Spherical Coordinates
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 5.5: “Triple Integrals in Cylindrical and Spherical Coordinates” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/5-5-triple-integrals-in-cylindrical-and-spherical-coordinates |
| Textbook used in class | Stewart, Calculus, Section 15.9: “Triple Integrals in Spherical Coordinates” |
Quick Reference
Spherical coordinates $(\rho, \phi, \theta)$:
- $\rho \geq 0$: distance from origin
- $0 \leq \phi \leq \pi$: polar angle from the positive $z$-axis
- $0 \leq \theta \leq 2\pi$: azimuthal angle (same as in cylindrical/polar)
Conversion: $$x = \rho\sin\phi\cos\theta, \quad y = \rho\sin\phi\sin\theta, \quad z = \rho\cos\phi.$$ $$\rho^2 = x^2+y^2+z^2, \quad r = \rho\sin\phi.$$
| Surface | Spherical equation |
|---|---|
| Sphere $x^2+y^2+z^2 = a^2$ | $\rho = a$ |
| Cone $z = \sqrt{x^2+y^2}$ | $\phi = \pi/4$ |
| Upper half-space $z > 0$ | $0 < \phi < \pi/2$ |
Motivation
Spherical coordinates are natural for problems involving spheres and cones. The key difference from cylindrical: $\rho$ measures the distance from the ORIGIN (not from the $z$-axis), and $\phi$ measures the angle down from the positive $z$-axis (like latitude measured from the North Pole, not from the equator).
The angle $\phi = 0$ gives the positive $z$-axis; $\phi = \pi/2$ gives the $xy$-plane; $\phi = \pi$ gives the negative $z$-axis.
Key Concept
Memorizing the formulas: Think of a “stacked” construction. First, $r = \rho\sin\phi$ is the horizontal distance from the $z$-axis (the cylindrical $r$). Then $x = r\cos\theta = \rho\sin\phi\cos\theta$ and $y = r\sin\theta = \rho\sin\phi\sin\theta$. And $z = \rho\cos\phi$.
The factor $\sin\phi$ appears in $x$ and $y$ because $\phi$ is measured from the $z$-axis; you must “project down” to the horizontal before applying the azimuthal angle $\theta$.
Worked Examples
Example 1. Convert the spherical point $(\rho,\phi,\theta) = (2, \pi/3, \pi/6)$ to Cartesian.
$x = 2\sin(\pi/3)\cos(\pi/6) = 2\cdot\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2} = \frac{3}{2}$.
$y = 2\sin(\pi/3)\sin(\pi/6) = 2\cdot\frac{\sqrt{3}}{2}\cdot\frac{1}{2} = \frac{\sqrt{3}}{2}$.
$z = 2\cos(\pi/3) = 2\cdot\frac{1}{2} = 1$.
Example 2. Describe the surface $\phi = \pi/4$ in Cartesian.
$z = \rho\cos(\pi/4) = \frac{\rho}{\sqrt{2}}$ and $r = \rho\sin(\pi/4) = \frac{\rho}{\sqrt{2}}$.
So $z = r = \sqrt{x^2+y^2}$, which is the upper half of the cone $z^2 = x^2+y^2$.
$\phi$ is the angle measured from the equator (like geographic latitude). In mathematics, $\phi$ is measured from the positive $z$-axis (the “North Pole”), NOT from the equator. So $\phi = 0$ is the North Pole, $\phi = \pi/2$ is the equator, and $\phi = \pi$ is the South Pole. Geographic latitude $L$ corresponds to $\phi = \pi/2 - L$.
Leveled Practice
Problem 1. What surface is described by $\rho = 2\cos\phi$?
Show answer
Multiply both sides by $\rho$: $\rho^2 = 2\rho\cos\phi = 2z$.
So $x^2+y^2+z^2 = 2z$, which gives $x^2+y^2+(z-1)^2 = 1$: a sphere of radius 1 centered at $(0,0,1)$.