Triple Integrals in Spherical Coordinates, multivariable and vector calculus lesson

Latitude and longitude locate any spot on the earth with two angles, and adding the distance from the center locates any point in space. Solids centered at a point, like a ball, a shell, or a scoop of ice cream on a cone, have one simple description in these coordinates. The price is a volume factor, and the most common mistakes come from mixing up which angle is which.

Key terms

The polar angle from the positive z-axis

Definition In a right-handed coordinate system, a point $P = (x, y, z)$ has spherical coordinates $(\rho, \theta, \phi)$ , where $\rho = \sqrt{x^2 + y^2 + z^2}$ is the distance from the origin, $\theta$ is the same angle as in cylindrical coordinates, measured counterclockwise from the positive $x$ -axis in the $xy$ -plane, and $\phi$ is the angle between the positive $z$ -axis and the segment from the origin to $P$ . The conventions are

\rho \ge 0, \quad 0 \le \theta < 2\pi, \quad 0 \le \phi \le \pi,

and the conversion is

\begin{gathered} x = \rho \sin \phi \cos \theta, \\[0.4em] y = \rho \sin \phi \sin \theta, \\[0.4em] z = \rho \cos \phi. \end{gathered}

The distance to the $z$ -axis is $r = \rho \sin \phi$ , which is never negative because $\sin \phi \ge 0$ on $[0, \pi]$ . Some books swap the names $\theta$ and $\phi$ , so always check the convention.

In plain words Stand at the origin. Turn by $\theta$ around the vertical axis, tip down from straight up by $\phi$ , then walk out a distance $\rho$ . An angle $\phi = 0$ points up, $\phi = \frac{\pi}{2}$ is horizontal and $\phi = \pi$ points down.

Example For the point $(1, 1, \sqrt{2})$ , the distance is $\rho = \sqrt{1 + 1 + 2} = 2$ . Then $\cos \phi = \frac{\sqrt{2}}{2}$ , so $\phi = \frac{\pi}{4}$ , and the shadow $(1, 1)$ gives $\theta = \frac{\pi}{4}$ . Check with $x = 2 \sin \frac{\pi}{4} \cos \frac{\pi}{4} = 1$ .

The volume element ρ² sin φ

Definition The map $(\rho, \phi, \theta) \mapsto (x, y, z)$ has Jacobian determinant

\frac{\partial(x, y, z)}{\partial(\rho, \phi, \theta)} = \rho^2 \sin \phi,

which is $\ge 0$ for $0 \le \phi \le \pi$ . Let $E$ be the solid given by

\begin{gathered} \alpha \le \theta \le \beta, \qquad c \le \phi \le d, \\[0.4em] g_1(\theta, \phi) \le \rho \le g_2(\theta, \phi), \end{gathered}

with $0 \le \beta - \alpha \le 2\pi$ , $0 \le c \le d \le \pi$ and $g_1$ , $g_2$ continuous with $0 \le g_1 \le g_2$ . If $f$ is continuous on $E$ , then

\begin{aligned} &\iiint_E f \, dV \\[0.4em] &= \int_\alpha^\beta \int_c^d \int_{g_1}^{g_2} f \, \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta, \end{aligned}

with $f$ evaluated at $(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi)$ . The map fails to be one-to-one only where $\rho = 0$ , where $\sin \phi = 0$ , and, when $\beta - \alpha = 2\pi$ , on the half-plane where $\theta = \alpha$ and $\theta = \beta$ meet. Those sets have zero volume, so they do not affect the integral.

In plain words A tiny spherical box has one side $\Delta \rho$ along the ray, one side $\rho \, \Delta \phi$ along a meridian, and one side $\rho \sin \phi \, \Delta \theta$ around a circle of latitude of radius $\rho \sin \phi$ . Multiplying gives $\rho^2 \sin \phi \, \Delta \rho \, \Delta \phi \, \Delta \theta$ . Boxes near the poles are thin, and the $\sin \phi$ accounts for that.

Example The ball of radius $2$ is $0 \le \rho \le 2$ , $0 \le \phi \le \pi$ , $0 \le \theta \le 2\pi$ . Including $\theta = 2\pi$ repeats only a half-plane, which has zero volume. The three integrals separate, so

\begin{aligned} V &= \int_0^{2\pi} d\theta \int_0^\pi \sin \phi \, d\phi \int_0^2 \rho^2 \, d\rho \\[0.4em] &= 2\pi \cdot 2 \cdot \frac{8}{3} = \frac{32\pi}{3}, \end{aligned}

which matches $\frac{4}{3}\pi \cdot 2^3$ .

Spherical regions

Definition The coordinate surfaces are the sphere $\rho = a$ for $a > 0$ , the half-plane $\theta = c$ bounded by the $z$ -axis, and the surface $\phi = c$ . For $0 < c < \frac{\pi}{2}$ it is the upper nappe of a cone with vertex at the origin, opening upward around the positive $z$ -axis. For $c = \frac{\pi}{2}$ it is the $xy$ -plane, for $\frac{\pi}{2} < c < \pi$ it is the lower nappe of a cone opening downward, and $c = 0$ or $c = \pi$ gives the nonnegative or nonpositive $z$ -axis. Typical spherical regions are the ball $\rho \le a$ , the shell $a \le \rho \le b$ , the solid cone $\phi \le c$ cut off by a sphere, and the ball $x^2 + y^2 + (z - a)^2 \le a^2$ with $a > 0$ , which touches the origin. Substituting the conversion formulas into its inequality gives $\rho^2 \le 2a\rho \cos \phi$ , that is

0 \le \rho \le 2a \cos \phi, \qquad 0 \le \phi \le \frac{\pi}{2}.

In plain words Spherical coordinates work best when the solid is cut out by spheres through or around the origin and by cones with vertex at the origin.

Example For the shifted ball $x^2 + y^2 + (z - 1)^2 \le 1$ , every ray from the origin with $\phi \le \frac{\pi}{2}$ starts at the origin, which lies on the sphere, runs through the ball and leaves it at $\rho = 2 \cos \phi$ . Rays with $\phi > \frac{\pi}{2}$ point below the $xy$ -plane and miss the ball. Nothing depends on $\theta$ , so the $\theta$ integral gives $2\pi$ and

\begin{aligned} V &= 2\pi \int_0^{\pi/2} \int_0^{2 \cos \phi} \rho^2 \sin \phi \, d\rho \, d\phi \\[0.4em] &= 2\pi \int_0^{\pi/2} \frac{8}{3} \cos^3 \phi \sin \phi \, d\phi \\[0.4em] &= 2\pi \cdot \frac{8}{3} \cdot \frac{1}{4} = \frac{4\pi}{3}, \end{aligned}

the volume of a ball of radius $1$ , as it should be.

Radial and angular bounds

Definition To set up bounds in the order $d\rho \, d\phi \, d\theta$ , first find the range of directions, the values of $\theta$ and $\phi$ for which the ray from the origin meets the solid. Then for each such direction find where the ray enters and leaves the solid. Those two values of $\rho$ are the radial bounds, and they may depend on $\phi$ and $\theta$ . Every point of the solid must be covered exactly once. A plane $z = h$ with $h > 0$ becomes $\rho = h \sec \phi$ for $0 \le \phi < \frac{\pi}{2}$ .

In plain words Shine a flashlight from the origin. The angles say which directions hit the solid, and the radial bounds say where along each beam the solid begins and ends.

Example The cap of the ball $\rho \le 2$ above the plane $z = 1$ is reached by rays with $\phi \le \frac{\pi}{3}$ , since the plane meets the sphere where $\cos \phi = \frac{1}{2}$ . Along such a ray the solid starts at the plane, $\rho = \sec \phi$ , and ends at the sphere, $\rho = 2$ . The $\theta$ integral again gives $2\pi$ , so

\begin{aligned} V &= 2\pi \int_0^{\pi/3} \int_{\sec \phi}^{2} \rho^2 \sin \phi \, d\rho \, d\phi \\[0.4em] &= \frac{2\pi}{3} \int_0^{\pi/3} \left( 8 - \sec^3 \phi \right) \sin \phi \, d\phi \\[0.4em] &= \frac{2\pi}{3} \left( 4 - \frac{3}{2} \right) = \frac{5\pi}{3}. \end{aligned}

The geometric cap formula, height $h = 1$ and sphere radius $R = 2$ , agrees.

\frac{\pi h^2 (3R - h)}{3} = \frac{\pi \cdot 1 \cdot 5}{3} = \frac{5\pi}{3}.

Symmetry

Definition If a solid $E$ and a continuous density $\delta$ are unchanged by rotation about the $z$ -axis, then $\delta$ does not depend on $\theta$ , the $\theta$ integral contributes $2\pi$ , and the center of mass lies on the $z$ -axis. If $E$ is symmetric under $(x, y, z) \mapsto (x, y, -z)$ , which is $\phi \mapsto \pi - \phi$ , and $f(x, y, -z) = -f(x, y, z)$ , then $\iiint_E f \, dV = 0$ . For a ball centered at the origin and an integrand that depends only on $\rho$ , both angular integrals give the total $4\pi$ .

In plain words Round solids let you skip whole integrals. Use symmetry to find the center of mass coordinates that must be zero, then compute only the one that is left.

Example The solid hemisphere $x^2 + y^2 + z^2 \le a^2$ , $z \ge 0$ , with constant density, is $0 \le \rho \le a$ , $0 \le \phi \le \frac{\pi}{2}$ . By symmetry $\bar{x} = \bar{y} = 0$ . Its volume is $\frac{2}{3}\pi a^3$ . The integrand for $M_{xy}$ is $z = \rho \cos \phi$ times the volume factor $\rho^2 \sin \phi$ , so

\begin{aligned} &M_{xy} \\[0.4em] &= 2\pi \int_0^{\pi/2} \int_0^a \rho^3 \cos \phi \sin \phi \, d\rho \, d\phi \\[0.4em] &= 2\pi \cdot \frac{1}{2} \cdot \frac{a^4}{4} = \frac{\pi a^4}{4}, \end{aligned}

so $\bar{z} = \frac{\pi a^4 / 4}{2\pi a^3 / 3} = \frac{3a}{8}$ . The balance point sits a bit below halfway up, since the hemisphere is wider at the bottom.

Check yourself

A ball of radius $2$ has density $\delta = \sqrt{x^2 + y^2 + z^2}$ . Find its mass.

Show the answer

The density is $\rho$ itself, and the angular integrals give $4\pi$ , so

m = 4\pi \int_0^2 \rho^3 \, d\rho = 16\pi.

Find the average distance from the center of a ball of radius $a$ to its points.

Show the answer

The integral of $\rho$ over the ball is $4\pi \cdot \frac{a^4}{4} = \pi a^4$ . Dividing by the volume $\frac{4}{3}\pi a^3$ gives $\frac{3a}{4}$ . Most of the volume is near the surface, so the average is well past $\frac{a}{2}$ .

Find the volume of the solid inside the sphere $x^2 + y^2 + z^2 = 9$ and inside the cone $z = \sqrt{3(x^2 + y^2)}$ .

Show the answer

On the cone, $\tan \phi = \frac{r}{z} = \frac{1}{\sqrt{3}}$ , so $\phi = \frac{\pi}{6}$ . The solid is $0 \le \rho \le 3$ , $0 \le \phi \le \frac{\pi}{6}$ , and

\begin{aligned} V &= 2\pi \left( 1 - \cos \frac{\pi}{6} \right) \cdot \frac{27}{3} \\[0.4em] &= 9\pi \left( 2 - \sqrt{3} \right). \end{aligned}

Triple Integrals in Spherical Coordinates

Key terms

The polar angle from the positive z-axis

The volume element ρ² sin φ

Spherical regions

Radial and angular bounds

Symmetry

Check yourself

How this lesson connects