Surface Integral Help - Calc 3 question...

Double Integral over S of (z + x^2y)dS where S is the part of the cylinder y^2 + z^2 = 1 that lies between x=0 and x=3.

I've parametrized the surface with respect to X and theta (cylindrical coordinates) taken the partial derrivatives with respect to x and theta, taken their cross product to reveal that dS = 1dA. If this is correct, where I am making my mistake is somewhere within setting up the bounds for dA. Help?

For a cylinder the best way to do this, is integrate over the surface directly .-
For getting the unit normal to the surface (Shell) , one of two .- The surface is
y^2 + z^2 = 1 , with x=x
As you know , if F(x,y,z) = y^2 + z^2 = 1 , a normal is Nabla F
N= 2y j+2z k and a unit normal is = N/INI
= 2 (yj j+zk) / 2sqrt (y^2+z^2)
= yj+zk ( Remember that y^2+z^2=1)

Taking cylindrical coordinates , at the surface , r=1 ,
y=1cosT
z=1sinT
x=x
so
= cos T j+ sinT k

The other way , the direction of the radius is normal to the surface , so the normal goes in direction of the radius over the shell
r= RcosT j +RsinT k , so = r/ IrI .- At the surface r=R , so
r/R == cosT j +sinT k
But, we don´t need this , because we will use dS directly , not its projection over any Region .-

dS= RdT dx ( Directly ) , R=1

So, INT_S of (z + x^2y)dS = INT _S (sinT +x^2cosT ) dx dT
0
INT (sinT +(x^3/3) cosT ) dT
INT (sinT +9cosT ) dT
0
= -cosT +9sinT
= - ( 0-0) +9(0-0) =0 ( Over the Shell only )

Over any cap , dS= i dA ,
dA= rdrdT
Note that it is r not R , because we are moving inside the Region y^2 + z^2 = 1
INTEGRAL = INT_A (sinT +x^2cosT )rdrdT
Bottom , x=0
Top , x=3

Bottom , INT _A sinT rdrdT
0 0 This is equal =0

Cap =0 too .-

The parameterization is the easy part:
r(s,t) = , 0 ≤ s ≤ 3, 0 ≤ t ≤ 2π

r_s = <1, 0, 0>
r_t = <0, -sin(t), cos(t)>

r_s × r_t = <0, -cos(t), -sin(t)>

which is indeed an outward normal and the magnitude of which is 1

So then we just have
∫[0,3] ∫[0,2π] (sin(t) + s²cos(t)) dt ds

which is easily integrated.