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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

    Defennder

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    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

    HallsofIvy

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    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

    HallsofIvy

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    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].
     
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