Spherical to rectangular

  • Thread starter gnome
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  • #1
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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?
 

Answers and Replies

  • #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.
 
  • #3
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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?
 
  • #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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