Given joint density $f_{X,Y}(x,y)$ find covariance, correlation, and specific expectations 
Given the joint density of $X$ and $Y$,
$$f_{X,Y}(x,y)=\begin{cases}e^{-y}&\text{for }0\le x<y<\infty\\0&\text{otherwise}\end{cases},$$
(a) find the covariance and correlation of $X$ and $Y$;
(b) find $E[X\mid Y=y]$ and $E[Y\mid X=x]$; and
(c) find $E[X]$ and $\mathrm{Var}[X]$.

I know that
$$\mathrm{Cov}[X,Y]=E[XY]-E[X]E[Y],$$
$$\mathrm{Corr}[X,Y]=\frac{\mathrm{Cov}[X,Y]}{\sqrt{\mathrm{Var}[X]\mathrm{Var}[Y]}},\text{ and}$$
$$E[XY]=\int_0^\infty\int_0^yxye^{-y}\,\mathrm dx\,\mathrm dy=3,$$
so theoretically, I should be able to compute the covariance by first finding the marginal densities for $X$ and $Y$, then computing the expectations of $X$ and $Y$. Alternatively, since
$$E[X]=E[E[X\mid Y=y]],$$
I can accomplish the same thing by finding the conditional densities and their expectations.
However, the ordering of the questions has me wondering:

(1) Can I compute the covariance and correlation directly knowing just the joint PDF?

As for finding the (necessary?) expectations, I've run into a complication...
The marginal density for $Y$ is
$$f_Y(y)=\int_0^ye^{-y}\,\mathrm dx=xe^{-y}\bigg|_0^y=\begin{cases}ye^{-y}&\text{for }y\ge0\\0&\text{otherwise}\end{cases}$$
so the conditional density of $X$ given $Y=y$ is
$$f_{X\mid Y}(x\mid y)=\begin{cases}\frac{e^{-y}}{ye^{-y}}=\frac1y&\text{for }0\le x<y<\infty\\0&\text{otherwise}\end{cases}$$
Then the conditional expectation of $X$ given $Y=y$ is
$$E[X\mid Y=y]=\int_0^y\frac xy\,\mathrm dx=\frac{x^2}{2y}\bigg|_0^y=\frac y2$$
and so
$$\boxed{E[X]=E[E[X\mid Y=y]]=E\left[\frac y2\right]=\frac y2}$$
Similarly, the marginal density for $X$ is
$$f_X(x)=\int_x^\infty e^{-y}\,\mathrm dy=-e^{-y}\bigg|_x^\infty=\begin{cases}e^{-x}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}$$
so the conditional density of $Y$ given $X=x$ is
$$f_{Y\mid X}(y\mid x)=\begin{cases}\frac{e^{-y}}{e^{-x}}=e^{x-y}&\text{for }0\le x<y<\infty\\0&\text{otherwise}\end{cases}$$
and the expectation of $Y$ given $X=x$ would be
$$E[Y\mid X=x]=\int_x^\infty ye^{x-y}\,\mathrm dy=-(y+1)e^{x-y}\bigg|_x^\infty=x+1$$
and so the expectation of $Y$ is
$$\boxed{E[Y]=E[E[Y\mid X=x]]=E[x+1]=x+1}$$
The complication is that I get different expectations for $X$ and $Y$ when trying to verify their values using the corresponding marginal densities:
$$\boxed{E[X]=\int_0^\infty xe^{-x}\,\mathrm dx=1\\E[Y]=\int_0^\infty y^2e^{-y}\,\mathrm dy=2}$$
I thought that perhaps the problem lies with the supports of $f_X$ and $f_Y$. We have $0\le x<y<\infty$ to start with, so the marginal densities could instead be
$$f_X(x)=\begin{cases}e^{-x}&\text{for }0\le x<y\\0&\text{otherwise}\end{cases}$$
$$f_Y(y)=\begin{cases}ye^{-y}&\text{for }x\le y\\0&\text{otherwise}\end{cases}$$
But even then,
$$\boxed{E[X]=\int_0^yxe^{-x}\,\mathrm dx=1-(y+1)e^{-y}\neq\frac y2\\E[Y]=\int_x^\infty y^2e^{-y}\,\mathrm dy=(x^2+2x+2)e^{-x}\neq x+1}$$

(2) Which expectations are correct? Why the discrepancy between methods?

 A: Thanks in large part to the commenters for getting me back on the right track. Posting the key points of my solution here for completeness.
$$E[XY]=\int_0^\infty\int_0^yxye^{-y}\,\mathrm dx\,\mathrm dy=3$$
$$E[X]=\int_0^\infty\int_0^yxe^{-y}\,\mathrm dx\,\mathrm dy=1$$
$$E[Y]=\int_0^\infty\int_0^yye^{-y}\,\mathrm dx\,\mathrm dy=2$$
$$E[X^2]=\int_0^\infty\int_0^yx^2e^{-y}\,\mathrm dx\,\mathrm dy=2$$
$$E[Y^2]=\int_0^\infty\int_0^yy^2e^{-y}\,\mathrm dx\,\mathrm dy=6$$
$$\boxed{\mathrm{Cov}[X,Y]=E[XY]-E[X]E[Y]=1}$$
$$\boxed{\mathrm{Corr}[X,Y]=\frac{E[XY]-E[X]E[Y]}{\sqrt{(E[X^2]-E[X]^2)(E[Y^2]-E[Y]^2)}}=\frac1{\sqrt2}}$$
$$\boxed{E[X\mid Y=y]=\int_0^y\frac xy\,\mathrm dx=\frac y2}$$
$$\boxed{E[Y\mid X=x]=\int_x^\infty y^2e^{x-y}\,\mathrm dy=x+1}$$
$$\begin{cases}
E[X]=E[E[X\mid Y]]=E\left[\frac Y2\right]=\frac{E[Y]}2\\[1ex]
E[Y]=E[E[Y\mid X]]=E[X+1]=E[X]+1
\end{cases}\implies\boxed{E[X]=1}$$
$$E[X^2\mid Y=y]=\int_0^y\frac{x^2}y\,\mathrm dx=\frac{y^2}3$$
$$\mathrm{Var}[X\mid Y]=E[X^2\mid Y]-E[X\mid Y]^2=\frac{Y^2}3-\left(\frac Y2\right)^2=\frac{Y^2}{12}$$
$$E[\mathrm{Var}[X\mid Y]]=E\left[\frac{Y^2}{12}\right]=\frac{E[Y^2]}{12}=\frac12$$
$$\mathrm{Var}[E[X\mid Y]]=\mathrm{Var}\left[\frac Y2\right]=\frac{\mathrm{Var}[Y]}4=\frac{E[Y^2]-E[Y]}4=\frac12$$
$$\boxed{\mathrm{Var}[X]=E[\mathrm{Var}[X\mid Y]]+\mathrm{Var}[E[X\mid Y]]=1}$$
