Triple Integrals over Boxes
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 5.4: “Triple Integrals” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/5-4-triple-integrals |
| Textbook used in class | Stewart, Calculus, Section 15.7: “Triple Integrals” (Example 1) |
Quick Reference
Triple integral over a box $B = [a,b]\times[c,d]\times[r,s]$:
By Fubini’s theorem, any of the six orderings of integration gives the same result: $$\iiint_B f\,dV = \int_a^b\int_c^d\int_r^s f(x,y,z)\,dz\,dy\,dx = \cdots$$
Volume: $\iiint_B 1\,dV = (b-a)(d-c)(s-r)$.
Motivation
The triple integral extends double integration to three dimensions. Its primary interpretation is as a weighted sum over a 3D region: if $f = 1$, the integral gives the volume of the box; if $f = \rho(x,y,z)$ is a mass density (in kg/m$^3$), the integral gives the total mass.
For a box (rectangular parallelepiped), the limits are all constants and Fubini’s theorem lets you integrate in any order.
Key Concept
For a box, the iterated integral has all constant limits. Integrate inside-out: the innermost integral is done first (treating the other two variables as constants), then the middle, then the outer. Since $f$ is continuous on a compact set, all six orderings give the same result -- choose the one that makes the arithmetic easiest.
Worked Example
Evaluate $\iiint_B xyz\,dV$ where $B = [0,1]\times[-1,1]\times[0,2]$. (Stewart 15.7, Example 1.)
$$\int_0^1\int_{-1}^1\int_0^2 xyz\,dz\,dy\,dx.$$
Inner (z): $\int_0^2 xyz\,dz = xy\left[\frac{z^2}{2}\right]_0^2 = 2xy$.
Middle (y): $\int_{-1}^1 2xy\,dy = 2x\left[\frac{y^2}{2}\right]_{-1}^1 = 2x(1/2-1/2) = 0$.
Outer (x): $\int_0^1 0\,dx = 0$.
The integral equals $0$. (This reflects the fact that $f = xyz$ is odd in $y$, and the box is symmetric in $y$ about $y=0$, so the positive and negative contributions cancel.)
the order of integration in $dz\,dy\,dx$ means integrate $x$ first. Read the differentials inside-out: $dz\,dy\,dx$ means integrate with respect to $z$ first (innermost), then $y$, then $x$ (outermost). The variable written closest to the integrand is the first variable of integration.
Leveled Practice
Problem 1. Evaluate $\iiint_B (x^2+y^2)\,dV$ over $B = [0,1]\times[0,1]\times[0,1]$.
Show answer
$$\int_0^1\int_0^1\int_0^1(x^2+y^2)\,dz\,dy\,dx = \int_0^1\int_0^1(x^2+y^2)\,dy\,dx$$
$= \int_0^1\left[x^2 y + \frac{y^3}{3}\right]_0^1 dx = \int_0^1\left(x^2+\frac{1}{3}\right)dx = \frac{1}{3}+\frac{1}{3} = \frac{2}{3}.$