Math Core

A double integral adds up a quantity spread over a flat region. A triple integral does the same thing over a solid region in space. It gives volume, mass, total charge, and average values of functions of three variables, and almost all of the technique carries over from double integrals.

From boxes to triple integrals

Start with a function $f(x, y, z)$ defined on a box $B = [a, b] \times [c, d] \times [r, s]$ . Cut the box into small sub-boxes of volume $\Delta V = \Delta x\,\Delta y\,\Delta z$ , pick a sample point $(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)$ in each one, and add:

\sum_{i}\sum_{j}\sum_{k} f\left(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*\right)\Delta V.

As the sub-boxes shrink, these Riemann sums approach the triple integral $\displaystyle\iiint_B f(x, y, z)\,dV$ .

If $f = 1$ , each term is just the volume of a sub-box, so $\iiint_E 1\,dV$ is the volume of $E$ . If $f$ is a density, the triple integral is a mass.

Fubini's Theorem for boxes

If $f$ is continuous on $B = [a, b] \times [c, d] \times [r, s]$ , then

\iiint_B f(x, y, z)\,dV = \int_r^s \int_c^d \int_a^b f(x, y, z)\,dx\,dy\,dz,

and the three integrations can be done in any of the six possible orders.

You work from the inside out: integrate with respect to the innermost variable while holding the other two constant, then the next, then the last.

Worked example: A separable integrand on a box

Evaluate $\displaystyle\iiint_B xyz^2\,dV$ , where $B = [0, 1] \times [-1, 2] \times [0, 3]$ .

Solution. The integrand is a product of a function of $x$ , a function of $y$ and a function of $z$ , and the limits are constants, so the integral splits into three single integrals:

\left(\int_0^1 x\,dx\right)\left(\int_{-1}^2 y\,dy\right)\left(\int_0^3 z^2\,dz\right) = \frac{1}{2} \cdot \frac{4 - 1}{2} \cdot 9 = \frac{1}{2} \cdot \frac{3}{2} \cdot 9 = \frac{27}{4}.

General regions

Most solids are not boxes. The most common way to describe a solid $E$ is to say it lies between two surfaces, one below and one above, over a region $D$ in the $xy$ -plane:

E = \{(x, y, z) : (x, y) \in D,\ u_1(x, y) \le z \le u_2(x, y)\}.

Here $D$ is the projection (shadow) of $E$ onto the $xy$ -plane. For such a region,

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

The inner integral runs from the bottom surface to the top surface. What is left is a double integral over $D$ , which you set up exactly as in the earlier lessons.

Setting up a triple integral

Decide which variable goes innermost (often $z$ ). Find the lower and upper surfaces: those are its limits.
Project the solid onto the plane of the other two variables to get $D$ .
Describe $D$ with limits, as for a double integral: the middle limits may depend on the outer variable, and the outer limits are constants.

You can just as well project onto the $yz$ -plane (with $x$ innermost) or the $xz$ -plane (with $y$ innermost). Choose whichever makes the bounding surfaces easiest to write.

Worked example: A tetrahedron

Let $E$ be the solid tetrahedron bounded by the coordinate planes and the plane $x + y + z = 1$ . Evaluate $\displaystyle\iiint_E z\,dV$ .

Solution. The bottom is $z = 0$ and the top is $z = 1 - x - y$ . The shadow $D$ on the $xy$ -plane is the triangle $x \ge 0$ , $y \ge 0$ , $x + y \le 1$ , which is $0 \le x \le 1$ , $0 \le y \le 1 - x$ . So

\iiint_E z\,dV = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} z\,dz\,dy\,dx = \int_0^1 \int_0^{1-x} \frac{(1 - x - y)^2}{2}\,dy\,dx.

For the middle integral, substitute $w = 1 - x - y$ : it equals $\displaystyle\int_0^{1-x}\frac{w^2}{2}\,dw = \frac{(1 - x)^3}{6}$ . Then

\int_0^1 \frac{(1 - x)^3}{6}\,dx = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}.

The shadow D of the solid in the next example: between y = x² and y = 1.Open in grapher →

Worked example: Volume under a slanted roof

Find the volume of the solid that lies above the $xy$ -plane, below the plane $z = y$ , and over the region $D$ between the parabola $y = x^2$ and the line $y = 1$ .

Solution. Volume is $\iiint_E 1\,dV$ . The limits for $z$ are $0 \le z \le y$ , and $D$ is $-1 \le x \le 1$ , $x^2 \le y \le 1$ :

V = \int_{-1}^1 \int_{x^2}^1 \int_0^y dz\,dy\,dx = \int_{-1}^1 \int_{x^2}^1 y\,dy\,dx = \int_{-1}^1 \frac{1 - x^4}{2}\,dx.

The integrand is even, so this is $\displaystyle\int_0^1 \left(1 - x^4\right)dx = 1 - \frac{1}{5} = \frac{4}{5}$ .

Common mistake

The limits must get simpler as you move outward. The innermost limits may involve both outer variables, the middle limits may involve only the outermost variable, and the outermost limits must be constants. If your final answer still contains a variable, a limit is in the wrong place.

Changing the order

Sometimes an integral is easier in a different order. To switch, describe the same solid a different way. For the tetrahedron $x, y, z \ge 0$ , $x + y + z \le 1$ , putting $y$ innermost gives $0 \le y \le 1 - x - z$ , and the shadow on the $xz$ -plane is the triangle $0 \le x \le 1$ , $0 \le z \le 1 - x$ :

\int_0^1 \int_0^{1-x} \int_0^{1-x-y} f\,dz\,dy\,dx = \int_0^1 \int_0^{1-x} \int_0^{1-x-z} f\,dy\,dz\,dx.

Sketching the solid, or at least its shadow, is the reliable way to do this.

Mass, center of mass and average value

Everything from the applications lesson moves up one dimension. If a solid $E$ has density $\rho(x, y, z)$ , then

m = \iiint_E \rho\,dV, \qquad \bar{x} = \frac{1}{m}\iiint_E x\,\rho\,dV,

and similarly for $\bar{y}$ and $\bar{z}$ . The average value of $f$ over $E$ is

f_{\text{avg}} = \frac{1}{V(E)}\iiint_E f\,dV.

Tip

Symmetry saves work here too. The tetrahedron $x + y + z \le 1$ looks the same if you swap any two variables, so $\iiint_E x\,dV = \iiint_E y\,dV = \iiint_E z\,dV = \dfrac{1}{24}$ .