Triple Integrals

Quick answer

A triple integral adds up f(x, y, z) times volume over a solid region E. With f = 1 it gives the volume of E; with f a density it gives the mass. Over a box, Fubini's theorem allows any of six orders. Over a solid between two surfaces z = u₁(x, y) and z = u₂(x, y), the innermost integral runs up a vertical column from the lower surface to the upper one, and the remaining double integral covers the shadow of E on the xy-plane. Dividing moments by the mass locates the center of mass.

What you'll learn

Evaluate a triple integral over a box
Set up limits for a solid between two surfaces
Compute volumes and masses with triple integrals
Find the center of mass of a solid

Adding up through a solid

A double integral adds up a quantity spread over a flat region. A triple integral adds up a quantity spread through a solid. Cut a solid region $E$ into small boxes of volume $\Delta V$ , multiply $f$ at a point of each box by $\Delta V$ , and add. The limit as the boxes shrink is

\iiint_E f(x, y, z)\,dV = \lim \sum f(x_i^*, y_j^*, z_k^*)\,\Delta V

With $f = 1$ , the sum counts boxes, and the integral is the volume of $E$ . When $f$ is a density, in mass per unit volume, each term is the mass of one box, and the integral is the mass of $E$ .

Over a box $[a, b] \times [c, d] \times [p, q]$ , Fubini’s theorem works as it did for rectangles: the triple integral equals an iterated integral in any of six orders, all with constant limits.

Why the innermost integral is a column

Suppose $E$ lies between a lower surface $z = u_1(x, y)$ and an upper surface $z = u_2(x, y)$ , over a region $D$ of the $xy$ -plane, its shadow. Group the small boxes into vertical columns. The column over the point $(x, y)$ runs from $u_1(x, y)$ up to $u_2(x, y)$ , so the column’s total is a single integral in $z$ . Adding the columns over $D$ is then a double integral:

\iiint_E f\,dV = \iint_D \left[\int_{u_1(x, y)}^{u_2(x, y)} f(x, y, z)\,dz\right] dA

Set up the limits from the inside out: first the column, from the bottom surface to the top one, then the shadow $D$ , described as in double integrals over general regions. With $f = 1$ , the inner integral is $u_2 - u_1$ , the height of the column, and the formula becomes the familiar volume between two surfaces.

The tetrahedron under 6x + 3y + 2z = 6

Worked examples

Example 2 — The volume of a tetrahedron

Find the volume of the tetrahedron bounded by the coordinate planes and the plane $6x + 3y + 2z = 6$ .

A column goes from $z = 0$ up to the plane, where $z = \tfrac{6 - 6x - 3y}{2}$ . The shadow is where that height is positive: the triangle $0 \le x \le 1$ , $0 \le y \le 2 - 2x$ . So

V = \int_0^1 \int_0^{2 - 2x} \int_0^{(6 - 6x - 3y)/2} dz\,dy\,dx = \int_0^1 \int_0^{2 - 2x} \frac{6 - 6x - 3y}{2}\,dy\,dx

Write $s = 2 - 2x$ , so $6 - 6x = 3s$ . The inner integral is $\int_0^s \tfrac{3s - 3y}{2}\,dy = \tfrac{3s^2}{4}$ , and $\int_0^1 \tfrac{3(2 - 2x)^2}{4}\,dx = 3\int_0^1 (1 - x)^2\,dx = 1$ . The pyramid formula agrees: $\tfrac{1}{3} \cdot \tfrac{1 \cdot 2}{2} \cdot 3 = 1$ .

Example 3 — Integrating x over a tetrahedron

Let $E$ be the tetrahedron $x, y, z \ge 0$ , $x + y + z \le 1$ . Evaluate $\iiint_E x\,dV$ .

Columns run from $z = 0$ to $z = 1 - x - y$ over the triangle $0 \le x \le 1$ , $0 \le y \le 1 - x$ :

\int_0^1 \int_0^{1 - x} \int_0^{1 - x - y} x\,dz\,dy\,dx = \int_0^1 x \cdot \frac{(1 - x)^2}{2}\,dx = \frac{1}{2} \cdot \frac{1}{12} = \frac{1}{24}

The middle step used $\int_0^{1 - x} (1 - x - y)\,dy = \tfrac{(1 - x)^2}{2}$ , and the last used $\int_0^1 (x - 2x^2 + x^3)\,dx = \tfrac{1}{2} - \tfrac{2}{3} + \tfrac{1}{4} = \tfrac{1}{12}$ .

Mass and center of mass

If a solid has density $\rho(x, y, z)$ , its mass is $m = \iiint_E \rho\,dV$ . Its center of mass is the balance point, $(\bar{x}, \bar{y}, \bar{z})$ , where

\bar{x} = \frac{1}{m}\iiint_E x\,\rho\,dV, \quad \bar{y} = \frac{1}{m}\iiint_E y\,\rho\,dV, \quad \bar{z} = \frac{1}{m}\iiint_E z\,\rho\,dV

Each coordinate is a weighted average, in the sense of the average value of a function, with density as the weight. For a constant density the center of mass is the centroid, a purely geometric point.

Common mistakes

Practice problems

Evaluate $\iiint_B (x + y + z)\,dV$ over $B = [0, 1] \times [0, 2] \times [0, 3]$ .

Answer

$18$

Full solution

The box has volume $6$ . The average of $x$ over it is $\tfrac{1}{2}$ , of $y$ is $1$ , and of $z$ is $\tfrac{3}{2}$ , so the integral is $6\left(\tfrac{1}{2} + 1 + \tfrac{3}{2}\right) = 18$ . Term by term: $3 + 6 + 9$ .
Evaluate $\displaystyle\int_0^1 \int_0^1 \int_0^1 xyz\,dz\,dy\,dx$ .

Answer

$\tfrac{1}{8}$

Full solution

The integral splits: $\tfrac{1}{2} \cdot \tfrac{1}{2} \cdot \tfrac{1}{2}$ .
Find the average value of $f(x, y, z) = z$ over the unit cube.

Answer

$\tfrac{1}{2}$

Full solution

The integral is $\int_0^1 z\,dz = \tfrac{1}{2}$ , and the cube has volume $1$ .
Use a triple integral to find the volume of the solid under $z = x + y$ and over the unit square $[0, 1] \times [0, 1]$ .

Answer

$1$

Full solution

$\int_0^1 \int_0^1 \int_0^{x + y} dz\,dy\,dx = \int_0^1 \int_0^1 (x + y)\,dy\,dx = \int_0^1 \left(x + \tfrac{1}{2}\right)dx = 1$ .
Evaluate $\iiint_E x\,dV$ , where $E$ is the wedge $0 \le x \le 2$ , $0 \le y \le 1$ , $0 \le z \le y$ .

Answer

$1$

Full solution

$\int_0^2 \int_0^1 \int_0^y x\,dz\,dy\,dx = \int_0^2 x\,dx \cdot \int_0^1 y\,dy = 2 \cdot \tfrac{1}{2} = 1$ .
Set up, without evaluating, a triple integral for the volume of the solid between the paraboloid $z = x^2 + y^2$ and the plane $z = 4$ .

Answer

$\displaystyle\int_{-2}^{2} \int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}} \int_{x^2 + y^2}^{4} dz\,dy\,dx$

Full solution

Columns run from the paraboloid up to the plane. They exist where $x^2 + y^2 \le 4$ , the disk of radius $2$ , so the shadow is that disk. The next lesson evaluates this volume, $8\pi$ , in cylindrical coordinates.
The tetrahedron of Example 2 has constant density. Find $\bar{x}$ .

Hint

The centroid of a tetrahedron is the average of its four corners.

Answer

$\tfrac{1}{4}$

Full solution

The corners are $(0, 0, 0)$ , $(1, 0, 0)$ , $(0, 2, 0)$ and $(0, 0, 3)$ , and the average of their $x$ -coordinates is $\tfrac{1}{4}$ . Example 4 confirmed this pattern for the tetrahedron with corners on the unit axes. By integration, $\iiint_E x\,dV = \int_0^1 x \cdot \tfrac{3(2 - 2x)^2}{4}\,dx = 3\int_0^1 x(1 - x)^2\,dx = \tfrac{1}{4}$ , and the volume is $1$ .
A student writes the volume of the tetrahedron $x + y + z \le 1$ as $\displaystyle\int_0^{1 - x - y} \int_0^{1 - x} \int_0^1 dx\,dy\,dz$ . What went wrong?

Answer

The limits are listed in reverse: the constant limits must go on the outer integral, not the inner one.

Full solution

With $dx$ innermost, the $x$ -limits come first, and here they are $0$ to $1$ , while the outer $z$ -limits contain $x$ and $y$ . A correct version is $\int_0^1 \int_0^{1 - x} \int_0^{1 - x - y} dz\,dy\,dx = \tfrac{1}{6}$ .

Frequently asked questions

What does a triple integral measure?

It adds up f(x, y, z) times volume over a solid. With f = 1 it is the volume of the solid; with f a density, in mass per unit volume, it is the total mass.

How do you set up the limits of a triple integral?

For a solid between surfaces z = u₁(x, y) below and z = u₂(x, y) above, the z-limits are u₁ and u₂. The remaining double integral runs over the shadow of the solid on the xy-plane, with limits found as for any double integral.

In how many orders can a triple integral be written?

Six: dz dy dx, dz dx dy, dy dz dx, dy dx dz, dx dz dy and dx dy dz. Over a box all six give the same value with constant limits; over other solids the limits change with the order.

How do you find the center of mass of a solid?

Divide each first moment by the mass. For example, x-bar is the triple integral of x times the density, divided by the triple integral of the density.

Why is a triple integral not a volume in four dimensions?

It can be read that way, but the useful readings are physical: a total amount of something spread through a solid, such as mass, charge or heat.

What to learn next

Triple Integrals in Cylindrical and Spherical CoordinatesDescribe cylinders, cones and balls with cylindrical and spherical coordinates, and integrate over them using dV = r dz dr dθ and dV = ρ² sin φ dρ dφ dθ.

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.