Let $U$ be $~U [0,1] $and let $Y = U^{\frac{1}{2}}$ Let $U$ be $\sim \mathcal{U}[0,1]$ and let $Y = U^{1/2}$.
I'm having trouble finding the $E(Y)$. How do I go about doing this? 

Hint: (1) Start with computing the distribution function of $Y$, we have for $y \in [0,1]$:
$$ F_Y(y) = \def\P{\mathbb P}\P(Y \le y) = \P(U^{1/2} \le y) = \P(U \le y^2) = \P(U \in [0,y^2])$$
Now use that $U$ is uniformly distributed on $[0,1]$ and hence the probiability of an interval is its lenght.
(2) The density of $U$ is given by $f_Y(y) = F_Y'(y)$.
(3) Now use 
$$ \mathbb E(Y) = \int_0^1 y \, d\P_Y = \int_0^1 yf_Y(y)\, dy $$
