$f_{X,Y}(x,y)=\frac{1}{2}xyI_{(0,x)}(y)I_{(0,2)}(x)$ Find the marginal distribution of $X\mbox{ and }Y$ Consider two random variables $X$ and $Y$ having a joint probability density function

$f_{X,Y}(x,y)=\frac{1}{2}xyI_{(0,x)}(y)I_{(0,2)}(x)$

(a) Find the marginal distribution of $X\mbox{ and }Y$

$f_y(y)=(y) $ 

http://www.wolframalpha.com/input/?i=+%28integral+%28xy%29%2F2+dx++from+0+to+2+%29

$f_x(x)=(x^3/4) $ 

http://www.wolframalpha.com/input/?i=+%28integral+%28xy%29%2F2+dy++from+0+to+x+%29

(b) Are $X$ and $Y$ independent? 

Given that $f_x(x) * f_y(y) = (x^3/4)* (y) \not=  (xy)/2 = f_{xy}(x,y) $ then $X$ and $Y$ aren't independents

Am I right??? 

(c)What is $E(Y)$?

Thanks for your time
I try to fix my previous answer, which was obviously wrong:

In order to calculate $E(y)$, one can use directly the joint distribution:

$$E(y) = \int_0^2 \int_0^x y f(x,y) dy dx = \int_0^2 \int_0^x \frac{xy^2}{2} dy dx = \int_0^2\frac{x^4}{6}dx = \frac{2^5}{30}=\frac{16}{15}$$
I try to fix my previous answer, which was obviously wrong:

In order to calculate $E(y)$, one can use directly the joint distribution:

$$E(y) = \int_0^2 \int_0^x y f(x,y) dy dx = \int_0^2 \int_0^x \frac{xy^2}{2} dy dx = \int_0^2\frac{x^4}{6}dx = \frac{2^5}{30}=\frac{16}{15}$$