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Converting Spherical Equations to Cylindrical and Rectangular

  1. Oct 20, 2004 #1

    (Note: p=rho and o=phi)
    Convert p(1-2cos^2(o))=-psin^2(o) into cylindrical and rectangular coordinates and describe or sketch the surface.

    The part that I don't know how to do is converting the spherical equation into cylindrical or rectangular coordinates. I know all the equations like x=psin(o)cos(theta) and y=psin(o)sin(theta) but I don't see how I can manipulate the given equation so that I could use those equations. Any help would be greatly appreciated!
  2. jcsd
  3. Oct 20, 2004 #2


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    Don't try manipulating them directly. Look at the geometry.Yeah...I'm working w/ spherical coord systems as well a lot in Electromagnetism right now.

    Think about what it means for something to be radially outward...sweeps out a sphere...every point equidistant ([itex] \rho [/itex]) no matter the direction:

    [tex] \rho = \sqrt{x^2 + y^2 + z^2} [/tex]

    For some reason our convention in physics class is opposite what we did in math (our thetas and phis are reversed from yours. No matter...I'll convert)

    See the projection of [itex] \rho [/itex] onto the xy plane? It represents a line "radially" outward in that plane i.e.:

    [tex] \rho\sin\phi = \sqrt{x^2 + y^2} [/tex]

    This projection into the plane forms a right triangle with z, the hypotenuese of which is [itex] \rho [/itex].

    The angle between z and [itex] \rho [/itex] is just [itex] \phi [/itex], so from the geometry of the right triangle:

    [tex] \phi = \tan^{-1}\left(\frac{\sqrt{x^2 + y^2 }}{z}\right) [/tex]

    It shouldn't be too hard to see that the azimuthal angle ([itex] \theta [/itex] in your case) is given by:

    [tex] \theta = \tan^{-1}\left(\frac{y}{x}\right) [/tex]

    After all that, cylindrical coords should be easy
  4. Oct 21, 2004 #3


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    You know x= ρcos(θ)sin(&phi), y= ρsin(θ)sin(φ), z= ρcos(φ) but you need to know them the other way:

    [tex]\rho= \sqrt{x^2+ y^2+ z^2}[/tex]
    [tex]\theta= arctan(\frac{y}{x})[/tex]
    [tex]\phi= arccos(\frac{z}{\sqrt{x^2+y^2+z^2})[/tex]

    Replace each occurance in your equation by the corresponding formula.
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