The Spherical Coordinate System

The page sketches three coordinate systems for the same point:

• Cartesian — $(x,y,z)$.
• Cylindrical — $(\rho ,\varphi ,z)$, with $\varphi$ measured in the base plane and $\rho$ the radial distance out from the $z$-axis.
• Spherical — $(r,\theta ,\varphi )$, with $\theta$ measured down from the $z$-axis, $r$ the radial line, $r\sin \theta$ its horizontal projection and $r\cos \theta$ its height.

Spherical → Cartesian

$x=r\sin \theta \cos \varphi$ $y=r\sin \theta \sin \varphi$ $z=r\cos \theta$

Cartesian → Spherical

$r=\sqrt{x^2+y^2+z^2}$ $\varphi =arctan2(x,y)$ $\theta =\arccos \frac{z}{r}$

$arctan2$ takes $(x,y)$ and returns a unique value in the range $[-\pi ,\pi ]$.

Convert from spherical to Cartesian

a) $\left(1,\frac{\pi }{2},\frac{\pi }{4}\right)$

$z=r\cos \theta \cos \theta =\cos \left(\frac{\pi }{2}\right)=0\therefore z=0$ $\sin \theta \left(\frac{\pi }{2}\right)=1\sin \varphi \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$ $y=1\cdot 1\cdot \frac{\sqrt{2}}{2}⟹y=\frac{\sqrt{2}}{2}$ $x=1\cdot 1\cdot \frac{\sqrt{2}}{2}\cos \varphi \therefore x=\frac{\sqrt{2}}{2}$

$\therefore \left(1,\frac{\pi }{2},\frac{\pi }{4}\right)$ translates to $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0\right)$ in the Cartesian system.

b) $\left(2,\frac{\pi }{3},\frac{\pi }{4}\right)$

$\sin \theta =\sin \left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}\cos \theta =\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}$ $\sin \varphi =\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\cos \varphi =\cos \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$ $x=2\cdot \frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{6}}{2}$ $y=2\cdot \frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{6}}{2}$ $z=2\cdot \frac{1}{2}=1$

$\therefore \left(2,\frac{\pi }{3},\frac{\pi }{4}\right)=\left(\frac{\sqrt{6}}{2},\frac{\sqrt{6}}{2},1\right)$

Convert from Cartesian to spherical

a) $(1,1,0)$

$r=\sqrt{1^2+1^2+0^2}=\sqrt{2}$ $\varphi =arctan2(1,1)=\frac{\pi }{4}$ $\theta =\arccos \left(\frac{0}{\sqrt{2}}\right)=\arccos (0)=\frac{\pi }{2}$

$⟹\left(\sqrt{2},\frac{\pi }{2},\frac{\pi }{4}\right)$

b) $(1,-1,-1)$

$r=\sqrt{1+1+1}=\sqrt{3}$ $\varphi =arctan2(y,x)=arctan2(-1,1)=-\frac{\pi }{4}$ $\theta =\arccos \left(\frac{z}{r}\right)=\arccos \left(\frac{-1}{\sqrt{3}}\right)=2.186\text{ rad}$

$⟹\left(\sqrt{3},2.186,-\frac{\pi }{4}\right)$