Really Confused on a surface area integral can't seem to finish the integral off. Basically the question asks to compute $\int \int_{S} ( x^{2}+y^{2}) dA$ where S is the portion of the sphere  $x^{2}  + y^{2}+  z^{2}= 4$ and $z \in [1,2]$ we start with a chnage of variables 
$x=x  $
$y=y$
$ z= 2 \cdot(4-(x^{2}  + y^{2}))^{1/2}$
$Det(u,v)= \begin{bmatrix}
i & j& k \\
1 & 0 & \frac {-x}{(4-(x^{2}  + y^{2}))^{1/2}} \\
0 & 1 & \frac {-y}{(4-(x^{2}  + y^{2}))^{1/2}} \\
\end{bmatrix}=(\frac {x}{(4-(x^{2}  + y^{2}))^{1/2}})i + (\frac {y}{(4-(x^{2}  + y^{2}))^{1/2}})j + k$
$dA=(\frac {x^{2}+y^{2}}{(4-(x^{2}  + y^{2}))} +1)^{1/2}$
$\int \int_{S} (\frac {(x^{2}+y^{2})^{3}}{(4-(x^{2}  + y^{2}))}+(x^{2}+y^{2})^{2})^{1/2}$
Projecting when z=1 and z=2 we have $x^{2}  + y^{2}= 4-1$ $\to r= 0,(3)^{1/2}$
going to polar we have:
$(\frac {r^{6}}{(4-r^{2})}+r^{4})^{1/2}rdrd\theta=(\frac {4r^{4}}{(4-r^{2})})^{1/2}rdrd\theta$ 
my problem is $2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} (\frac {4r^{4}}{(4-r^{2})})^{1/2}rdrd\theta$ there is no nice way i can think of to integrate this. it can also be written as:
$2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}rdrd\theta$ 

It seems that spherical coordinates are more appropriate for calculating  $\iint\limits_{S} ( x^{2}+y^{2}) dA$.
In the last integral 
$$
2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}rdrd\theta=
{4\pi}\int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}r\,dr$$
you can make the substitution $r=2\sin{t}.$
