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Proving a form ##z=f(r)## to be a surface of revolution

  1. Sep 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Suppose that a surface has an equation in cylindrical coordinates of the form ##z=f(r)##. Explain why it must be a surface of revolution.

    2. Relevant equations


    3. The attempt at a solution
    I consider ##z=f(r)## in terms of spherical coordinates.

    ## p cosφ = f \sqrt{(p sinφcosθ)^2 + (p sinφsinθ)^2} ##

    ## p cosφ= f\sqrt{(p sinφ)^2} ##

    ## p cosφ=f(p sinφ)##

    ##cosφ= f (sinφ)##

    ##∴φ= \cos^{-1} f(sinφ)##

    Since equation is independent of ##\theta##, it describes a surface of revolution about the ##z## axis.

    Is my prove right or acceptable?
     
  2. jcsd
  3. Sep 17, 2016 #2

    Ray Vickson

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    You are using spherical, not cylindreical coordinates. Also: in LaTeX, put a "\" before sin, cos, etc. Without it, the results are ugly and hard to read, like ##sin \phi cos \theta##; with it, they look good, as in ##\sin \phi \cos \theta##.
     
  4. Sep 17, 2016 #3
    Okay! Thanks for the advice! Just started using it, so sorry for the ugly text. I don't have any idea of starting the proof using cylindrical coordinates, that's why I converted it to spherical ones. But I will give it a try now.

    The cylindrical coordinates are in the form ##(r,\theta, z)##. I assume that ##r \geq 0##
    So ##r= \sqrt{(x^2 + y^2)}##

    and ##z=f(\sqrt{(x^2 + y^2)})##, which gives a unique value. These gives a set of points that form a line in ##3D## space.

    Since equation is independent of ##\theta##, line is the same for any value of ##\theta##. These similar set of lines form a surface of revolution.

    Is it right in any way at all? I'm guessing here.
     
  5. Sep 17, 2016 #4

    LCKurtz

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    Why introduce spherical coordinates, which have nothing to do with this question?
    If you have a surface of revolution revolved about the ##z## axis, that would mean ##z(r,\theta_1) = z(r,\theta_2)## for any ##\theta_1## and ##\theta_2## wouldn't it? Is that true in your case?
     
  6. Sep 17, 2016 #5
    Yes. So, is what I'm saying above right? But I'm not sure if having a unique set of points will form a line, though.
    It depends on ##f##, right?
     
  7. Sep 17, 2016 #6
    Oh, I think I see now why ##z=f(r)## represents a surface of revolution. Using the explanation on ##zr## planes given by @LCKurtz , each value of ##r## gives a value of ##z##, and the set of values of ##r## gives its respective values of ##z##. Since equation is independent of ##\theta##, these set of points are the same for all values of ##\theta##, and this is the reason why it forms a surface of revolution.

    Am I right?
     
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