral directly, as this expected value can be easily split in easier terms and leads to the following identity:
$$\begin{align}
cov(X,Y) &= E((X-\mu_x)(Y-\mu_y) \\
&= E(XY - X\mu_y - Y\mu_x + \mu_y\mu_y) \\
&= E(XY) - E(X\mu_y) - E(Y\mu_x) + E(\mu_x\mu_y)\\
&= E(XY) - \mu_yE(X) - \mu_xE(Y) + \mu_x\mu_y\\
&= E(XY) - \mu_x\mu_y - \mu_x\mu_y + \mu_x\mu_y\\
&= E(XY) - \mu_x\mu_y\\
\end{align}
$$
If you are taking a course on probability you may want to memorize the previous identity also. 
Then the correlation is simply:
$$
\rho(X,Y)=\frac{cov(X,Y)}{\mu_x\mu_y}
$$
