I have to find the covariance of the following PDF
$$f(x,y)=\begin{cases} 3\min\{x,y\}&, \text{ if }\ 0<x,y<1 \\ 0 &, \text{ otherwise} \end{cases}$$
Therefore I need to find $E(X)$ and $E(Y)$.
I tried turning the PDF into this
$$f(x,y)=\begin{cases} 3x &, \text{ if }\ 0<x<y<1 \\ 3y &, \text{ if }\ 0<y<x<1 \\ 0 &, \text{ otherwise} \end{cases}$$
After I changed it I tried to find the marginal PDFs for $X$ and $Y$ and they look like this, but I don't think they are right. Is it okay? How do I keep going?
$$f(x)=\begin{cases} 3x(1-x)&, \text{ if }\ 0<x<y \\\left(\frac{3}{2}\right)x^2 &, \text{ if }\ y<x<1 \\ 0 &, \text{ otherwise} \end{cases}$$
$$f(y)=\begin{cases} \left(\frac{3}{2}\right)y^2 &, \text{ if }\ x<y<1 \\3y(1-y) &, \text{ if }\ y<x<1 \\ 0 &,\text{ otherwise} \end{cases}$$
The marginal densities are the following
$$f_X(x)=\int_0^x 3ydy+\int_x^1 3x dy=[\frac{3}{2}x^2+3x(1-x)]\mathbb{1}_{(0;1)}(x)$$
and similarly....
$$f_Y(y)=\int_0^y 3x dx+\int_y^1 3y dx=[\frac{3}{2}y^2+3y(1-y)]\mathbb{1}_{(0;1)}(y)$$
Now you can go on calculating $\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$
Note that the covariance is not "of the PDF" but it is "between X and Y"