A coordinate system is defined as a space where an object is located. The points defining the position of an object are known as coordinates. There are different type of coordinates, such as - Cartesian coordinates, polar coordinates, cylindrical coordinates and spherical coordinates. We are going to learn about spherical coordinates in this page. These are quite similar to polar coordinates.

Spherical coordinates exist in three dimensional spaces, in which a point is represented by three numbers:

We represent a general spherical coordinate system as (d, $\theta$, $\phi$), where d equals radius, $\theta$ equals inclination and $\phi$ equals azimuth. These values can easily be obtained from the x y z coordinate system of the point. Consider a point R (x, y, z). Then the spherical coordinates of R will be given by (d,$\theta$, $\phi$), where

d = $\sqrt {(x^2 + y^2 + z^2)}$

$\theta$ = $cos^{-1}$ $(\frac{z}{\sqrt (x^2 + y^2 + z^2)})$ = $cos^{-1}$ $(\frac{z}{d})$

$\phi$ = $tan^{-1}$ $(\frac{y}{x})$

Let us see the converse: if given spherical coordinates and we need to find Cartesian coordinates we can use the following:

x = $d\ sin \theta \ cos \phi$

y = $d\ sin \theta \ sin \phi$

z = $d\ cos \theta$

Here, the restrictions that are to be followed are d $\geq$ 0, 0 $\leq \theta < 2 \pi$, and

$0 \leq \theta \leq \pi$

If d is a constant value, then we get a sphere of radius ‘d’ with origin being the centre with changing $\theta$ and $\phi$ in the plane.

When $\theta$ is a constant value, we always get a cone all of whose points are at an angle of $\theta$ from the z axis. When $\theta$ is a constant value, we get a vertical plane which is forming an angle of $\theta$ with the x axis.

So, we use following formulas to find the spherical coordinates of the given system.

d = $\sqrt (x^2 + y^2 + z^2)$

$\Rightarrow d = \sqrt {1^{2} + (-1)^{2} + (-2)^{2}}$

$\Rightarrow d = \sqrt{1 + 1 + 4}$

$\Rightarrow d = \sqrt{6}$

$\theta$ = $cos^{-1}$ $(\frac{z}{\sqrt {x^{2} + y^{2} + z^{2}}})$

$\Rightarrow \theta$ = $cos^{-1}$ $(\frac{z}{d})$

$\Rightarrow \theta$ = $cos^{-1}$ $(\frac{-2}{\sqrt {6}})$

$\Rightarrow \theta$ = $cos^{-1}$ $(\frac{-2}{\sqrt {6}}$ $\times$ $\frac{\sqrt {6}}{\sqrt {6}})$

$\Rightarrow \theta$ = $cos^{-1}$ $(-\frac{\sqrt{6}}{3})$

$\phi$ = $tan^{-1}$ $(\frac{y}{x})$

$\Rightarrow \phi$ = $tan^{-1}$ $(\frac{-1}{1})$

$\Rightarrow \phi$ = $tan^{-1} (-1)$

$\Rightarrow tan \phi = -1$

$\Rightarrow tan \phi$ = $\tan$ $\frac{5 \pi}{4}$ or$\ tan$ $\frac{7 \pi}{4}$

$\Rightarrow \phi$ = $\frac{5\pi}{4}$ or $\phi$ = $\frac{7 \pi}{4}$

Thus the given Cartesian system (1, -1, -2) can be rewritten in spherical coordinate system as ($\sqrt {6},\ cos^{-1}$ (- $\frac{\sqrt {6}}{3}$), $\frac{5}{4})$ or can also be written in another form as ($\sqrt {6},\ cos^{-1}$ (- $\frac{\sqrt {6}}{3})$, $\frac{7}{4})$.