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Spherical to rectangular

  1. Oct 10, 2003 #1
    This equation of a sphere in spherical coordinate form:
    ρ = 4sinφcosθ converts very readily to (x-2)2 + y2 + z2 = 4 with very little effort.

    Now this similar equation looks to me like it should also be a sphere, but I can't seem to get anywhere with it:
    ρ = 4sinθcosφ

    I just end up with a very ugly
    x2 + y2 + z2 = 4yz/(√(x2+y2)
    and I have no idea what to do with that.

    Is this a dead end? Is the second equation not a sphere?
     
  2. jcsd
  3. Oct 10, 2003 #2

    HallsofIvy

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    I can only ask what makes you think the second equation would be a sphere. Spherical coordinates are not "symmetric" in θ and φ.
     
  4. Oct 10, 2003 #3
    I guess I can only plead insanity on this one.

    When it comes to spherical coordinates I'm an absolute greenhorn. The only reason I thought it might be a sphere is that I thought that was the equation that was given on my calc 3 exam last night, and I guess I was "mis-remembering".

    But actually, I did wake up this morning thinking that the 2nd equation probably isn't a sphere; I realized that the sinφ and cosθ are "out of phase", i.e. φ is approaching its max when θ is approaching its min, and I was going to post that as a "supplementary" question. So thanks for answering my second question before I even asked it.

    And thanks for pointing out so succinctly what characterizes a sphere's equation in spherical coordinates.

    So, do you have any idea what "my" equation looks like on a graph?
     
  5. Oct 10, 2003 #4

    selfAdjoint

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    It's a quartic of some kind. Clear the square roots and fractions and I believe you will have a fourth degree equation.
     
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