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Circle to cylindrical coordinates

  1. Mar 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Transform to cylindrical coordinates:
    x[tex]^{2}[/tex]+y[tex]^{2}[/tex]=R[tex]^{2}[/tex]

    Doesn't look like a problem at all first... :smile:

    2. Relevant equations
    .. after all I know that is a circle (2d) and we can forget the z-axis (=0) and transform it to just polar coords.
    Also I know, that for polar coordinates
    x = r cos[tex]\theta[/tex]
    y = r sin[tex]\theta[/tex]

    3. The attempt at a solution
    But I don't know how to put it out as an answer. Am I supposed to give a parametric equation in terms of x=... and y=... or should it be like r=... and [tex]\theta[/tex]=... and z=0 or can I express it without parametres? :confused:
     
  2. jcsd
  3. Mar 13, 2009 #2
    :surprised
    hey, do I get the correct equation just by replacing x with r cos[tex]\theta[/tex] and y with r sin[tex]\theta[/tex]? So that cartesian eq
    x[tex]^{2}[/tex]+y[tex]^{2}[/tex]=R[tex]^{2}[/tex]
    is
    (r cos[tex]\theta[/tex])[tex]^{2}[/tex]+(r sin[tex]\theta[/tex])[tex]^{2}[/tex]=R[tex]^{2}[/tex]
    in cylindrical?

    Or should it be something else, like parametric form of some kind?
     
  4. Mar 13, 2009 #3
    :bugeye:
    (r cos[tex]\theta[/tex])[tex]^{2}[/tex]+(r sin[tex]\theta[/tex])[tex]^{2}[/tex]=R[tex]^{2}[/tex]
    and it all goes down to r=R?

    I guess it goes the same way with the spherical coordinates.

    I'm sorry for this monoloque. :blushing:

    I hope someone would correct me if I'm wrong. thank you.
     
  5. Mar 13, 2009 #4

    HallsofIvy

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    Science Advisor

    Yes, [itex]x^2+ y^2= R^2[/itex] is a cylinder (not a circle) of radius R so it is particulary simple in cylindrical coordinates.

    No, it does not "go that way" with spherical coordinates. In spherical coordinates [itex]x= \rho cos(\theta) sin(\phi)[/itex] and [itex]\phi= \rho sin(\theta) sin(\phi)[/itex] so [itex]x^2+ y^2= \rho^2 cos^2(\theta)sin^2(\phi)+ \rho^2 sin^2(\theta)sin^2(\phi)[/itex][itex]=
    \rho^2 sin^2(\phi)= R^2[/itex].
     
    Last edited: Mar 13, 2009
  6. Mar 13, 2009 #5
    I think (and still do) that it is a circle, because it has no z. :eek:

    Isn't the equation for circle R=r in both cylindrical and spherical coordinates?
     
  7. Mar 13, 2009 #6
    It is a cylinder, as a matter of fact. Because, it is saying that for any z at all, we have x^2+y^2=R^2.:tongue:
     
  8. Mar 13, 2009 #7
    CylindricalCoordinates_1001.gif

    If the z is zero, aren't we only left with a circle?

    man.. understanding one thing requires not understanding two.. :yuck:

    I hope I know how to turn cartesian equations to cylindrical and spherical now.

    I'm really trusting on these tricks:

    For cylindrical: x^2+y^2=r^2
    For spherical: x^2+y^2+z^2=[tex]\rho[/tex]^2

    So if I must for example turn this cartesian
    x^2+y^2-z=0
    into a speherical equation, i just add both sides z^2 and
    x^2+y^2+z^2=z+z^2
    [tex]\rho[/tex]^2=z+z^2
    [tex]\rho[/tex]=[tex]\sqrt{z+z^2}[/tex] and because z= [tex]\rho[/tex] cos [tex]\phi[/tex]
    [tex]\rho[/tex]=[tex]\sqrt{\rho cos \phi +(\rho cos \phi)^2}[/tex]

    Most likely I'm wrong. Once again, help is really appreciated. :frown:
     

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  9. Mar 13, 2009 #8

    That is one way of doing it. You could have also just substitute the values of x,y and z in terms of ro, cos, sin and corresponding angles, and you would have ended up with the same thing.

    Note: the reason you can take the square root of [tex]\rho^2[/tex] is because it cannot be negative. Otherwise, you would have stepped on the red line...:biggrin:
     
  10. Mar 13, 2009 #9
    Well, as a matter of fact, the levle curves of this surface projected in any plane z=k, where k is a constant, are circles.
     
  11. Mar 13, 2009 #10

    HallsofIvy

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    Yes, but you are NOT told that z= 0. Because "it has no z", there is no restriction on z: z can be any number.

     
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