Evaluating a surface integral $\iint A.dS$ Suppose $A=6z\hat i+(2x+y)\hat j-x\hat k$ .Evaluate $$\iint A.dS$$ Over the entire surface S of the region bounded by the cylinder $x^2+z^2=9,x=0,y=0,z=0$ and $y=8$.I split it into three surface 1.Upper circle part, $S_1$ 2.Lower circle part, $S_2$ 3.cylindrical part, $S_3$.I couldn't do surface integral for  $S_3$.Since i am familiar with parametrization of a cylinder and cylindrical coordinates but failed to approach to the answer. The solution provided by the book is $18\Pi$.Can anyone help me to explain how to get $\iint A.n dS$ please.Thanks in Advanced.
 A: I don't know if you want to do this way, but in case you are allowed to do by the divergence theorem here it goes (I'm a little bit lazy to do it by surface integrals).
$$\int_{\partial\Omega}{\vec{A}\cdot d\vec{S}}=\int_{\Omega}{\textrm{div}(\vec{A})\,dV}$$
Since $\textrm{div}(\vec{A})=1$ in your case, you must calculate the volume of your body. In this case is a quarter of the volume of a cylinder of radius equal to $R=3$ and a height of $h=8$, giving 
$$V = \int_{\Omega}{\,dV}=\frac{1}{4}\pi R^2h = 18\pi $$
For the cylindrical part, make a change of coordinates:
$$x = R\sin{\theta}\qquad y=y\qquad z = R\cos{\theta}\qquad $$
In the cylindrical surface , $\vec{A}=(6R\cos{\theta},2R\sin{\theta}+y,-R\sin{\theta})$, and $d\vec{S}=(\sin{\theta},0,\cos{\theta})Rd\theta dy$. Multiply and integrate! $\theta\in[0,\pi/2]$ and $y\in[0,8]$
A: If you don't want to use the divergence theorem.
Your volume has 5 surfaces.
$S_1$ is the quarter disk in the plane $y = 0$
$S_2$ is the quarter dis in the plane $y = 8$
$S_3$ is the rectangle in the plane $x = 0$
$S_4$ is the rectangle in the plane $z = 0$
$S_5$ is the surface of the cylinder.
The normal vectors to these respective surfaces are $(0,-1,0), (0,1,0), (-1,0,0),(0,0,-1), (\cos\theta, 0, \sin \theta) $ respectively.
I am going evaluate the surfaces $S_1, S_2$ together for reasons that hopefully become apparent.
$\iint -2x\ dS_1 + \iint 2x + 8 \ dS_2\\
\int_0^3\int_0^\sqrt{9-x^2} -2x\ dz\ dx + \int_0^3\int_0^\sqrt{9-x^2} 2x + 8 \ dz\ dx\\
\int_0^3\int_0^\sqrt{9-x^2} 8\ dz\ dx\\
18\pi$
$\iint -6z\ dS_3\\
\int_0^3\int_0^8 -6z\ dy\ dz\\
-216$
$\iint x\ dS_4\\
\int_0^3\int_0^8 x\ dy\ dx\\
36$
$S_5$ We will parameterize the surface.
$x = 3\cos \theta\\
y = y\\
z = 3\sin\theta$ 
$dS = (3\cos\theta, 0,3\sin\theta)\ dy\ d\theta\\
F(y,\theta)\cdot dS = 45\sin\theta\cos\theta \ dy\ d\theta$
$\int_0^{\frac{\pi}{2}}\int_0^8 45\sin\theta\cos\theta \ dy\ d\theta\\
\int_0^{\frac{\pi}{2}} 180\sin 2\theta d\theta\\
-90\cos 2\theta|_0^{\frac {\pi}2}\\
180$
Add them together we get $18\pi$
There are some reasons we might deduce that flux over $S_3, S_4$ and $S_5$ will cancel each other out.
