Triple Integrals in Spherical
Triple Integrals in 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” (Examples 1, 2, 3) |
Quick Reference
Volume element: $$dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta.$$
Full sphere of radius $a$: $0 \leq \rho \leq a$, $0 \leq \phi \leq \pi$, $0 \leq \theta \leq 2\pi$.
Upper half of sphere: replace $0 \leq \phi \leq \pi$ with $0 \leq \phi \leq \pi/2$.
“Ice cream cone” (sphere above cone $\phi = \phi_0$): $0 \leq \phi \leq \phi_0$, $0 \leq \rho \leq a$.
Motivation
Spherical coordinates are optimal when the solid is a sphere, part of a sphere, or bounded by a cone. The expression $\rho^2 = x^2+y^2+z^2$ collapses a complicated Cartesian integrand into a simple function of $\rho$. The limits for a sphere become three constant ranges: $\rho \in [0,a]$, $\phi \in [0,\pi]$, $\theta \in [0,2\pi]$.
The factor $\rho^2\sin\phi$ in $dV$ is the Jacobian of the spherical transformation. It reflects that spherical “wedges” near the origin are small (small $\rho^2$) and that slices near the poles ($\phi \approx 0$ or $\pi$) are also small (small $\sin\phi$).
Key Concept
Do NOT forget $\rho^2\sin\phi$. The two most common errors in spherical integrals:
- Writing $dV = d\rho\,d\phi\,d\theta$ (missing the entire Jacobian factor).
- Writing $dV = \rho^2\,d\rho\,d\phi\,d\theta$ (forgetting the $\sin\phi$).
Derive it once from the Jacobian, then memorize it.
Worked Examples
Example 1. Find the volume of the solid ball of radius $a$.
$$V = \int_0^{2\pi}\int_0^\pi\int_0^a \rho^2\sin\phi\,d\rho\,d\phi\,d\theta.$$
$$= \int_0^{2\pi}d\theta\cdot\int_0^\pi\sin\phi\,d\phi\cdot\int_0^a\rho^2\,d\rho = 2\pi\cdot[-\cos\phi]_0^\pi\cdot\frac{a^3}{3} = 2\pi\cdot 2\cdot\frac{a^3}{3} = \frac{4\pi a^3}{3}.$$
This confirms the familiar formula $V = \frac{4}{3}\pi r^3$.
Example 2. Evaluate $\iiint_E e^{(x^2+y^2+z^2)^{3/2}}\,dV$ over the ball $\rho \leq 1$. (Stewart 15.9, Example 2.)
In spherical: $(x^2+y^2+z^2)^{3/2} = \rho^3$.
$$\int_0^{2\pi}\int_0^\pi\int_0^1 e^{\rho^3}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot 2\cdot\int_0^1 e^{\rho^3}\rho^2\,d\rho.$$
Let $u = \rho^3$: $= 4\pi\cdot\frac{1}{3}[e^u]_0^1 = \frac{4\pi(e-1)}{3}$.
the $\phi$ limit for the upper half of a sphere is $\pi$ (not $\pi/2$). For the FULL sphere, $\phi$ runs from $0$ to $\pi$. For the UPPER hemisphere (where $z \geq 0$), $\phi$ runs from $0$ to $\pi/2$ (since $\phi = \pi/2$ is the equator). For the lower hemisphere ($z \leq 0$), $\phi$ runs from $\pi/2$ to $\pi$. Getting these limits wrong reflects confusion about whether $\phi$ is measured from the pole or the equator.
Leveled Practice
Problem 1. Find the volume of the region inside the sphere $\rho = 2$ and above the cone $\phi = \pi/3$.
Show answer
Limits: $0 \leq \rho \leq 2$, $0 \leq \phi \leq \pi/3$, $0 \leq \theta \leq 2\pi$.
$$V = \int_0^{2\pi}\int_0^{\pi/3}\int_0^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta.$$
$$= 2\pi\cdot[-\cos\phi]_0^{\pi/3}\cdot\frac{8}{3} = 2\pi\cdot(1-1/2)\cdot\frac{8}{3} = 2\pi\cdot\frac{1}{2}\cdot\frac{8}{3} = \frac{8\pi}{3}.$$