A ball, a radial density, and a volume cell, integrals from easy to advanced lesson

A solid unit ball has density $x^2 + y^2 + z^2$ , the squared distance from its center, so it is light near the center and densest at the surface. What is its total mass? In rectangular coordinates every inner bound of the triple integral is a square root, but the ball and the density both depend only on the distance from the center, which suggests coordinates built from that distance and two angles. This page sets up spherical coordinates, derives the volume of a small spherical cell from its three edge lengths, and finds the mass $\tfrac{4\pi}{5}$ .

Key terms

Triple integral

Definition Let $E$ be a closed bounded solid whose boundary has zero volume, for example a solid bounded by finitely many smooth surfaces, and let $f$ be continuous on $E$ . Enclose $E$ in a box, set $f = 0$ outside $E$ , cut the box into small boxes with volumes $\Delta V_1, \dots, \Delta V_n$ and pick a point $P_i$ in each. The triple integral of $f$ over $E$ is the limit of the Riemann sums

\begin{aligned} &\iiint_E f\,dV \\[0.4em] &\quad = \lim_{h \to 0} \sum_{i=1}^{n} f(P_i)\,\Delta V_i, \end{aligned}

where $h$ is the largest diameter of the small boxes. Under these conditions the limit exists and does not depend on the choice of the points. When $E$ is described by bounds $a \le x \le b$ , $g_1(x) \le y \le g_2(x)$ and $k_1(x, y) \le z \le k_2(x, y)$ with continuous bounding functions, Fubini’s theorem lets the triple integral be computed as the iterated integral with those bounds. With $f = 1$ it gives the volume of $E$ .

In plain words Cut the solid into tiny pieces, multiply the value of $f$ in each piece by the volume of the piece, and add. When $f$ is a density, each product is the mass of a piece, and the sum is the total mass.

Example The unit ball $x^2 + y^2 + z^2 \le 1$ is described by

\begin{gathered} -1 \le x \le 1, \\[0.4em] \lvert y \rvert \le \sqrt{1 - x^2}, \\[0.4em] \lvert z \rvert \le \sqrt{1 - x^2 - y^2}. \end{gathered}

The innermost integral of $1$ gives the length $2\sqrt{1 - x^2 - y^2}$ of a vertical segment through the ball. The middle integral adds these lengths across the slice at a fixed $x$ , which is a disk of radius $\sqrt{1 - x^2}$ , and so it gives the area $\pi(1 - x^2)$ of that disk. Then

\begin{aligned} V &= \int_{-1}^{1} \pi (1 - x^2)\,dx \\[0.4em] &= \frac{4\pi}{3}. \end{aligned}

With the density $x^2 + y^2 + z^2$ in place of $1$ , the same bounds lead to a much longer calculation.

Spherical coordinates

Definition Spherical coordinates locate a point $P = (x, y, z)$ by its distance $r = \sqrt{x^2 + y^2 + z^2}$ from the origin, its polar angle $\phi$ in $[0, \pi]$ and its azimuth $\theta$ in $[0, 2\pi)$ . The next card describes the two angles. The conversion is

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

and $x^2 + y^2 + z^2 = r^2$ . Each point off the $z$ -axis has exactly one such triple, with $r > 0$ and $0 < \phi < \pi$ . On the $z$ -axis the azimuth can be anything, and at the origin both angles can be anything. This page writes $r$ for the distance from the origin. Many books call it $\rho$ and keep $r$ for the distance from the $z$ -axis.

In plain words The segment from the origin to $P$ has length $r$ . Tilting it by the angle $\phi$ away from the positive $z$ -axis gives it the height $z = r\cos\phi$ and a shadow in the $xy$ -plane of length $r\sin\phi$ , which is never negative because $\sin\phi \ge 0$ for $0 \le \phi \le \pi$ . The azimuth then splits the shadow into $x$ and $y$ exactly as polar coordinates do. Squaring and adding gives $r^2 \sin^2\phi\,(\cos^2\theta + \sin^2\theta) + r^2 \cos^2\phi = r^2$ .

Example The unit ball is the coordinate box $0 \le r \le 1$ , $0 \le \phi \le \pi$ , $0 \le \theta \le 2\pi$ , where including $\theta = 2\pi$ repeats only a half-plane, a set of zero volume. The density $x^2 + y^2 + z^2$ becomes $r^2$ . The upper half of the ball is $0 \le r \le 1$ , $0 \le \phi \le \tfrac{\pi}{2}$ , $0 \le \theta \le 2\pi$ , and the squared distance from the $z$ -axis, $x^2 + y^2$ , becomes $r^2 \sin^2\phi$ .

Polar angle and azimuth

Definition The polar angle $\phi$ of a point $P$ other than the origin $O$ is the angle between the positive $z$ -axis and the segment from $O$ to $P$ , so $\cos\phi = z/r$ and $0 \le \phi \le \pi$ . It is $0$ on the positive $z$ -axis, $\tfrac{\pi}{2}$ in the $xy$ -plane and $\pi$ on the negative $z$ -axis. The azimuth $\theta$ of a point off the $z$ -axis is the angle of its shadow $(x, y)$ in the $xy$ -plane, measured counterclockwise from the positive $x$ -axis as in polar coordinates, so $\cos\theta = x/(r\sin\phi)$ and $\sin\theta = y/(r\sin\phi)$ . Some books, especially in physics, swap the names and use $\theta$ for the polar angle and $\phi$ for the azimuth. The formulas on this page assume the convention described here.

In plain words Picture a globe with its north pole on the positive $z$ -axis. The polar angle says how far the point is from the north pole, from $0$ at the north pole through $\tfrac{\pi}{2}$ at the equator to $\pi$ at the south pole. The azimuth says how far around the axis the point is, like longitude. Geographic latitude is $\tfrac{\pi}{2} - \phi$ , measured up from the equator.

Example Take $P = (\tfrac{\sqrt3}{2}, \tfrac32, -1)$ . Its distance from the origin is $r = \sqrt{\tfrac34 + \tfrac94 + 1} = 2$ , and $\cos\phi = -\tfrac12$ gives $\phi = \tfrac{2\pi}{3}$ . The polar angle is larger than $\tfrac{\pi}{2}$ because $P$ lies below the $xy$ -plane. The shadow $(\tfrac{\sqrt3}{2}, \tfrac32)$ has length $\sqrt3 = 2\sin\tfrac{2\pi}{3}$ and lies in the first quadrant with $\tan\theta = \sqrt3$ , so $\theta = \tfrac{\pi}{3}$ . As a check, $z = 2\cos\tfrac{2\pi}{3} = -1$ and $x = \sqrt3 \cos\tfrac{\pi}{3} = \tfrac{\sqrt3}{2}$ .

Spherical volume element

Definition In spherical coordinates with the polar angle measured from the positive $z$ -axis, the volume element is

dV = r^2 \sin\phi\,dr\,d\phi\,d\theta.

Its factor is the Jacobian determinant of the map $(r, \phi, \theta) \mapsto (x, y, z)$ ,

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

which is never negative for $0 \le \phi \le \pi$ , so it equals its own absolute value. This is the three-dimensional case of the change of variables with the absolute Jacobian. Precisely, let $S$ be a closed bounded set of triples $(r, \phi, \theta)$ with $r \ge 0$ , $0 \le \phi \le \pi$ and $\theta$ in an interval of length at most $2\pi$ , whose boundary has zero volume, let $E$ be its image, and let $f$ be continuous on $E$ . Then

\begin{aligned} &\iiint_E f\,dV \\[0.4em] &= \iiint_S f\,r^2 \sin\phi\,dr\,d\phi\,d\theta, \end{aligned}

with $f$ evaluated at $(r\sin\phi\cos\theta, r\sin\phi\sin\theta, r\cos\phi)$ on the right. The map fails to be one-to-one only at the origin, on the $z$ -axis and, when $\theta$ runs over a full turn, on the half-plane where its two end values meet. Those sets have zero volume, so they do not change the integral.

In plain words A small cell between $r$ and $r + dr$ , $\phi$ and $\phi + d\phi$ , and $\theta$ and $\theta + d\theta$ is nearly a box with three perpendicular edges. The radial edge has length $dr$ . The edge along a meridian is an arc of a circle of radius $r$ , with length about $r\,d\phi$ . The edge along a circle of latitude is an arc of a circle of radius $r\sin\phi$ , the distance from the $z$ -axis, with length about $r\sin\phi\,d\theta$ . The product of the three lengths is $r^2 \sin\phi\,dr\,d\phi\,d\theta$ , and the curved shape of the cell changes its volume only by an amount that is small compared with this product as the steps shrink. The factor $\sin\phi$ makes cells with the same angular steps thin near the poles and largest at the equator.

Example Let $B$ be the unit ball $x^2 + y^2 + z^2 \le 1$ , including its interior, with density $x^2 + y^2 + z^2$ . Its mass $M$ is

\iiint_B (x^2 + y^2 + z^2)\,dV.

The ball is the coordinate box $0 \le r \le 1$ , $0 \le \phi \le \pi$ , $0 \le \theta \le 2\pi$ , with $\phi$ measured down from the positive $z$ -axis and $\theta$ the azimuth, and the density is continuous and nonnegative. The density contributes $r^2$ and the volume element contributes $r^2 \sin\phi$ , so $M$ is the iterated integral

\int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^1 r^4 \sin\phi\,dr\,d\phi\,d\theta.

The bounds are constants and the integrand is a function of $r$ times a function of $\phi$ , so the integral is the product of three single integrals,

\begin{gathered} \int_0^1 r^4\,dr = \frac15, \\[0.4em] \int_0^{\pi} \sin\phi\,d\phi = 2, \\[0.4em] \int_0^{2\pi} d\theta = 2\pi. \end{gathered}

Therefore

M = \frac15 \cdot 2 \cdot 2\pi = \frac{4\pi}{5}.

The two factors $r^2$ in $r^4$ play different roles. One is the density of the material, and the other is the size of the cell.

Radial density

Definition A density on a solid is radial about a point $C$ when its value at each point $P$ depends only on the distance from $P$ to $C$ , that is, when it equals $g(d)$ for a function $g$ of one variable, where $d$ is that distance. With $C$ at the origin, a radial density is $g(r)$ in spherical coordinates. For the ball of radius $R > 0$ centered at the origin and a continuous function $g \ge 0$ on $[0, R]$ , the angular integrals give the area of the unit sphere,

\int_0^{2\pi} \int_0^{\pi} \sin\phi\,d\phi\,d\theta = 4\pi,

and the mass is

M = 4\pi \int_0^R g(r)\,r^2\,dr.

In plain words All material at the same distance from the center has the same density, so the ball can be cut into thin spherical shells. The shell between $r$ and $r + dr$ has volume about $4\pi r^2\,dr$ and density about $g(r)$ , and the mass adds up the shells. The shortcut needs a ball or a shell centered at the same point as the density.

Example The density $x^2 + y^2 + z^2$ is radial with $g(r) = r^2$ , and the formula gives $4\pi \int_0^1 r^4\,dr = \tfrac{4\pi}{5}$ again. The density $1 - (x^2 + y^2 + z^2)$ is also radial, with $g(r) = 1 - r^2$ , and it is highest at the center. On the unit ball it gives the mass

\begin{aligned} M &= 4\pi \int_0^1 (1 - r^2)\,r^2\,dr \\[0.4em] &= 4\pi\Bigl(\frac13 - \frac15\Bigr) = \frac{8\pi}{15}. \end{aligned}

The density $x^2 + y^2$ is not radial. The points $(1, 0, 0)$ and $(0, 0, 1)$ are both at distance $1$ from the origin, but their densities are $1$ and $0$ .

Mass from density

Definition Let a solid $E$ with positive volume $V$ carry a continuous density $\delta \ge 0$ , measured as mass per unit volume. A small piece of volume $\Delta V$ around a point $P$ has mass about $\delta(P)\,\Delta V$ , and the Riemann sums of $\delta$ add these masses, so the mass of $E$ is

M = \iiint_E \delta\,dV.

The average density is $M / V$ . It lies between the smallest and the largest value of $\delta$ on $E$ , which gives a quick check on a computed mass.

In plain words Density says how much mass a unit of volume holds near a point. Integrating it over the solid adds up the masses of all the small pieces. Dividing the total by the volume gives the constant density that would produce the same mass.

Example For the ball $B$ , $M = \tfrac{4\pi}{5}$ and $V = \tfrac{4\pi}{3}$ , so the average density is $\tfrac35$ . The density $r^2$ runs from $0$ at the center to $1$ on the surface, and it is less than $1$ inside, so $0 < M < \tfrac{4\pi}{3}$ , as found. The average $\tfrac35$ is well above $\tfrac13$ , the average of $r^2$ when the radius is chosen uniformly from $[0, 1]$ , because the outer shells hold most of the volume. The outer half of the radius, $\tfrac12 \le r \le 1$ , contains $\tfrac78$ of the volume and $\tfrac{31}{32}$ of the mass. As another check, the densities $r^2$ and $1 - r^2$ add to $1$ , and their masses $\tfrac{4\pi}{5}$ and $\tfrac{8\pi}{15}$ add to $\tfrac{4\pi}{3}$ , the volume of the ball.

Check yourself

The spherical shell $1 \le x^2 + y^2 + z^2 \le 4$ has density $1/(x^2 + y^2 + z^2)$ . Find its mass.

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The shell is $1 \le r \le 2$ , $0 \le \phi \le \pi$ , $0 \le \theta \le 2\pi$ , and the density is $1/r^2$ , which is continuous on the shell because the origin is not in it. The density cancels the $r^2$ of the volume element, so the mass is

\int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_1^2 \sin\phi\,dr\,d\phi\,d\theta.

The $r$ interval has length $1$ , the $\phi$ integral is $2$ and the $\theta$ integral is $2\pi$ , so the mass is $1 \cdot 2 \cdot 2\pi = 4\pi$ . Every thin shell of thickness $dr$ carries the same mass $4\pi\,dr$ , since its volume $4\pi r^2\,dr$ is multiplied by the density $1/r^2$ .

A classmate computes the volume of the unit ball with $\phi$ running from $0$ to $2\pi$ and $\theta$ from $0$ to $\pi$ , as the integral below, and gets $0$ . What went wrong?

\int_0^{\pi}\!\!\int_0^{2\pi}\!\!\int_0^1 r^2 \sin\phi\,dr\,d\phi\,d\theta

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The integral over $\phi$ is $\int_0^{2\pi} \sin\phi\,d\phi = 0$ , so the result is $0$ , which cannot be the volume of a ball. These ranges do reach every point, since the triple $(r, \phi, \theta)$ with $\pi < \phi < 2\pi$ names the same point as $(r, 2\pi - \phi, \theta + \pi)$ , and apart from a set of zero volume each point is reached once. The trouble is the sign. For $\pi < \phi < 2\pi$ the Jacobian $r^2 \sin\phi$ is negative, while the change of variables formula uses its absolute value, and $r^2 \sin\phi$ equals its absolute value only when $0 \le \phi \le \pi$ . With the convention $0 \le \phi \le \pi$ and $0 \le \theta \le 2\pi$ the volume is

\int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^1 r^2 \sin\phi\,dr\,d\phi\,d\theta,

which equals $\tfrac13 \cdot 2 \cdot 2\pi = \tfrac{4\pi}{3}$ .

Find the integral of $z^2$ over the unit ball $B$ , and check the answer with the symmetry of the ball and the mass $\tfrac{4\pi}{5}$ found above.

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In spherical coordinates $z^2 = r^2 \cos^2\phi$ , so on the box $0 \le r \le 1$ , $0 \le \phi \le \pi$ , $0 \le \theta \le 2\pi$ the integrand times the volume factor is $r^4 \cos^2\phi \sin\phi$ . The bounds are constants and this is a function of $r$ times a function of $\phi$ , so the integral is the product of three single integrals,

\begin{gathered} \int_0^1 r^4\,dr = \frac15, \\[0.4em] \int_0^{\pi} \cos^2\phi \sin\phi\,d\phi = \frac23, \\[0.4em] \int_0^{2\pi} d\theta = 2\pi. \end{gathered}

The middle one comes from the substitution $w = \cos\phi$ , with $dw = -\sin\phi\,d\phi$ . It sends $\phi = 0$ to $w = 1$ and $\phi = \pi$ to $w = -1$ and turns the integral into $\int_{-1}^{1} w^2\,dw = \tfrac23$ . The result is

\frac15 \cdot \frac23 \cdot 2\pi = \frac{4\pi}{15}.

The ball looks the same from every axis, so the integrals of $x^2$ , $y^2$ and $z^2$ over it are equal. Their sum is the integral of $x^2 + y^2 + z^2$ , which is $\tfrac{4\pi}{5}$ , so each one is $\tfrac13 \cdot \tfrac{4\pi}{5} = \tfrac{4\pi}{15}$ , in agreement.

A ball, a radial density, and a volume cell

Key terms

Triple integral

Spherical coordinates

Polar angle and azimuth

Spherical volume element

Radial density

Mass from density

Check yourself

How this lesson connects

Builds on

Used later in