Evaluate the surface intgral

$\displaystyle \iint_S y^2 \ dS$

where S is the part of the sphere $\displaystyle x^2+y^2+z^2=4$ that lies inside the cylinder $\displaystyle x^2 +y^2=1$ and above the xy-plane

so far I have:

$\displaystyle 1+z^2=4 \rightarrow \ z=\sqrt{3}$

$\displaystyle \iint_D (4-x^2-z^2) \sqrt{ \left( \frac{\partial y}{\partial x } \right) ^2 + \left( \frac{\partial y}{\partial z } \right) ^2 +1 } \ dA

differentiating and converting to polar coordinates:

$\displaystyle \iint_D 4-x^2-z^2 \sqrt{ \frac{x^2}{4-x^2-z^2} + \frac{z^2}{4-x^2-z^2} +1 } \ dA$ $\displaystyle \Rightarrow \iint_D 4-x^2-z^2 \sqrt{ \frac{x^2+z^2}{4-x^2-z^2} +1 } \ dA$

$\displaystyle x=r\cdot \cos(\theta) \ z=r\cdot \sin(\theta)$

putting it all together:

$\displaystyle \int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=1} (4-r^2) \left(\sqrt{ \frac{r^2}{4-r^2}-1} \right)\cdot r \ dr \ d\theta$