Triple Integrals over a Rectangular Region

Core Definition

A triple integral over a rectangular box $B = [a, b] \times [c, d] \times [r, s] \subset \mathbb{R}^3$ is the formal three-dimensional limit of a triple Riemann sum:

\iiint_{B} f(x, y, z) \, dV = \lim_{\substack{l \to \infty \\ m \to \infty \\ n \to \infty}} \sum_{i=1}^{l} \sum_{j=1}^{m} \sum_{k=1}^{n} f(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*) \, \Delta V_{ijk}

where

\Delta V_{ijk} = \Delta x_i \, \Delta y_j \, \Delta z_k

is the volume of sub-box

B_{ijk}

, and

(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)

is any sample point in

B_{ijk}

Concept Explanation

A triple integral extends single and double integration into three-dimensional space, accumulating a continuous scalar property $f(x, y, z)$ over a spatial volume. When the domain is a rectangular box (or rectangular parallelepiped) $B$ , all boundary limits for $x$ , $y$ , and $z$ are fixed constants:

a \le x \le b, \quad c \le y \le d, \quad r \le z \le s

The differential volume element $dV$ represents an infinitesimal rectangular cuboid with side lengths $dx$ , $dy$ , and $dz$ :

dV = dz \, dy \, dx = dx \, dy \, dz = \dots

Fubini's Theorem for Triple Integrals

If $f(x, y, z)$ is continuous on the rectangular box $B$ , Fubini's Theorem guarantees that the triple integral is equal to the iterated integral evaluated in any of the $3! = 6$ possible variable permutations:

\iiint_{B} f(x, y, z) \, dV = \int_{a}^{b} \int_{c}^{d} \int_{r}^{s} f(x, y, z) \, dz \, dy \, dx

Because all limits are constants, we evaluate the iterated integral sequentially from the inside layer out, treating all outer variables as constants during each integration pass.

Algebraic Separability

When the integrand factors completely into independent single-variable functions:

f(x, y, z) = g(x) \cdot h(y) \cdot k(z)

the triple integral over a rectangular box factors directly into the scalar product of three single-variable definite integrals:

\iiint_{B} g(x)h(y)k(z) \, dV = \left( \int_{a}^{b} g(x) \, dx \right) \left( \int_{c}^{d} h(y) \, dy \right) \left( \int_{r}^{s} k(z) \, dz \right)

Geometric and Physical Interpretations

Geometric Volume: If $f(x, y, z) = 1$ , the integral evaluates the spatial volume of the box: $V(B) = \iiint_{B} 1 \, dV = (b - a)(d - c)(s - r)$
Average Value: The average value $\overline{f}$ of a continuous function across the box is: $\overline{f} = \frac{1}{V(B)} \iiint_{B} f(x, y, z) \, dV$
Total Mass and Enclosed Charge: If $\rho(x, y, z)$ represents mass density (in $\text{kg/m}^3$ ) or charge density (in $\text{C/m}^3$ ), the triple integral computes total mass $M = \iiint_B \rho \, dV$ or total charge $Q = \iiint_B \rho \, dV$ .

Engineering Applications

Computer & Mechanical Engineering: Evaluating total volumetric heat generation $Q = \iiint q(x, y, z) \, dV$ and spatial temperature profiles across microchip heat sinks and thermoelectric cooling blocks.
Electrical & Semiconductor Engineering: Computing total enclosed electrostatic charge in 3D doped semiconductor substrates and MOS transistor channels governed by Gauss's Law: $\iiint (\nabla \cdot \vec{E}) \, dV = \frac{Q_{\text{enc}}}{\varepsilon_0}$ .
Materials Science & Solid Mechanics: Determining total mass and gravitational centers of uniform composite structural blocks, acoustic dampening panels, and additive manufacturing test coupons.

Example Problems

Example 1: Iterated Integration over a 3D Box

Evaluate the triple integral:

I = \iiint_{B} 12xyz^2 \, dV \quad \text{where } B = [0, 1] \times [1, 2] \times [0, 2]

Solution

Apply Fubini's Theorem for Triple Integrals:

I = \int_{0}^{1} \int_{1}^{2} \int_{0}^{2} 12xyz^2 \, dz \, dy \, dx

Step 1: Integrate innermost layer with respect to $z$ (holding $x, y$ constant):

\int_{0}^{2} 12xyz^2 \, dz = \left[ 12xy \frac{z^3}{3} \right]_{0}^{2} = \left[ 4xyz^3 \right]_{0}^{2} = 4xy(2)^3 - 0 = 32xy

Step 2: Integrate middle layer with respect to $y$ (holding $x$ constant):

\int_{1}^{2} 32xy \, dy = \left[ 16xy^2 \right]_{1}^{2} = 16x(2)^2 - 16x(1)^2 = 64x - 16x = 48x

Step 3: Integrate outermost layer with respect to $x$ :

\int_{0}^{1} 48x \, dx = \left[ 24x^2 \right]_{0}^{1} = 24(1)^2 - 0 = 24

Thus, $I = 24$ .

Example 2: Average Value of a Spatial Function

Find the average value $\overline{f}$ of the scalar field $f(x, y, z) = x + 2y - z$ over the box $B = [0, 2] \times [0, 3] \times [1, 2]$ .

Solution

Step 1: Compute box volume $V(B)$ :

V(B) = (2 - 0)(3 - 0)(2 - 1) = 2 \times 3 \times 1 = 6

Step 2: Formulate and evaluate the triple integral:

I = \int_{0}^{2} \int_{0}^{3} \int_{1}^{2} (x + 2y - z) \, dz \, dy \, dx

Inner $z$ -integral:

\int_{1}^{2} (x + 2y - z) \, dz = \left[ (x + 2y)z - \frac{z^2}{2} \right]_{1}^{2} = \left(2(x + 2y) - 2\right) - \left(1(x + 2y) - \frac{1}{2}\right) = x + 2y - \frac{3}{2}

Middle $y$ -integral:

\int_{0}^{3} \left(x + 2y - \frac{3}{2}\right) \, dy = \left[ xy + y^2 - \frac{3}{2}y \right]_{0}^{3} = 3x + 9 - \frac{9}{2} = 3x + \frac{9}{2}

Outer $x$ -integral:

\int_{0}^{2} \left(3x + \frac{9}{2}\right) \, dx = \left[ \frac{3}{2}x^2 + \frac{9}{2}x \right]_{0}^{2} = \frac{3}{2}(4) + \frac{9}{2}(2) = 6 + 9 = 15

Step 3: Calculate the average value:

\overline{f} = \frac{I}{V(B)} = \frac{15}{6} = 2.5

Connections & References

Parent Concepts: Multiple Integrals, Double Integrals over a Rectangular Region
Sub-concepts & Extensions:
- Triple Integrals over a General Region
- Changing the Order of Integration in Triple Integrals
Course Links:
- Calc-3 Session 16 - Triple Integrals

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