Calc III Surface Integral Question

Hi everyone. I've been trying this out for a while and it hasn't been working. Can anyone give me a hand please?

Evaluate the surface integral ∫∫S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.

F(x, y, z) = y i + x j + z2 k

S is the helicoid (with upward orientation) with vector equation r(u, v) = ucos(v)i + usin(v)j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 3π.

Thanks!

r_u =
r_v = <-u sin v, u cos v, 1>.

So, r_u x r_v = .

Hence, ∫∫s F · dS
= ∫∫ · dA
= ∫(u = 0 to 2) ∫(v = 0 to 3π) (-u cos(2v) + uv^2) dv du, via double angle identity ·
= ∫(u = 0 to 2) (-u sin(2v)/2 + uv^3/3) {for v = 0 to 3π} du
= 9π^3 * ∫(u = 0 to 2) u du
= 9π^3 * (1/2)u^2 {for u = 0 to 2}
= 18π^3.

I hope this helps!