Integrable Functions of Three Variables

In earlier sections, we discussed the double integral of a function [latex]f(x,y)[/latex] of two variables over a rectangular region in the plane. In this section we define the triple integral of a function [latex]f(x,y,z)[/latex] of three variables over a rectangular solid box in space, [latex]\mathbb{R}^{3}[/latex].

We can define a rectangular box [latex]B[/latex] in [latex]\mathbb{R}^3[/latex] as [latex]B = {(x,y,z)\mid a \leq x \leq b, c \leq y \leq d, e \leq z \leq f }[/latex]. We follow a similar procedure to what we did for double integrals over rectangular regions.

We divide the interval [latex][a,b][/latex] into [latex]l[/latex] subintervals [latex][x_{i-1},x_i][/latex] of equal length [latex]\Delta x = \frac{x_i-x_{i-1}}{l}[/latex], divide the interval [latex][c,d][/latex] into [latex]m[/latex] subintervals [latex][y_{j-1},y_j][/latex] of equal length [latex]\Delta y = \frac{y_j-y_{j-1}}{m}[/latex], and divide the interval [latex][e,f][/latex] into [latex]n[/latex] subintervals [latex][z_{k-1},z_k][/latex] of equal length [latex]\Delta z = \frac{z_k-z_{k-1}}{n}[/latex]. Then the rectangular box [latex]B[/latex] is subdivided into [latex]lmn[/latex] subboxes [latex]B_{ijk} = [x_{i-1},x_i] \times [y_{j-1},y_j] \times [z_{k-1},z_k][/latex], as shown in Figure 1.

For each [latex]i[/latex], [latex]j[/latex], and [latex]k[/latex], consider a sample point [latex]{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}[/latex] in each sub-box [latex]B_{ijk}[/latex]. We see that its volume is [latex]\Delta V = \Delta x\Delta y\Delta z[/latex]. Form the triple Riemann sum:

[latex]{\displaystyle\sum_{i=1}^{l}}{\displaystyle\sum_{j=1}^{m}}{\displaystyle\sum_{k=1}^{n}}{f}{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}{{\Delta}{x}}{{\Delta}{y}}{{\Delta}{z}}[/latex]

We define the triple integral in terms of the limit of a triple Riemann sum, as we did for the double integral in terms of a double Riemann sum.

When the triple integral exists on [latex]B[/latex], the function [latex]f(x, y, z)[/latex] is said to be integrable on [latex]B[/latex]. The triple integral exists if [latex]f(x, y, z)[/latex] is continuous on [latex]B[/latex], so we will use continuous functions for our examples. However, continuity is sufficient but not necessary; in other words, [latex]f[/latex] is bounded on [latex]B[/latex] and continuous except possibly on the boundary of [latex]B[/latex] The sample point [latex]{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}[/latex] can be any point in the rectangular sub-box [latex]{B}_{ijk}[/latex] and all the properties of a double integral apply to a triple integral.

Now that we have developed the concept of the triple integral, we need to know how to compute it. Just as with double integrals, we can use an iterated triple integral. Fubini’s theorem extends to triple integrals as well.

For [latex]a[/latex], [latex]b[/latex], [latex]c[/latex], [latex]d[/latex], [latex]e[/latex], and [latex]f[/latex] real numbers, the iterated triple integral can be expressed in six different orderings:

For a rectangular box, the order of integration does not make any significant difference in the level of difficulty in computation. We compute triple integrals using Fubini’s Theorem rather than using the Riemann sum definition. We follow the order of integration in the same way as we did for double integrals (that is, from inside to outside).

Evaluate the triple integral [latex]\underset{B}{\displaystyle\iiint}{x^2}{yz}{dV}[/latex] where[latex]{B} = {\left \{{(x,y,z)}{\mid}{-2} \ {\leq} \ {x} \ {\leq} \ {1,0} \ {\leq} \ {y} \ {\leq} \ {3,1} \ {\leq} \ {z} \ {\leq} \ {5} \right \}}[/latex] as shown in the following figure.

Show Answer

The order is not specified, but we can use the iterated integral in any order without changing the level of difficulty. Choose, say, to integrate [latex]y[/latex] first, then [latex]x[/latex], and then [latex]z[/latex].

[latex]\hspace{2cm}\begin{align} \underset{B}{\displaystyle\iiint}{x^2}{yz}{dV}&=\displaystyle\int_1^5\displaystyle\int_{-2}^1\displaystyle\int_0^3[x^2yz]dy \ dx \ dz =\displaystyle\int_1^5\displaystyle\int_{-2}^1\left[x^2\frac{y^2}2z\bigg|_0^3\right]dx \ dz \\ &=\displaystyle\int_1^5\displaystyle\int_{-2}^1\frac92x^2z \ dx \ dz = \displaystyle\int_1^5\left[\frac92\frac{x^3}3z\bigg|_{-2}^1\right]dz=\displaystyle\int_1^5\frac{27}2\frac{z^2}2\bigg|_1^5=162. \end{align}[/latex]

Now try to integrate in a different order just to see that we get the same answer. Choose to integrate with respect to [latex]x[/latex] first, then [latex]z[/latex], and then [latex]y[/latex].

[latex]\hspace{2cm}\begin{align} \underset{B}{\displaystyle\iiint}{x^2}{yz}{dV}&=\displaystyle\int_0^3\displaystyle\int_1^5\displaystyle\int_{-2}^1[x^2yz]dx \ dz \ dy =\displaystyle\int_0^3\displaystyle\int_1^5\left[\frac{x^3}3yz\bigg|_{-2}^1\right]dz \ dy \\ &=\displaystyle\int_0^3\displaystyle\int_1^53yz \ dz \ dy =\displaystyle\int_0^3\left[3y\frac{z^2}2\bigg|_1^5\right]dy=\displaystyle\int_0^336y \ dy = 36\frac{y^2}2\bigg|_0^3=18(9-0)=162. \end{align}[/latex]