Show that Cov(X,Y)=0 while X and Y are dependent Problem:
Show that $Cov(X,Y)=0$ while X and Y are dependent.$$f_{X,Y}(x,y)=\begin{cases}1 \ &\text{ for } -y<x<y,0<y<1\\0\ &\text{ otherwise } \end{cases}$$ 
Attempt:
Starting by drawing the domain for the joint density function I derive a triangle with corners in (0,0), (1,0) and (1,1). 
My solution strategy hence becomes to compute the corresponding marginal density functions from:
$$f_{X}(x)=\int_{\in D} f_{X,Y}(x,y) \text{ }dy $$ and to compute the covariance from:
$$Cov(X,Y)=E[XY]-E[X]E[Y].$$ The expected value for X is derived from:
$$E[X]=\int_{-\infty}^\infty x f_{X}(x) \text{ }dx $$ and E[Y] is similarly computed. Likewise
$$E[XY]=\int_{-\infty}^\infty\int_{-\infty}^\infty xy f_{X,Y}(x,y) \text{ }dxdy. $$ 
Lastly, dependent variables would not fulfil the independent criterion:
$$f_{X,Y}(x,y)=f_X(x)f_Y(y).$$
I suspect that my marginal density functions are incorrect since I fail to get the equality: 
$$E[XY]=E[X]E[Y]$$
Which in turn implies that I have misinterpreted the domain or the boundaries from which the marginal densities are computed. 
I am again solving this as an exercise in my probability course and any help would be greatly appreciated!
 A: The support comprises a triangle with vertices $(0,0)$, $(1,1)$, and $(-1,1)$.  Thus, the proper calculation should show that $\operatorname{E}[X] = 0$, $\operatorname{E}[Y] = 2/3$, and $$\operatorname{E}[XY] = \int_{y=0}^1 \int_{x=-y}^y xy \, dx \, dy = 0.$$  Thus $$\operatorname{Cov}[X,Y] = 0$$ as claimed, but clearly $X$ and $Y$ are not independent.
A: The support is the triangle $\triangle (0,0)(-1,1)(1,1)$ because, notice for any $y\in(0;1)$ that $x\in(-y;y)$.   Thus establishing the bounds: $$\mathsf E(g(X,Y)) =\int_0^1\int_{-y}^y g(x,y) \operatorname d x\operatorname d y$$
However, you can demonstrate the required result without integration.
The support is : $\{(x,y):−y<x<y,0<y<1\}$ and the density is uniform within the support, so we may use the known properties of uniform distributions.   Mainly: $$\begin{align}\because~&X\mid Y~\sim~\mathcal U(-Y;Y)\\[2ex]\therefore ~&\mathsf E(X\mid Y) ~=~ \dfrac{Y-(-Y)}{2}~=~0\end{align}$$
Now $\mathsf {Cov}(X,Y) = \mathsf E(Y\,\mathsf E(X\mid Y))- \mathsf E(Y)\,\mathsf E(\mathsf E(X\mid Y))$  by the tower property. (Law of Iterated Expectation). So $$\mathsf {Cov}(X,Y)=0$$
There is clearly a relation between $X,Y$, as the larger the value of $Y$ the wider the range of $X$, however the relation is not linear as $X$ is symmetrically distributed around zero regardless of the value of $Y$.
