I'm trying to solve the following problem:
Integrate $f(x,y,z)=(x,y,z)$ over the surface $z=12$ $x^2 + y^2 \leq 25$
I parametrized the surface with $\sigma (r, \theta) = r \sin(\theta), r \cos(\theta), 12)$ with $0 \leq r \leq \sqrt(25) $ and $0 \leq \theta \leq 2\pi$.
Then since $\sigma _{r} \times \sigma_{\theta} = (0,0,-r)$ I get that
$\int_S f ds = \int _{0} ^{\sqrt(25)} \int _0 ^{2\pi} (r\sin\theta, r\cos\theta,12).(0,0,-r) d\theta dr = \int _{0} ^{\sqrt(25)} \int _0 ^{2\pi} (-12 r) d\theta dr = -300\pi $
Problem is that my book lists the answer as $300\pi$.
My thinking is that the unitary outwards pointing normal of the surface is $(0,0,1)$ so my parametrization doesn't preserve the orientation, then I should multiply by -1 when solving the integral.
But
My question is, is there a way to do this without going through the outwards normal vector? I went back to the definition of surface integrals and it makes no mention of orientation really.