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I have a problem that requires me to convert the equation of the sphere $x^2 + y^2 + z^2 = a^2$ into spherical coordinates. When writing out my reasoning for this problem, I want to represent the spherical coordinates for this sphere in set-builder notation. I have never taken a class that focuses specifically on set theory, so my knowledge of the subject is only whatever rudimentary understanding I was taught when learning other fields such as linear algebra.

My understanding is that set-builder notation is $\{$ elements in set $\mid$ condition $\}$ where $\mid$ means such that. In implementing this understanding, I came up with $S = \{ \rho \ge 0, 2\pi \ge \theta \ge 0, \pi \ge \phi \ge 0 \mid \rho^2 \cos^2(\theta) \sin^2(\phi) + \rho^2 \sin^2(\theta) \sin^2(\phi) + \rho^2 \cos^2(\phi) \le a^2 \}$. Is this the correct formal representation of the sphere using set-builder notation? If not, then what are the errors and what is the correct representation?

I would greatly appreciate it if people could please take the time to clarify this.

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  • $\begingroup$ $$\rho^2 \cos^2(\theta) \sin^2(\phi) + \rho^2 \sin^2(\theta) \sin^2(\phi) + \rho^2 \cos^2(\phi) \\ = \rho^2 (\cos^2(\theta) \sin^2(\phi) + \sin^2(\theta) \sin^2(\phi) + \cos^2(\phi)) \\ = \rho^2 ((\cos^2(\theta) + \sin^2(\theta)) \sin^2(\phi) + \cos^2(\phi)) \\ = \rho^2 (\sin^2(\phi) + \cos^2(\phi)) \\ = \rho^2 $$ $\endgroup$ Aug 1 '17 at 1:33
  • $\begingroup$ @MichaelSeifert Thanks for the response. I'm aware of this fact, but I left it out because I'm interesting in the set-builder notation. $\endgroup$ Aug 1 '17 at 1:34
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    $\begingroup$ The sphere described by your initial equation is a surface, so you should have $\rho=a$ instead of any inequality involving that coordinate. In your attempt at a solution, however, it appears that you’re trying to describe a ball of radius $a$, which includes the volume bounded by the sphere. Which is it? $\endgroup$
    – amd
    Aug 1 '17 at 5:28
  • $\begingroup$ @amd You're correct -- I should have said the ball. Thanks. $\endgroup$ Aug 1 '17 at 5:49
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This isn't quite right - in this case $\rho$ represents the radius of the sphere, so it would be simpler to write: $$ S = \{\rho, \theta, \phi : \rho = a, 0 \leq \theta \leq 2\pi, 0 \leq \phi \leq \pi\}$$.

Otherwise, your conceptual understanding of set-builder notation seems good. Keep trying to write other surfaces as sets.

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  • $\begingroup$ Thanks for the response. This seems to be what I was looking for. I hope someone can confirm that it is correct, because it looks correct to me. $\endgroup$ Aug 1 '17 at 1:46
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    $\begingroup$ Correct, I'll fix it. Thanks. $\endgroup$
    – smb3
    Aug 1 '17 at 1:58
  • $\begingroup$ @ThePointer This describes the ball of radius $a$, not the sphere given by the equation in the question. $\endgroup$
    – amd
    Aug 1 '17 at 5:25
  • $\begingroup$ @amd Aren't they the same thing? $\endgroup$ Aug 1 '17 at 5:27
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    $\begingroup$ @ThePointer no, they are not. A sphere is a surface, and is what your initial equation describes. A ball is a solid that includes the volume bounded by a sphere. $\endgroup$
    – amd
    Aug 1 '17 at 5:29
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When using set-builder notation, we want to define our set by describing what each element must satisfy. It's better to describe each element as a single object. Fox example, instead of writing $\rho, \theta, \phi$ and viewing them as three values, it's better to write them as $(\rho, \theta, \phi)$ and view them as a triple. Also, since your condition is quite long, I would define a function $f: [0,\infty)\times[0,2\pi]\times[0,\pi]\to\mathbb{R}$ by $$f(\rho,\theta,\phi)=\rho^2\cos^2(\theta)\sin^2(\theta)+\rho^2\sin^2(\theta)\sin^2(\phi)+\rho^2\cos^2(\phi),$$ and then write the set as $$ S = \{ (\rho, \theta, \phi) \in [0,\infty)\times[0,2\pi]\times[0,\pi] \mid f(\rho,\theta,\phi)\leq a^2\}. $$ Since $f$ only makes sense for triples in $[0,\infty)\times[0,2\pi]\times[0,\pi]$, some would simply write $$ S = \{ (\rho, \theta, \phi) \mid f(\rho,\theta,\phi)\leq a^2\}. $$

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  • $\begingroup$ Thanks for the response. This seems to be more elaborate than what I was looking for but enlightening nonetheless. $\endgroup$ Aug 1 '17 at 1:45

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