If X and Y are uncorrelated, are X^2 and Y also uncorrelated? If two random variables $X$ and $Y$ are uncorrelated, can we also know that $X^2$ and $Y$ uncorrelated? My hypothesis is yes.
$X, Y$ uncorrelated means $E[XY]=E[X]E[Y]$, or
$$
E[XY]=\int xy f_X(x)f_Y(y)dxdy=\int xf_X(x)dx\int yf_Y(y)dy=E[X]E[Y]
$$
Does that also mean the following?
$$
E[X^2Y]=\int x^2y f_X(x)f_Y(y)dxdy=\int x^2f_X(x)dx\int yf_Y(y)dy=E[X^2]E[Y]
$$
 A: The error in your reasoning is that you write the following about $E[h(X,Y)]$:
$$E[h(X,Y)]=\int h(x,y) f_X(x)f_Y(y)dxdy$$
while in general
$$E[h(X,Y)]=\int h(x,y) f_{XY}(x,y)dxdy.$$
The two coincide if $f_{XY}(x,y)=f_X(x)f_Y(y)$, i.e. if $X$ and $Y$ are independent. Being uncorrelated is a necessary but not sufficient condition for being independent. So if two variables $X$ and $Y$  are uncorrelated but dependent, then $f(X)$ and $g(Y)$ may be correlated.
A: No. A counterexample:
Let $X$ be uniformly distributed on $[-1, 1]$, $Y = X^2$.
Then $E[X]=0$ and also $E[XY]=E[X^3]=0$ ($X^3$ is odd function), so $X,Y$ are uncorrelated.
But $E[X^2Y] = E[X^4] = E[{X^2}^2] > E[X^2]^2 = E[X^2]E[Y]$ 
The last inequality follows from Jensen's inequality. It also follows from the fact that 
$E[{X^2}^2] - E[X^2]^2 = Var(X) > 0$ since $X$ is not constant.

The problem with your reasoning is that $f_X$ might depend on $y$ and vice versa, so your penultimate equality is invalid.
A: Even if $\operatorname{Corr}(X,Y)=0$, not only is it possible that $X^2$ and $Y$ are correlated, but they may even be perfectly correlated, with $\operatorname{Corr}(X^2,Y)=1$:
> x <- c(-1,0,1); y <- c(1,0,1)
> cor(x,y)
[1] 0
> cor(x^2,y)
[1] 1

Or $\operatorname{Corr}(X^2,Y)=-1$:
> x <- c(-1,0,1); y <- c(-1,0,-1)
> cor(x,y)
[1] 0
> cor(x^2,y)
[1] -1

In case you cannot read R code, the first example is equivalent to considering two random variables $X$ and $Y$ with a a joint distribution such that $(X,Y)$ is equally likely to be $(-1,1)$, $(0,0)$ or $(1,1)$. In the perfectly negatively correlated example, $(X,Y)$ is equally likely to be $(-1,-1)$, $(0,0)$ or $(1,-1)$.
Nevertheless, we can also construct $X$ and $Y$ such that $\operatorname{Corr}(X^2,Y)=0$, so all extremes are possible:
> x <- c(-1,-1,0,1,1); y <- c(1,-1,0,1,-1)
> cor(x,y)
[1] 0
> cor(x^2,y)
[1] 0

