I get to this bit
$\displaystyle \int^{\frac{\pi}{2}}_{0} \int^{1}_{0} \sqrt{r^{2} +1} r dr d\theta $
from here, can I not simplify the expression $\displaystyle \sqrt{r^{2}+1} r $
=
$\displaystyle r^{2} + r $ ??
Back to algebra class for you: you can't take the square root of each term like that. For example $\displaystyle \sqrt{3^2+1}$ is $\displaystyle \sqrt{10}$, not 4. Or looking the other way, $\displaystyle (r+1)^2$ is $\displaystyle r^2+2r+1$, not $\displaystyle r^2+1$.
When you make HallsofIvy's substitution, you get
$\displaystyle \int^{\frac{\pi}{2}}_{0} \int^{2}_{1} \frac{1}{2} u^{1/2}\, du \,d\theta$
which you should be able to integrate directly.
- Hollywood