How do I evaluate the surface integral

Double integral,S, of (z+x^2y) dS
S is the part of the cylinder
y^2 + z^2 = 16
that lies between the planes
x = 0 and x = 6
in the first octant
My answer was around 196, but it was wrong. What is the correct answer and how to get to it?

Using cylindrical coordinates with x playing the usual role of z, we have
R(u, v) = with u in [0, 6] and v in [0, π/2].

R_u x R_v = <0, -4 cos v, -4 sin v>
==> ||R_u x R_v|| = 4.

So, ∫∫s (z + x^2 y) dS
= ∫∫ (4 sin v + u^2 * 4 cos v) * 4 dA
= 16 * ∫(u = 0 to 6) ∫(v = 0 to π/2) (sin v + u^2 cos v) dv du
= 16 * ∫(u = 0 to 6) (-cos v + u^2 sin v) {for v = 0 to π/2} du
= 16 * ∫(u = 0 to 6) (u^2 + 1) du
= 16(u^3/3 + u) {for u = 0 to 6}
= 1248.

I hope this helps!