The covariance between $X$ and $Y$. Suppose that $X$ and $Y$ are both continuous random variables that have a joint probability density that is uniform over the rectangle given by the four $(x,y)$ coordinates 
$(0,0)$ , $(2.46,0)$ , 
$(0,6.99)$ and $(2.46,6.99)$ , and which vanishes outside of this rectangle. 
What is the covariance between $X$ and $Y$? The answer is $0$ but I don't have idea how to calculate it.
I found out that the mean of $X$ is $1.23$ and the mean of Y is $3.495$  
The covariance$(X+Y) = E(XY) - (E(X)*E(Y))$
But how can i find the $E(XY)$?  
Thank you for helping me.
 A: Let the rectangle $K$ have corners at $(0,0)$, $(a,0)$, $(a,b)$, and $(0,b)$. Then 
$$E(XY)=\iint_K xy\,dy\,dx.$$
This can be expressed as the iterated integral
$$\int_{x=0}^a\left(\int_{y=0}^b \frac{xy}{ab}\,dy\right)\,dx.$$
Integrate. We get $\frac{ab}{4}$. (A "general" approach was used in order to prepare you for more complicated problems.) 
A: We're given that the joint density is uniform over that rectangle $\;R\;$, which means $\;f_{XY}(x,y)=C\;$ over $\;R\;$, zero outside of it, but then
$$1=\int\int_RC\;dA=CA\implies C=\frac1A\;,\;\;A:=\text{the area of}\;\;R$$
and then the marginal densities are
$$f_X(x)=\int_{R_y}\frac1Ady=\frac{\ell(y)}A\;,\;\;f_Y(y)=\int_{R_x}\frac1Adx=\frac{\ell(x)}A$$
with $\;\ell(x)\,,\,\,\ell(y)\;$ the lengths of the respective sides of the rectangle (observe the rectangle's side are parallel to the axis)
Finally, we get at once
$$f_X(x)f_Y(y)=\frac{\ell(x)\ell(y)}{A^2}=\frac1A=f_{XY}(x,y)\implies X,\,Y\;\;\text{are independent}$$
and thus
$$E(XY)=E(X)E(Y)\implies \text{Cov}(X,Y)=0$$
