Surface integral over a cylinder bounded by 2 planes From Schaum's vector analysis:

My attempt:
$\vec n = \nabla S = 2x \hat i + 2z \hat k$
$ \hat n = \frac{1}{3} x \hat i + \frac{1}{3} z \hat k $
$ \vec A . \hat n = 2xz - \frac{xz}{3} = \frac {5}{3} xz$
$dS = \frac {dxdy}{ \hat n . \hat k}$  , $ \hat n . \hat k = \frac {z}{3}$
$dS = \frac {3}{z} dxdy$
$ \iint_S \vec A . \hat n dS = 5 \iint_R x dxdy$
I know $y$ ranges from $0$ to $8$ then 
$5 \iint_R x dxdy = 40 \int x dx$
This is where I stop, I can't integrate $x$ from $0$ to $3$ directly and I can't substitute it with the equation $x^2 + z^2 = 9$ , how do I continue? Also without making use of the divergence theorem please.


This is a Divergence Theorem problem.  The surface integral is equal to the the triple integral over the solid of the divergence of the vector field.  Since the divergence equals $1$, the answer is the volume of the quarter cylinder (which is $18\pi.$
