Rectangular coordinates are not always the easiest way to describe a shape. You already know polar coordinates for circles in the plane. Space has two natural extensions: cylindrical coordinates (polar in the floor, plus height) for things with an axis, like pipes and cans, and spherical coordinates (distance from the origin and two angles) for balls and cones.
Key idea
Cylindrical $(r, \theta, z)$: $x = r\cos\theta$, $y = r\sin\theta$, $z = z$, with $r^2 = x^2 + y^2$.
Spherical $(\rho, \theta, \varphi)$: $\rho$ is the distance from the origin, $\theta$ the same angle as in polar coordinates, and $\varphi$ the angle measured down from the positive $z$-axis, with $\rho \ge 0$ and ${}0 \le \varphi \le \pi$. $$x = \rho\sin\varphi\cos\theta, \quad y = \rho\sin\varphi\sin\theta, \quad z = \rho\cos\varphi, \quad \rho^2 = x^2 + y^2 + z^2.$$
Spherical $(\rho, \theta, \varphi)$: $\rho$ is the distance from the origin, $\theta$ the same angle as in polar coordinates, and $\varphi$ the angle measured down from the positive $z$-axis, with $\rho \ge 0$ and ${}0 \le \varphi \le \pi$. $$x = \rho\sin\varphi\cos\theta, \quad y = \rho\sin\varphi\sin\theta, \quad z = \rho\cos\varphi, \quad \rho^2 = x^2 + y^2 + z^2.$$
Why it works
Picture the vertical half-plane that contains the point and the $z$-axis. In it, the segment from the origin to the point has length $\rho$ and makes angle $\varphi$ with the $z$-axis. That right triangle has vertical leg $z = \rho\cos\varphi$ and horizontal leg $r = \rho\sin\varphi$, the distance from the $z$-axis. The horizontal leg points in the direction $\theta$, so polar coordinates finish the job: $x = r\cos\theta$ and $y = r\sin\theta$.
| Surface | Cylindrical | Spherical |
|---|---|---|
| cylinder $x^2 + y^2 = 9$ | $r = 3$ | |
| sphere $x^2 + y^2 + z^2 = 9$ | $\rho = 3$ | |
| cone $z = \sqrt{x^2 + y^2}$ | $z = r$ | $\varphi = \frac{\pi}{4}$ |
| half-plane through the $z$-axis | $\theta = \frac{\pi}{3}$ | $\theta = \frac{\pi}{3}$ |
| plane $z = 2$ | $z = 2$ | $\rho\cos\varphi = 2$ |
Worked examples
Worked example: Cylindrical to rectangular
Convert $(r, \theta, z) = \left(4, \frac{2\pi}{3}, -1\right)$.
- $x = 4\cos\frac{2\pi}{3} = 4\left(-\frac{1}{2}\right) = -2$.
- $y = 4\sin\frac{2\pi}{3} = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}$.
- $z = -1$ stays.
Answer: $\left(-2, 2\sqrt{3}, -1\right)$
Worked example: Spherical to rectangular
Convert $(\rho, \theta, \varphi) = \left(6, \frac{\pi}{4}, \frac{\pi}{3}\right)$.
- $z = 6\cos\frac{\pi}{3} = 3$ and $r = 6\sin\frac{\pi}{3} = 3\sqrt{3}$.
- $x = r\cos\frac{\pi}{4} = 3\sqrt{3} \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{6}}{2}$, and $y$ is the same.
Answer: $\left(\frac{3\sqrt{6}}{2}, \frac{3\sqrt{6}}{2}, 3\right)$
Worked example: Rectangular to spherical
Convert $\left(1, -1, \sqrt{2}\right)$.
- $\rho = \sqrt{1 + 1 + 2} = 2$.
- $\cos\varphi = \frac{z}{\rho} = \frac{\sqrt{2}}{2}$, so $\varphi = \frac{\pi}{4}$.
- $(x, y) = (1, -1)$ is in the fourth quadrant: $\theta = \frac{7\pi}{4}$.
Answer: $\left(2, \frac{7\pi}{4}, \frac{\pi}{4}\right)$
Common mistake
Mixing up $\theta$ and $\varphi$, or measuring $\varphi$ from the $xy$-plane. $\theta$ goes around the $z$-axis, exactly as in polar coordinates; $\varphi$ tilts down from the north pole. A point above the $xy$-plane has $\varphi < \frac{\pi}{2}$. Quick check: the sign of $z = \rho\cos\varphi$ must match your point. If you measured from the $xy$-plane (latitude), you would get $\frac{\pi}{2} - \varphi$ instead, and $z$ and $r$ swap roles.