---
title: Triple Integrals and Coordinate Systems
module: Multiple Integrals and Vector Calculus
moduleNumber: 12
lessonNumber: 2
order: 1202
summary: >
  The triple integral integrates a function of three variables over a solid, as
  a limit of Riemann sums evaluated by three nested single integrations.
  Cylindrical coordinates add the factor r to handle axial symmetry, spherical
  coordinates add rho-squared sine-phi for radial symmetry, and the general
  change of variables shows both volume elements are Jacobian determinants of
  the coordinate map. Surface area for a graph completes the measurement toolkit.
topics: [Multiple Integrals and Vector Calculus]
draft: false
sources:
  - book: Stewart
    ref: "Ch. 15 — Multiple Integrals; §15.5 Surface Area, §15.6 Triple Integrals, §15.7 Triple Integrals in Cylindrical Coordinates"
  - book: Stewart
    ref: "§15.8 Triple Integrals in Spherical Coordinates, §15.9 Change of Variables in Multiple Integrals"
---

The [double integral](/calculus/multiple-integrals-and-vector-calculus/double-integrals)
sums a density over a plane region; the triple integral sums a density over a
solid. The definition, Fubini's Theorem, and the region descriptions all carry
over with one more nesting. The new material is three coordinate systems
for three-dimensional space and the general rule that ties them together: under
a change of variables, the volume element scales by the absolute value of a
determinant, the **Jacobian**. Cylindrical and spherical coordinates are the two
most useful cases, and both are instances of that rule.

## Surface area of a graph

Before the triple integral, one loose end from the double integral: the area of a
curved surface $z = f(x,y)$ lying above a region $D$. Cut $D$ into subrectangles;
above each, the surface is nearly a slanted parallelogram, the piece of the
tangent plane. Its two edge vectors are $(\Delta x, 0, f_x\,\Delta x)$ and
$(0, \Delta y, f_y\,\Delta y)$, and the area of the parallelogram they span is the
magnitude of their cross product, $\sqrt{f_x^2 + f_y^2 + 1}\;\Delta x\,\Delta y$.
Summing and taking the limit gives the area.

> **Definition (Surface area of a graph).** If $f$ has continuous partial
> derivatives on $D$, the area of the surface $z = f(x,y)$ over $D$ is
> $$
> A(S) = \iint\limits_D \sqrt{\,[f_x(x,y)]^2 + [f_y(x,y)]^2 + 1\,}\;\d A .
> $$

The integrand exceeds $1$ wherever the surface tilts, so the surface is always at
least as large as its shadow $D$, with equality only for a flat horizontal piece.
This formula is the two-variable analog of the arc-length integral
$\int \sqrt{1 + (f')^2}\,\d x$; the
[parametric-surface](/calculus/multiple-integrals-and-vector-calculus/surface-integrals)
form generalizes it to surfaces that are not graphs.

## The triple integral

Let $f(x,y,z)$ be defined on a rectangular box
$B = [a,b] \times [c,d] \times [r,s]$. Partition each edge, cutting $B$ into
sub-boxes $B_{ijk}$ of volume $\Delta V = \Delta x\,\Delta y\,\Delta z$, choose a
sample point in each, and form the triple Riemann sum. The limit defines the
integral.

> **Definition (Triple integral).** The **triple integral** of $f$ over the box
> $B$ is
> $$
> \iiint\limits_B f(x,y,z)\,\d V = \lim_{l,m,n \to \infty} \sum_{i=1}^{l} \sum_{j=1}^{m} \sum_{k=1}^{n} f(x_{ijk}^\ast, y_{ijk}^\ast, z_{ijk}^\ast)\,\Delta V,
> $$
> when the limit exists. It always exists if $f$ is continuous on $B$.

Fubini's Theorem again reduces the integral to nested single integrations, in any
of the six orders:

$$
\iiint\limits_B f(x,y,z)\,\d V = \int_a^b \int_c^d \int_r^s f(x,y,z)\,\d z\,\d y\,\d x .
$$

Over a general bounded solid $E$, enclose $E$ in a box, extend $f$ by zero, and
the inner limits become surfaces. A solid $E$ is of **type 1** if it sits between
two graphs $z = u_1(x,y)$ and $z = u_2(x,y)$ over a plane region $D$:

$$
\iiint\limits_E f\,\d V = \iint\limits_D \left[ \int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z)\,\d z \right] \d A .
$$

The outer $\iint\limits_D$ is itself a double integral, evaluated as type I or type II.
So the three limits are read off in order: the innermost from the two bounding
surfaces, the middle two from the shadow $D$ of the solid on a coordinate plane.

$$
% caption: A type 1 solid sits between a lower surface $z=u_1$ and an upper
% surface $z=u_2$ over its shadow $D$; the inner $z$-integral runs between the
% surfaces, the outer double integral sweeps $D$.
\begin{tikzpicture}[scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% shadow D in the base plane (oblique)
\fill[black] (0.6,0.4) -- (3.4,0.1) -- (4.6,1.0) -- (1.8,1.3) -- cycle;
\draw[black] (0.6,0.4) -- (3.4,0.1) -- (4.6,1.0) -- (1.8,1.3) -- cycle;
\node[font=\small] at (2.6,0.7) {$D$};
% lower surface u1 (near base) and upper surface u2 (raised)
\draw[acc, thick] (0.6,1.1) .. controls (2.3,0.8) and (3.6,1.0) .. (4.6,1.7);
\draw[acc, thick] (0.6,2.9) .. controls (2.3,3.4) and (3.6,2.9) .. (4.6,3.3);
\node[acc, font=\small, anchor=south] at (4.4,3.3) {$z=u_2$ (upper)};
\node[acc, font=\small, anchor=north] at (4.4,1.65) {$z=u_1$ (lower)};
% a vertical z-strip between surfaces
\draw[black, very thick, ->] (2.5,1.15) -- (2.5,3.05);
\node[font=\small, anchor=west] at (2.55,2.1) {$dz$};
% projection dashed
\draw[black, dashed] (2.5,1.15) -- (2.5,0.55);
\fill[black] (2.5,0.55) circle (1.3pt);
\node[font=\small, anchor=north] at (3.15,0.5) {$dA$ in $D$};
\end{tikzpicture}
$$

The most useful reading of $\iiint\limits_E 1\,\d V$ is the volume of $E$, since the
constant integrand $1$ counts $\Delta V$ over every sub-box. With a density
$\rho(x,y,z)$, the same three definitions from the plane repeat: mass
$m = \iiint\limits_E \rho\,\d V$, moments about the coordinate planes, and center of mass
$(\bar{x},\bar{y},\bar{z})$ with $\bar{x} = \tfrac{1}{m}\iiint\limits_E x\rho\,\d V$, and
similarly for the other two coordinates.

> **Worked example.** Find the volume of the tetrahedron $E$ bounded by the plane
> $x + y + z = 1$ and the coordinate planes. Its shadow $D$ on the $xy$-plane is the
> triangle $0 \le x \le 1$, $0 \le y \le 1 - x$; above it, $z$ runs from $0$ to
> $1 - x - y$:
>
> $$
> V = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} \d z\,\d y\,\d x
> = \int_0^1 \int_0^{1-x} (1 - x - y)\,\d y\,\d x .
> $$
>
> The inner integral is $\big[(1-x)y - \tfrac{y^2}{2}\big]_0^{1-x} = \tfrac{1}{2}(1-x)^2$,
> so $V = \int_0^1 \tfrac{1}{2}(1-x)^2\,\d x = \tfrac{1}{6}$.

> **Worked example (center of mass).** Take the same tetrahedron with constant
> density $\rho$. By symmetry of the region and the density under permuting
> $x, y, z$, the three coordinates of the center of mass are equal, so it is enough
> to compute one. The mass is $m = \rho V = \tfrac{1}{6}\rho$, and the moment about
> the $yz$-plane is
>
> $$
> M_{yz} = \rho\int_0^1 \int_0^{1-x} \int_0^{1-x-y} x\,\d z\,\d y\,\d x
> = \rho\int_0^1 x\cdot\tfrac{1}{2}(1-x)^2\,\d x = \tfrac{1}{24}\rho,
> $$
>
> using $\int_0^1 x(1-x)^2\,\d x = \tfrac{1}{12}$. Then
> $\bar{x} = M_{yz}/m = \tfrac{1/24}{1/6} = \tfrac{1}{4}$, and likewise
> $\bar{y} = \bar{z} = \tfrac{1}{4}$. The center of mass is
> $\big(\tfrac14, \tfrac14, \tfrac14\big)$, the average of the four vertices, as a
> constant density requires.

## Cylindrical coordinates

Cylindrical coordinates keep $z$ and put the other two variables in polar form:
$x = r\cos\theta$, $y = r\sin\theta$, $z = z$. They fit solids with an axis of
symmetry — cylinders, cones, paraboloids of revolution — because the axis becomes
$r = 0$ and circular cross-sections become $r = \text{const}$. The volume element
inherits the polar factor: the base patch has area $r\,\d r\,\d\theta$ and the box
above it has height $\d z$, so

$$
\d V = r\,\d z\,\d r\,\d\theta .
$$

$$
% caption: The cylindrical volume element is a wedge of base $r\,dr\,d\theta$ and
% height $dz$; its volume $r\,dz\,dr\,d\theta$ carries the same factor $r$ as the
% polar area element.
\begin{tikzpicture}[scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% axes
\draw[->, black] (0,0) -- (4.6,0) node[right, font=\small] {$x$};
\draw[->, black] (0,0) -- (1.4,1.0) node[right, font=\small] {$y$};
\draw[->, black] (0,0) -- (0,4.2) node[above, font=\small] {$z$};
% base wedge (polar patch) on the xy plane, in oblique
\fill[acc!12] (2.1,0.35) -- (2.9,0.5) -- (3.15,0.95) -- (2.3,0.8) -- cycle;
\draw[acc, thick] (2.1,0.35) -- (2.9,0.5) -- (3.15,0.95) -- (2.3,0.8) -- cycle;
% raise the wedge to height dz
\fill[acc!8] (2.1,2.35) -- (2.9,2.5) -- (3.15,2.95) -- (2.3,2.8) -- cycle;
\draw[acc, thick] (2.1,2.35) -- (2.9,2.5) -- (3.15,2.95) -- (2.3,2.8) -- cycle;
\draw[acc, thick] (2.1,0.35) -- (2.1,2.35);
\draw[acc, thick] (2.9,0.5) -- (2.9,2.5);
\draw[acc, thick] (3.15,0.95) -- (3.15,2.95);
\draw[acc, thick] (2.3,0.8) -- (2.3,2.8);
% labels
\draw[<->, black] (3.25,0.6) -- (3.25,2.6);
\node[font=\small, anchor=west] at (3.3,1.6) {$dz$};
\node[font=\small, anchor=north] at (2.5,0.3) {base wedge};
% r and theta at origin
\draw[black, dashed] (0,0) -- (2.5,0.6);
\node[font=\small, anchor=south] at (1.55,0.5) {$r$};
\draw[black] (0.7,0) arc (0:14:0.7);
\node[font=\small, anchor=west] at (0.85,-0.12) {angle};
\end{tikzpicture}
$$

> **Formula (Cylindrical triple integral).** If $E = \{(r,\theta,z) : \alpha \le \theta \le \beta,\ h_1(\theta) \le r \le h_2(\theta),\ u_1(r,\theta) \le z \le u_2(r,\theta)\}$,
> then
> $$
> \iiint\limits_E f(x,y,z)\,\d V = \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} \int_{u_1(r,\theta)}^{u_2(r,\theta)} f(r\cos\theta, r\sin\theta, z)\,r\,\d z\,\d r\,\d\theta .
> $$

> **Worked example.** Find the volume of the solid bounded below by the paraboloid
> $z = x^2 + y^2$ and above by the plane $z = 4$. In cylindrical coordinates
> $z = r^2$ below and $z = 4$ above; the shadow is the disk $r \le 2$:
>
> $$
> V = \int_0^{2\pi} \int_0^2 \int_{r^2}^{4} r\,\d z\,\d r\,\d\theta
> = \int_0^{2\pi}\int_0^2 (4 - r^2)\,r\,\d r\,\d\theta
> = 2\pi\left[ 2r^2 - \tfrac{r^4}{4} \right]_0^2 = 2\pi(8 - 4) = 8\pi .
> $$

## Spherical coordinates

Spherical coordinates measure a point by its distance from the origin and two
angles: $\rho \ge 0$ the distance, $\theta$ the same longitude angle as in
cylindrical coordinates, and $\phi \in [0,\pi]$ the angle down from the positive
$z$-axis (the colatitude). The conversions are

$$
x = \rho\sin\phi\cos\theta, \qquad y = \rho\sin\phi\sin\theta, \qquad z = \rho\cos\phi,
$$

and $\rho^2 = x^2 + y^2 + z^2$. Spheres are $\rho = \text{const}$, cones from the
origin are $\phi = \text{const}$, half-planes off the $z$-axis are
$\theta = \text{const}$. A spherical volume element is a box with edges $\d\rho$
radially, $\rho\,\d\phi$ along a meridian, and $\rho\sin\phi\,\d\theta$ around a
circle of latitude (radius $\rho\sin\phi$). Their product is the volume element.

$$
% caption: The spherical volume element has radial edge $d\rho$, meridian edge
% $\rho\,d\phi$, and latitude edge $\rho\sin\phi\,d\theta$, giving
% $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$.
\begin{tikzpicture}[scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (4.4,0) node[right, font=\small] {$x$};
\draw[->, black] (0,0) -- (1.3,0.95) node[right, font=\small] {$y$};
\draw[->, black] (0,0) -- (0,4.2) node[above, font=\small] {$z$};
% a spherical patch (small curved box) at mid-latitude
\fill[acc!12] (2.5,2.4) -- (3.15,2.15) -- (3.35,2.6) -- (2.7,2.85) -- cycle;
\draw[acc, thick] (2.5,2.4) -- (3.15,2.15) -- (3.35,2.6) -- (2.7,2.85) -- cycle;
% radial line from origin
\draw[black] (0,0) -- (3.0,2.5);
\node[font=\small, anchor=south east] at (1.55,1.35) {radius};
% phi angle from z-axis
\draw[black] (0,1.1) arc (90:39:1.1);
\node[font=\small] at (0.5,1.4) {colatitude};
% theta angle in base
\draw[black, dashed] (0,0) -- (2.9,0.7);
\draw[black] (0.85,0) arc (0:13:0.85);
\node[font=\small] at (1.15,0.1) {angle};
% edge labels
\draw[<->, black] (2.62,2.55) -- (3.28,2.3);
\node[font=\small, anchor=south] at (3.4,2.6) {radial};
\node[font=\small, anchor=west] at (3.4,2.3) {meridian};
\node[font=\small, anchor=north] at (2.95,2.02) {latitude};
\end{tikzpicture}
$$

> **Formula (Spherical triple integral).** Over a spherical region
> $E = \{(\rho,\theta,\phi) : a \le \rho \le b,\ \alpha \le \theta \le \beta,\ c \le \phi \le d\}$,
> $$
> \iiint\limits_E f\,\d V = \int_c^d \int_\alpha^\beta \int_a^b f(\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi)\,\rho^2\sin\phi\,\d\rho\,\d\theta\,\d\phi .
> $$

The factor $\rho^2\sin\phi$ is the spherical volume element; $\rho^2$ because area
grows with the square of the radius, $\sin\phi$ because circles of latitude
shrink toward the poles.

> **Worked example.** The volume of a ball of radius $a$: the region is
> $0 \le \rho \le a$, $0 \le \theta \le 2\pi$, $0 \le \phi \le \pi$, so
>
> $$
> V = \int_0^\pi \int_0^{2\pi} \int_0^a \rho^2\sin\phi\,\d\rho\,\d\theta\,\d\phi
> = \left(\int_0^a \rho^2\,\d\rho\right)\left(\int_0^{2\pi} \d\theta\right)\left(\int_0^\pi \sin\phi\,\d\phi\right)
> = \tfrac{a^3}{3}\cdot 2\pi \cdot 2 = \tfrac{4}{3}\pi a^3 .
> $$

The three factors separate because every limit is constant, recovering the
familiar ball volume.

## The general change of variables and the Jacobian

Polar, cylindrical, and spherical coordinates are three instances of one theorem.
A change of variables is a map $T$ from a region $S$ in $uv$-space to a region $D$
in $xy$-space, $x = x(u,v)$, $y = y(u,v)$. A small rectangle in $uv$-space with
sides $\Delta u$ and $\Delta v$ maps to a small parallelogram in $xy$-space
spanned by the vectors $(x_u, y_u)\,\Delta u$ and $(x_v, y_v)\,\Delta v$. The area
of that parallelogram is the absolute value of a cross product, which is the
determinant

$$
\frac{\partial(x,y)}{\partial(u,v)} = \det \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = x_u y_v - x_v y_u .
$$

> **Definition (Jacobian).** The **Jacobian** of a transformation $x = x(u,v)$,
> $y = y(u,v)$ is the determinant
> $$
> \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[2mm] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix} .
> $$
> Its absolute value is the local area-scaling factor of the map.

$$
% caption: A transformation carries a unit cell in $uv$-space to a parallelogram
% in $xy$-space; the Jacobian determinant is the ratio of the two areas, the
% factor $dA$ picks up under the change of variables.
\begin{tikzpicture}[scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% uv square (left)
\draw[->, black] (-0.3,0) -- (2.6,0) node[right, font=\small] {$u$};
\draw[->, black] (0,-0.3) -- (0,2.6) node[above, font=\small] {$v$};
\fill[acc!12] (0.7,0.7) rectangle (1.6,1.6);
\draw[acc, thick] (0.7,0.7) rectangle (1.6,1.6);
\node[font=\small] at (1.15,1.15) {$du\,dv$};
\node[font=\small, anchor=north] at (1.15,-0.35) {$uv$-space $S$};
% arrow T
\draw[->, acc, very thick] (3.0,1.3) -- (4.4,1.3);
\node[acc, font=\small, anchor=south] at (3.7,1.4) {$T$};
% xy parallelogram (right)
\begin{scope}[xshift=5.0cm]
  \draw[->, black] (-0.3,0) -- (3.2,0) node[right, font=\small] {$x$};
  \draw[->, black] (0,-0.3) -- (0,2.6) node[above, font=\small] {$y$};
  \fill[acc!12] (0.6,0.6) -- (2.0,0.9) -- (2.5,1.9) -- (1.1,1.6) -- cycle;
  \draw[acc, thick] (0.6,0.6) -- (2.0,0.9) -- (2.5,1.9) -- (1.1,1.6) -- cycle;
  \node[font=\small, align=center] at (1.55,1.15) {stretched\\cell};
  \node[font=\small, anchor=north] at (1.4,-0.35) {$xy$-space $D$};
\end{scope}
\end{tikzpicture}
$$

> **Theorem (Change of variables).** If $T$ is a one-to-one $C^1$ transformation
> with nonzero Jacobian mapping $S$ onto $D$, and $f$ is continuous, then
> $$
> \iint\limits_D f(x,y)\,\d A = \iint\limits_S f\big(x(u,v), y(u,v)\big)\left| \frac{\partial(x,y)}{\partial(u,v)} \right| \d u\,\d v .
> $$

Polar coordinates are the case $x = r\cos\theta$, $y = r\sin\theta$. The Jacobian
is

$$
\frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r,
$$

matching the factor $r$ in $\d A = r\,\d r\,\d\theta$. The three-variable
version replaces the $2\times 2$ determinant with a $3\times 3$ one; computing it
for the spherical map $x = \rho\sin\phi\cos\theta$, etc., gives
$\partial(x,y,z)/\partial(\rho,\theta,\phi) = \rho^2\sin\phi$, and for the
cylindrical map it gives $r$. Each volume element is a Jacobian.

| Coordinate map | Element | Jacobian | Best for |
| --- | --- | --- | --- |
| Cartesian | $\d A = \d x\,\d y$ | $1$ | rectangles, straight boundaries |
| polar $(r,\theta)$ | $\d A = r\,\d r\,\d\theta$ | $r$ | disks, sectors, $x^2+y^2$ integrands |
| cylindrical $(r,\theta,z)$ | $\d V = r\,\d z\,\d r\,\d\theta$ | $r$ | axial symmetry: cylinders, cones |
| spherical $(\rho,\theta,\phi)$ | $\d V = \rho^2\sin\phi\,\d\rho\,\d\theta\,\d\phi$ | $\rho^2\sin\phi$ | radial symmetry: balls, spherical caps |

> **Worked example (custom substitution).** Evaluate $\iint\limits_D e^{(x+y)/(x-y)}\,\d A$
> over the trapezoid $D$ with vertices $(1,0)$, $(2,0)$, $(0,-2)$, $(0,-1)$. The
> integrand suggests $u = x + y$, $v = x - y$; then $x = \tfrac{1}{2}(u+v)$,
> $y = \tfrac{1}{2}(u-v)$, and the Jacobian is
>
> $$
> \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \tfrac12 & \tfrac12 \\ \tfrac12 & -\tfrac12 \end{vmatrix} = -\tfrac{1}{2}, \qquad \left|\frac{\partial(x,y)}{\partial(u,v)}\right| = \tfrac{1}{2}.
> $$
>
> Each edge maps to a line in $uv$-space: the edges on $y = 0$ give $u = v$, the
> edges on $x = 0$ give $u = -v$, and the two short slanted edges give $v = 1$ and
> $v = 2$. So $D$ becomes the trapezoid $1 \le v \le 2$, $-v \le u \le v$, and
>
> $$
> \iint\limits_D e^{(x+y)/(x-y)}\,\d A
>   = \int_1^2 \int_{-v}^{v} e^{u/v}\cdot \tfrac{1}{2}\,\d u\,\d v
>   = \tfrac12 \int_1^2 v\bigl(e - e^{-1}\bigr)\,\d v
>   = \tfrac{3}{4}\bigl(e - e^{-1}\bigr).
> $$

The inner integral is $\int_{-v}^{v} e^{u/v}\,\d u = v\bigl(e - e^{-1}\bigr)$, and
the outer one contributes $\tfrac12\!\int_1^2 v\,\d v = \tfrac{3}{4}$. Choosing
$u$ and $v$ to straighten the boundary or simplify the integrand, then including
the Jacobian, is the general method behind every named coordinate system.

## Reading off the limits

Setting up a triple integral is entirely about describing the solid. The
integrand rarely causes trouble; the limits do. The reliable procedure:

- **Sketch the solid and its shadow.** Project $E$ onto a coordinate plane to get
  the double-integral region $D$; the two outer limits come from $D$.
- **Innermost variable spans the two surfaces.** For a type 1 solid the inner
  $z$-limits are the lower and upper surfaces $u_1(x,y)$ and $u_2(x,y)$.
- **Match the coordinate system to the symmetry.** Axis of symmetry -> cylindrical;
  center of symmetry -> spherical; otherwise Cartesian or a custom Jacobian.
- **Never drop the Jacobian.** Every non-Cartesian setup multiplies by $r$,
  $\rho^2\sin\phi$, or $|\partial(x,y,z)/\partial(u,v,w)|$.[^stewart15]

[^stewart15]: Stewart, §15.5 — Surface Area; §15.6 — Triple Integrals; §15.7 — Cylindrical Coordinates; §15.8 — Spherical Coordinates; §15.9 — Change of Variables in Multiple Integrals. The volume elements $r\,\d z\,\d r\,\d\theta$ and $\rho^2\sin\phi\,\d\rho\,\d\theta\,\d\phi$ and their derivation as Jacobian determinants.