Help with parameterization of surface

[Kuma]

Homework Statement

If I have been given a surface x = 12 − y^2 − z^2 between x = 3 and x = 8, oriented by the unit normal which points away from the x–axis.

I want to find an orientation preserving parameterization.

Homework Equations

The Attempt at a Solution

I know orientation preserving means that the normal vector is pointing outward. I'm not sure how to apply this to parameterize this surface however.

 

[LCKurtz]

Homework Statement

If I have been given a surface x = 12 − y^2 − z^2 between x = 3 and x = 8, oriented by the unit normal which points away from the x–axis.

I want to find an orientation preserving parameterization.

Homework Equations

The Attempt at a Solution

I know orientation preserving means that the normal vector is pointing outward. I'm not sure how to apply this to parameterize this surface however.

Parameterization and orientation are separate issues. Try cylindrical like coordinates only on y and z instead of x and y.

 

[Kuma]

Parameterization and orientation are separate issues. Try cylindrical like coordinates only on y and z instead of x and y.

I figure I can parameterize it no problem but the question literally asks what I said. Find an orientation preserving parameterization. What does that mean?

 

[LCKurtz]

I figure I can parameterize it no problem but the question literally asks what I said. Find an orientation preserving parameterization. What does that mean?

I have seen instances when textbooks say the parameterization itself determines the orientation. For example, if your surface is parameterized as $\vec R =\vec R(u,v)$, then the direction of $\vec R_u \times \vec R_v$ determines the positive orientation of the surface. So, if you parameterize your surface using $r$ and $\theta$, one or the other of $\vec R_r\times \vec R_\theta$ or $\vec R_\theta\times\vec R_r$ will point in the direction that was specified by the problem. If it is the first, then write your parameterization as $\vec R = \vec R(r,\theta)= \ ...$ and if it is the second write it as $\vec R = \vec R(\theta,r)=\ ...$. Personally, I don't care for that notion because, as in your problem, the orientation is given separately. Anyway, that's my best guess what it might mean.