<s>[INST] 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:
$2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}rdrd\theta$ 
 [/INST] Although the sine substitution probably is the easiest method, here's another one:  
If the integrand is a fraction with a square root in the denominator, see if you can write the integrand as the derivative of the square root times another function to pave the way for an integration by parts:
$I = \int \frac{r^3}{\sqrt{4-r^2}}\, \mathrm{d}r = \int -r^2 \frac{-r}{\sqrt{4-r^2}}\, \mathrm{d}r  $
$\: = -r^2 \sqrt{4-r^2} + 2 \int r \sqrt{4-r^2}\, \mathrm{d}r$
$\: = -r^2 \sqrt{4-r^2} - \frac{2}{3} (4-x^2)^{\frac{3}{2}}$
</s>