Is the second equation a sphere?

[gnome]

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?

 

[HallsofIvy]

I can only ask what makes you think the second equation would be a sphere. Spherical coordinates are not "symmetric" in θ and φ.

 

[gnome]

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?

 

[selfAdjoint]

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