Simple stokes - why is this the correct orientation?

Question

We want to calculate $\iint_{S}\text{curl}(\vec F)dS$ where $\vec{F}(x,y,z)=(y^2z, xz,x^2y^2)$ and $S$ is the part of the paraboloid $z=x^2+y^2$ that's inside the cylinder $x^2+y^2= 1$ with an outward facing normal.

My answer:

The boundary of $S$ is simply the unit circle $x^2+y^2 = 1$ at height $z=1$.

A parametrization of $\partial S$ would be $r(\theta) = (\cos(\theta), \sin(\theta), 1)$, as $\theta$ goes from $0$ to $2\pi$.

This is because the normal is facing outwards I believe, so for $\partial S$ to be positively oriented, $\theta$ goes from $0$ to $2\pi$.

Calculating, we get $$\int_{\partial S}\vec{F}ds = \int_{0}^{2\pi}\vec{F}(r(\theta)) \cdot\nabla r(\theta)d\theta = \int_{0}^{2\pi}\cos^2(\theta)-\sin^3(\theta)d\theta = \pi$$

However, the correct answer seems to be $-\pi$, so I got the orientation wrong, but I'm not sure why.

Answer 1

The induced orientation on $\partial S$ can be described this way:

If you walk along $\partial S$ with your head pointing in the direction of the normal vector, the induced orientation is the one that keeps $S$ “on your left.”

So if the orientation on $S$ is the one with downward $z$-component, you're walking around $\partial S$ with your head pointing down. Somebody looking downwards on the $xy$-plane would see you going clockwise around the boundary circle.

Answer 2

Look at this image. The normal points outwards meaning that the positive side of the surface is the outer/lower side of the surface. For the boundary to have positive orientation you must walk it in the clockwise direction, not in the anticlockwise as you have done.