# Homework Help: Is this parameterization correct?

1. Dec 11, 2008

### andrassy

I have to parameterize the elliptic paraboloid z = 16 - x^2 - y^2 from z = 12 to z = 16. Is the correct parameterization just X(s,t) = (s*cos(t), s*sin(t), 16 - s^2 - t^2) where s ranges from 0 to 2 and t ranges from 0 to 2pi?

2. Dec 11, 2008

### Defennder

I don't think so, since $$16-(s \cos t)^2 - (s \sin t)^2 \neq 16 - s^2 - t^2$$.

Since you already have z=f(x,y), you can just let x,y be your parametric variables s,t.

3. Dec 11, 2008

### HallsofIvy

Defennder's suggestion is the simplest.

If you really are determined to use polar coordinates (which is what you appear to be doing), then x= s cos t, y= s sin t and then

z= 16- x2- y2= what?

4. Dec 11, 2008

### andrassy

I had thought about doing it that way as well with (s, t, 16 - s^2 - t^2), but I don't know what s and t range from then. Is it just from 0 to 2 for both?

whoops also, before I meant to make my parameterization (s cost, s sint, 16 - s^2). Would this be right doing it that way? Because 16 - (s cost)^2 - (s sint)^2 = 16 - s^2(cos^2 + sin^2) = 16 - s^2. Then the s would range from 0 to 2. Is this right?

Last edited: Dec 11, 2008
5. Dec 11, 2008

### HallsofIvy

You said "z = 12 to z = 16" which would project to the xy-plane as the circle from r= 0 to r= 2. Yes, taking x= s cos(t), y= s sin(t), z= 16- s2, essentially polar coordinates, s ranges from 0 to 2 and t from 0 to $2\pi$.