Expected value of a marginal distribution is a function of $x$? 
Let $f$ be the joint PDF of the random vector $(X, Y)$.
$f(x, y) = \displaystyle\frac{x(x-y)}{8}$ for $0 < x < 2$ and $-x < y < x$, otherwise it's zero.
Calculate the correlation between $X$ and $Y$.

The problem I'm struggling with here is that to compute the correlation, I need $E(X)$, $E(Y)$, $E(XY)$, etc.
I already calculated the $E(X)$, but when trying to calculate $E(Y)$, I get:
$E(Y) = \displaystyle\int_{-x}^xyf_y(y)dy$
Where $f_y(y)$ is the marginal distribution of y, which I computed as:
$f_y(y) = \displaystyle\int_0^2\frac{x(x-y)}{8}dx = \frac{1}{3}-\frac{1}{4}y$
Because of the limits of integration when computing $E(Y)$, I assumed that the integral was going to be in terms of $x$, and in fact, I computed it and got $E(Y) = -\displaystyle\frac{x^3}{6}$.
Why is my expected value in terms of $x$? Is this the proper way to calculate this? Shouldn't be Y a random variable of it's own?
 A: No, indeed the expectation of $Y$ should not be a function over $X$ (nor over $x$).   It shall be a constant.
Be careful.   You are using the support for the conditional pdf for $Y$ given $X=x$, not that for the marginal pdf for $Y$.
Examine the support for the joint pdf:
Because $\{(x,y): 0\leqslant x\leqslant 2, -x\leqslant y\leqslant x\}=\{(x,y):-2\leqslant y\leqslant 2, \lvert y\rvert\leqslant x\leqslant 2\}$
Therefore, by Fubini's Theorem : $\quad\displaystyle\int_0^2 \int_{-x}^x f_{X,Y}(x,y)\,\mathrm d y\,\mathrm d x ~=~ \int_{-2}^2\int_{\lvert y\rvert}^2 f_{X,Y}(x,y)\,\mathrm d x\,\mathrm d y $
And here the inner integrals represent the marginal pdf for the relevant random variable, and the bound on the outer integrals cover their support.
So we have: $\displaystyle f_X(x) ~{= \mathbf 1_{(0\leqslant x\leqslant 2)} \int_{-x}^x \tfrac{x(x-y)}8\mathrm d y \\ = \tfrac 14{x^3} \mathbf 1_{(0\leqslant x\leqslant 2)}}\\ \displaystyle f_Y(y) ~{= \mathbf 1_{(-2\leqslant y\leqslant 2)}\int_{\lvert y\rvert}^2 \tfrac{x(x-y)}8\mathrm d x \\ = \tfrac 1{48} (- 2 \lvert y\rvert^3 + 3 y^3 - 12 y + 16)\mathbf 1_{(-2\leqslant y\leqslant 2)}}$
And hence: $\displaystyle \mathsf E(Y) = \int_{-2}^2 y\,f_Y(y)\,\mathrm d y\\ \displaystyle \mathsf E(X) = \int_0^2 x\,f_X(x)\mathrm d x$

Although it mayhap be easier to avoid the absolute values and use: $$\mathsf E(Y)=\int_0^2 \int_{-x}^x y\,f_{X,Y}(x,y)\,\mathrm d y\,\mathrm d x \\ \mathsf E(X)=\int_0^2 x \int_{-x}^x f_{X,Y}(x,y)\,\mathrm d y\,\mathrm d x \\ \mathsf E(XY)=\int_0^2 x\int_{-x}^x y\,f_{X,Y}(x,y)\,\mathrm d y\,\mathrm d x$$
A: Note that we can calculate $E(Y)$ directly from the original joint density function, without the use of marginal density functions. 
$$\int_0^2 \int_{-x}^x y \cdot f(x,y) \text{ }dy \text{ } dx  $$
This will eliminate any $x$ terms in the final answer of $E(Y)$.
