1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Is this parameterization correct?

  1. Dec 11, 2008 #1
    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. jcsd
  3. Dec 11, 2008 #2


    User Avatar
    Homework Helper

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

    Since you already have z=f(x,y), you can just let x,y be your parametric variables s,t.
  4. Dec 11, 2008 #3


    User Avatar
    Science Advisor

    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?
  5. Dec 11, 2008 #4
    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
  6. Dec 11, 2008 #5


    User Avatar
    Science Advisor

    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 [itex]2\pi[/itex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook