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I am currently learning about the spherical coordinate system in class, but I do not know its advantages or even if it is advantageous in using this coordinate system over another.

I am very comfortable in using the rectangular coordinate system and the cylindrical coordinate system (polar coordinate system but just in 3D), as the rectangular coordinate system is just the cartesian coordinate system with another dimension and the cylindrical coordinate system is just the polar coordinate system with an additional dimension.

However, the concept of spherical coordinates come out of nowhere (that I know of) and I am unable to see its advantages. For example, if wanting to calculate an integral in the first octant, you can just restrict to $x>0$, $y>0$, and $z>0$ for the rectangular coordinate system. And for cylindrical coordinates, you can restrict $z>0$, $0<\theta<\frac{\pi}{2}$, and $r$ its corresponding boundary conditions.

My question is are there ever cases when using spherical coordinates are more intuitive than using cylindrical or rectangular coordinates?

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  • $\begingroup$ The equivalent to 2D polars in 3D and higher is spherical, not cylindrical (think of open balls in Euclidean space). Anyway, it doesn't come out of nowhere, for example, you describe your location on Earth in terms of latitude, longitude and altitude above MSL and that is essentially (up to moving altitude back to distance to the center of the Earch and approximating Earth as a sphere) spherical coordinates. $\endgroup$ Jul 28, 2020 at 2:25
  • $\begingroup$ Some regions are described most simply using spherical coordinates, and some functions have simpler formulas when using spherical coordinates. For example I think integrating the function $f(x,y,z) = x^2 + y^2 + z^2$ over the region region $\Omega = \{ (x,y,z) \mid x^2 + y^2 + z^2 \leq R^2, z \geq 0 \}$ is easier if we use spherical coordinates. $\endgroup$
    – littleO
    Jul 28, 2020 at 2:26
  • $\begingroup$ Yes, most problem can be done using either cylindrical or spherical polars, but depending on the symmetry of the problem one is easier than the other. For example, if $K\subseteq S^2=\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2+z^2=1\}$ is a spherical triangle with area $A$, then how do you propose to calculate the volume of the set $\{tv\mid t\in[0,1], v\in K\}\subset\mathbb{R}^3$ using cylindrical coordinates? With spherical polars the answer jumps out immediately, but for cylindrical polars you really have to work. $\endgroup$ Jul 28, 2020 at 2:33

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You want to choose a coordinate system that matches symmetry of the problem at hand. That makes everything easier. Spherical coordinates work well for situations with spherical symmetry, like the field of a point charge. Cylindrical coordinates work well for situations with cylindrical symmetry, like the field of a long wire. Usually it is obvious which set (and there are many more for more complex geometries) you should use.

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In one word - for sphere.

In more words - in spherical coordinates we obtain parallelepiped from sphere: $$\begin{array}{} x = r \sin \phi \cos \theta; \\ y = r \sin \phi \sin \theta; \\ z = r \cos \phi \end{array} $$ $$\{(x,y,z): x^2+y^2+z^2 \leqslant R^2\} \to \{(r,\theta,\phi): 0 \leqslant r \leqslant R, \theta \in [0, \pi], \phi \in [0, 2 \pi)\}$$

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    $\begingroup$ Even when some type of problem involves a sphere, we can use cylindrical coordinates. $x^2+y^2=r^2$, $x=r\cos\theta$,$y=r\sin\theta$,$z=z$ $\endgroup$
    – Jaden Lee
    Jul 28, 2020 at 2:06
  • $\begingroup$ What do you mean obtaining parallelopiped from a sphere? $\endgroup$
    – Jaden Lee
    Jul 28, 2020 at 2:08
  • $\begingroup$ parallelopiped added to answer. $\endgroup$
    – zkutch
    Jul 28, 2020 at 2:12
  • $\begingroup$ Sure. As always devil in details. You take specific decision in each special case. Spherical coordinates are one more powerful tool plus others. $\endgroup$
    – zkutch
    Jul 28, 2020 at 2:14

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