Cartesian, cylindrical, spherical conversion

Coordinate systems

• Cartesian $(x, y, z)$: three mutually perpendicular axes; simplest for prismatic shapes and used internally by CAD.
• Cylindrical $(\rho, \phi, z)$: radial distance in the XY plane, azimuth angle from +X, and height; suited to shafts, holes and revolved parts.
• Spherical $(r, \theta, \phi)$: distance from origin, polar angle from +Z, azimuth from +X; suited to spheres and radial scanning.

Relations:

$$\rho=\sqrt{x^2+y^2},\quad \phi=\tan^{-1}\frac{y}{x},\quad r=\sqrt{x^2+y^2+z^2},\quad \theta=\cos^{-1}\frac{z}{r}$$ $$x=r\sin\theta\cos\phi,\quad y=r\sin\theta\sin\phi,\quad z=r\cos\theta$$

(a) Point (3, 4, 5)

$$\begin{aligned} \rho &= \sqrt{3^2+4^2} = 5 \\ \phi &= \tan^{-1}(4/3) = 53.13^\circ \\ r &= \sqrt{9+16+25} = 7.071 \\ \theta &= \cos^{-1}(5/7.071) = 45^\circ \end{aligned}$$

Cylindrical: $(5,\ 53.13^\circ,\ 5)$. Spherical: $(7.071,\ 45^\circ,\ 53.13^\circ)$.

(b) Spherical to Cartesian

$$\begin{aligned} x &= 10\sin 60^\circ\cos 45^\circ = 10(0.8660)(0.7071) = 6.124 \\ y &= 10\sin 60^\circ\sin 45^\circ = 6.124 \\ z &= 10\cos 60^\circ = 5.000 \end{aligned}$$

Answer: (a) cylindrical (5, 53.13 deg, 5); spherical (7.071, 45 deg, 53.13 deg). (b) Cartesian (6.124, 6.124, 5.000) in the same length units.