Find $E(|X-Y|^a)$ where $X$ and $Y$ are independent uniform on $(0,1)$ 
Let $X,Y$ be independent $Uniform(0,1)$ random variables. Find $E(|X-Y|^a)$ where $a>0$.

My working:

Define $W=1$ if $X>Y$ and $W=0$ if $X<Y$. We seek $E(|X-Y|^a)=E[E(|X-Y|^a|W)]=E(|X-Y|^a|W=1)P(W=1)+E(|X-Y|^a|W=0)P(W=0)=E((X-Y)^a)P(X>Y)+E((Y-X)^a)P(X<Y)$.
Now $P(X<Y)=\dfrac{1}{2}$ and $P(X>Y)=\dfrac{1}{2}$ by symmetry.
$E((X-Y)^a)=\int_{0}^{1}\int_{0}^{x}(x-y)^af_{X,Y}(x,y)dydx=\int_{0}^{1}\int_{0}^{x}(x-y)^adydx=\dfrac{1}{(a+1)(a+2)}$.
Similarly $E((Y-X)^a)=\dfrac{1}{(a+1)(a+2)}$ by symmetry when $Y>X$.
Required expectation=$\dfrac{1}{(a+1)(a+2)}\dfrac{1}{2}+\dfrac{1}{(a+1)(a+2)}\dfrac{1}{2}=\dfrac{1}{(a+1)(a+2)}$

But correct answer is $\dfrac{2}{(a+1)(a+2)}$.
Where am I going wrong?

You should have something like:
$$
E(|X-Y|^a)=\iint_{x>y}(x-y)^a\;dxdy+\iint_{y>x}(y-x)^a\;dxdy=2\iint_{x>y} (x-y)^a\;dxdy
$$
Now:
$$
\iint_{x>y}(x-y)^a\;dxdy=\int_{y=0}^1\int_{x=y}^1(x-y)^a\;dxdy=\frac{1}{(a+1)(a+2)} 
$$
So:
$$
E(|X-Y|^a)=2\iint_{x>y}(x-y)^a\;dxdy=\frac{2}{(a+1)(a+2)}
$$
