Orientability of surface r(u,v)=(ucos(v),usin(v),v)

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Say one wanted to take the surface integral $\iint_S F\cdot dS$ of the vector field $F(x,y,z)=\langle z,y,x \rangle$ over the surface $S$ parametrized by $r(u,v)=\langle u\cos(v),u\sin(v),v\rangle$, $0 \leq u \leq 1$, $0 \leq v \leq \pi$ with $\textit{upward orientation}$. This is an example from Stewart 7th ed.

We first see if $r_u \times r_v=\langle \sin v, -\cos v, u \rangle$ agrees with the requested upwards orientation. It does, since the $k$ component is nonnegative as $u$ ranges from $0$ to $1$. So we proceed to set up the integral and evaluate without any sign-change.

However, my question is, if we were asked to take $\iint_S F\cdot dS$ all the same, but the surface was slightly altered by changing the bounds on $u$ to $\mathbf{-1 \leq u \leq 1}$, what happens in terms of orientation? I have plotted $S$, and see it gives both "sides" of the helicoid rather than just one (when $0 \leq u \leq 1$).

Is this surface orientable? As $u$ varies, the normal vector field to our surface will have both positive and negative $k$ components. I'm not sure which orientation this would induce (if any) on $S$. Let me know if I can help clarify anything.

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The surface is orientable. It's important not to get too hung up on (the sign of) the components of the normal vector along any particular coordinate axis; a sphere, for instance, is orientable but the 'oriented' normal vector field will inevitably have vectors with both positive and negative values of any coordinate.

What orientable means in this context is that if we start at some point $p$ on the surface with a normal $\hat{n}_p$ and trace a loop along the surface continuously varying the normal, we can never get back to $p$ with the normal $-\hat{n}_p$. (For instance, imagine taking a Möbius strip and going once around it; you'll see that the normal continuously varies, but when you return to your original position it will be the inverse of the original normal.) In the case of the helicoid, it's true that the normal won't always be pointing upwards, but that doesn't mean that any path will ever flip it to the 'other side' of the surface.