Learning Objectives
- Recognize when a function of three variables is integrable over a rectangular box.
- Evaluate a triple integral by expressing it as an iterated integral.
- Recognize when a function of three variables is integrable over a closed and bounded region.
- Simplify a calculation by changing the order of integration of a triple integral.
Integrable Functions of Three Variables
We can define a rectangular box [latex]B[/latex] in [latex]{{\mathbb{R}}^3}[/latex] as [latex]{B} = {\left \{{(x,y,z)}{\mid}{a} \ {\leq} \ {x} \ {\leq} \ {b,c} \ {\leq} \ {y} \ {\leq} \ {d,e} \ {\leq} \ {z} \ {\leq} \ {f} \right \}}[/latex]. We follow a similar procedure to what we did in 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}_{i-1}},{y_i}][/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}_{i-1}},{z_i}][/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}_{i-1}},{y_i}] \ {\times} \ [{{z}_{i-1}},{z_i}] [/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]\Large{\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.
definition
The triple integral of a function [latex]f(x, y, z)[/latex] over a rectangular box [latex]B[/latex] is defined as
[latex]\Large{\displaystyle\lim_{{l,m,n}{\rightarrow}{\infty}}} \ {\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}} = \underset{B}{\displaystyle\iiint}{f}{(x,y,z)}{dV}[/latex]
if this limit exists.
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]. Also, the triple integral exists if [latex]f(x, y, z)[/latex] is continuous on [latex]B[/latex]. Therefore, 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. Just as the double integral has many practical applications, the triple integral also has many applications, which we discuss in later sections.
Now that we have developed the concept of the triple integral, we need to know how to compute it. Just as in the case of the double integral, we can have an iterated triple integral, and consequently, a version of Fubini’s thereom for triple integrals exists.
theorem: fubini’s theorem for triple integrals
If [latex]f(x, y, z)[/latex] is continuous on a rectangular box [latex]{B} = {[a,b]} \ {\times} \ {[c,d]} \ {\times} \ {[e,f]}[/latex], then
[latex]\Large{\underset{B}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = {\displaystyle\int_{e}^{f}}{\displaystyle\int_{c}^{d}}{\displaystyle\int_{a}^{b}}{f}{(x,y,z)}{dx} \ {dy} \ {dz}}[/latex].
This integral is also equal to any of the other five possible orderings for the iterated triple integral.
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:
[latex]\large{\begin{align} {\displaystyle\int_{e}^{f}}{\displaystyle\int_{c}^{d}}{\displaystyle\int_{a}^{b}}{f}{(x,y,z)}{dx} \ {dy} \ {dz}&={\displaystyle\int_{e}^{f}}\left({\displaystyle\int_{c}^{d}}\left({\displaystyle\int_{a}^{b}}{f}{(x,y,z)}{dx}\right){dy}\right){dz}={\displaystyle\int_{c}^{d}}\left({\displaystyle\int_{e}^{f}}\left({\displaystyle\int_{a}^{b}}{f}{(x,y,z)}{dx}\right){dz}\right){dy} \\ &={\displaystyle\int_{a}^{b}}\left({\displaystyle\int_{e}^{f}}\left({\displaystyle\int_{c}^{d}}{f}{(x,y,z)}{dy}\right){dz}\right){dx}={\displaystyle\int_{e}^{f}}\left({\displaystyle\int_{a}^{b}}\left({\displaystyle\int_{c}^{d}}{f}{(x,y,z)}{dy}\right){dx}\right){dz} \\ &={\displaystyle\int_{c}^{e}}\left({\displaystyle\int_{a}^{b}}\left({\displaystyle\int_{e}^{f}}{f}{(x,y,z)}{dz}\right){dx}\right){dy}={\displaystyle\int_{a}^{b}}\left({\displaystyle\int_{c}^{e}}\left({\displaystyle\int_{e}^{f}}{f}{(x,y,z)}{dz}\right){dy}\right){dx}. \end{align}}[/latex]
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).
Example: evaluating a triple integral 1
Evaluate the triple integral [latex]{\displaystyle\int^{z=1}_{z=0}}{\displaystyle\int^{y=4}_{y=2}}{\displaystyle\int^{x=5}_{x=-1}}{(x+yz^2)}{dx} \ {dy} \ {dz}[/latex].
Example: evaluating a triple integral 2
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.
try it
Evaluate the triple integral [latex]\underset{B}{\displaystyle\iiint}{z} \ {\sin \ x} \ {\cos \ y} \ {dV}[/latex] where
[latex]{B} = {\left \{{(x,y,z)}{\mid}{0} \ {\leq} \ {x} \ {\leq} \ {{\pi},{\frac{3{\pi}}{2}}} \ {\leq} \ {y} \ {\leq} \ {2{\pi},1} \ {\leq} \ {z} \ {\leq} \ {3} \right \}}[/latex].
Watch the following video to see the worked solution to the above Try It
Try It
Triple Integrals over a General Bounded Region
We now expand the definition of the triple integral to compute a triple integral over a more general bounded region [latex]E[/latex] in [latex]{\mathbb{R}^3}[/latex]. The general bounded regions we will consider are of three types. First, let [latex]D[/latex] be the bounded region that is a projection of [latex]E[/latex] onto the [latex]xy[/latex]-plane. Suppose the region [latex]E[/latex] in [latex]{\mathbb{R}^3}[/latex] has the form
[latex]\Large{E} = \left \{{(x,y,z)}{\mid}{(x,y)} \ {\in} \ {{D},{u_1}{(x,y)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(x,y)} \right \}[/latex].
For two functions [latex]z=u_1(x, y)[/latex] and [latex]z=u_2(x,y)[/latex], such that [latex]{u_1}{(x,y)} \ {\leq} \ {u_2}{(x,y)}[/latex] for all [latex](x, y)[/latex] in [latex]D[/latex] as shown in the following figure.
theorem: triple integral over a general region
The triple integral of a continuous function [latex]f(x, y, z)[/latex] over a general three-dimensional region
[latex]\large{E} = \left \{{(x,y,z)}{\mid}{(x,y)} \ {\in} \ {{D},{u_1}{(x,y)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(x,y)} \right \}[/latex]
in [latex]{{\mathbb{R}}^3}[/latex], where [latex]D[/latex] is the projection of [latex]E[/latex] onto the [latex]xy[/latex]–plane, is
[latex]\large{\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = \underset{D}{\displaystyle\iint}{\left [{\displaystyle\int^{{u_2}{(x,y)}}_{{u_1}{(x,y)}}}{f}{(x,y,z)}{dz} \right ]}{dA}}[/latex].
Similarly, we can consider a general bounded region [latex]D[/latex] in the [latex]xy[/latex]-plane and two functions [latex]{y} = {u_1}{(x,z)}[/latex] and [latex]{y} = {u_2}{(x,z)}[/latex] such that [latex]{u_1}{(x,z)} \ {\leq} \ {u_2}{(x,z)}[/latex] for all [latex](x, z)[/latex] in [latex]D[/latex]. Then we can describe the solid region [latex]E[/latex] in [latex]{{\mathbb{R}}^3}[/latex] as
[latex]{E} = \left \{{(x,y,z)}{\mid}{(x,z)} \ {\in} \ {{D},{u_1}{(x,z)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(x,z)} \right \}[/latex]
where [latex]D[/latex] is the projection of [latex]E[/latex] onto the [latex]xy[/latex]-plane and the triple integral is
[latex]\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = \underset{D}{\displaystyle\iint}{\left [{\displaystyle\int^{{u_2}{(x,z)}}_{{u_1}{(x,z)}}}{f}{(x,y,z)}{dy} \right ]}{dA}[/latex].
Finally, if [latex]D[/latex] is a general bounded region in the [latex]yz[/latex]-plane and we have two functions [latex]{x} = {u_1}{(y,z)}[/latex] and [latex]{x} = {u_2}{(y,z)}[/latex] such that [latex]{u_1}{(y,z)} \ {\leq} \ {u_2}{(y,z)}[/latex] for all [latex](y, z)[/latex] in [latex]D[/latex], then the solid region [latex]E[/latex] in [latex]{{\mathbb{R}}^3}[/latex] can be described as
[latex]{E} = \left \{{(x,y,z)}{\mid}{(y,z)} \ {\in} \ {{D},{u_1}{(y,z)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(y,z)} \right \}[/latex]
where [latex]D[/latex] is the projection of [latex]E[/latex] onto the [latex]yz[/latex]-plane and the triple integral is
[latex]\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = \underset{D}{\displaystyle\iint}{\left [{\displaystyle\int^{{u_2}{(y,z)}}_{{u_1}{(y,z)}}}{f}{(x,y,z)}{dx} \right ]}{dA}[/latex].
Note that the region [latex]D[/latex] in any of the planes may be of Type I or Type II as described in Double Integrals over General Regions. If [latex]D[/latex] in the [latex]xy[/latex]-plane is of Type I (Figure 4), then
[latex]{E} = \left \{{(x,y,z)}{\mid}{a} \ {\leq} \ {x} \ {\leq} \ {b,{g_1}{(x)}} \ {\leq} \ {y} \ {\leq} \ {g_2{(x)}},{{u_1}{(x,y)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(x,y)} \right \}[/latex].
Then the triple integral becomes
[latex]\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = {\displaystyle\int_{a}^{b}} \ {\displaystyle\int_{{g_1}{(x)}}^{{g_2}{(x)}}} \ {\displaystyle\int_{{u_1}{(x,y)}}^{{u_2}{(x,y)}}}{f}{(x,y,z)}{dz}{dy}{dx}[/latex].
If [latex]D[/latex] in the [latex]xy[/latex]-plane is of Type II (Figure 5), then
[latex]{E} = \left \{{(x,y,z)}{\mid}{c} \ {\leq} \ {x} \ {\leq} \ {d,{h_1}{(x)}} \ {\leq} \ {y} \ {\leq} \ {h_2{(x)}},{{u_1}{(x,y)}} \ {\leq} \ {z} \ {\leq} \ {u_2}{(x,y)} \right \}[/latex].
Then the triple integral becomes
[latex]\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = {\displaystyle\int^{y=d}_{y=c}}{\displaystyle\int^{x={{h_2}{(y)}}}_{x={h_1}{(y)}}}{\displaystyle\int^{z={{u_2}{(x,y)}}}_{z={u_1}{(x,y)}}}{f}{(x,y,z)}{dz}{dx}{dy}[/latex].
Example: evaluating a triple integral over a general bounded region
Evaluate the triple integral of the function [latex]f(x, y, z)=5x-3y[/latex] over the solid tetrahedron bounded by the planes [latex]x=0, \ y=0, \ z=0,[/latex] and [latex]x+y+z=1[/latex].
Just as we used the double integral [latex]\underset{D}{\displaystyle\iint}{1}{dA}[/latex] to find the area of a general bounded region [latex]D[/latex], we can use [latex]\underset{E}{\displaystyle\iiint}{1}{dV}[/latex] to find the volume of a general solid bounded region [latex]E[/latex]. The next example illustrates the method.
Example: finding a volume by evaluating a triple integral
Find the volume of a right pyramid that has the square base in the [latex]xy[/latex]-plane [latex][-1,1] \ {\times} \ [-1,1][/latex] and vertex at the point [latex](0, 0, 1)[/latex] as shown in the following figure.
try it
Consider the solid sphere [latex]{E} = {(x,y,z)}{\mid}\left \{{{x^2}+{y^2}+{z^2}} = {9} \right \}[/latex]. Write the triple integral [latex]\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV}[/latex] for an arbitrary function [latex]f[/latex] as an iterated integral. Then evaluate this triple integral with [latex]f(x, y, z)=1[/latex]. Notice that this gives the volume of a sphere using a triple integral.
Watch the following video to see the worked solution to the above Try It
Changing the Order of Integration
As we have already seen in double integrals over general bounded regions, changing the order of the integration is done quite often to simplify the computation. With a triple integral over a rectangular box, the order of integration does not change the level of difficulty of the calculation. However, with a triple integral over a general bounded region, choosing an appropriate order of integration can simplify the computation quite a bit. Sometimes making the change to polar coordinates can also be very helpful. We demonstrate two examples here.
Example: changing the order of integration
Consider the iterated integral
[latex]{\displaystyle\int^{x=1}_{x=0}} \ {\displaystyle\int^{y=x^2}_{y=0}} \ {\displaystyle\int^{z=y^2}_{z=0}}{f}{(x,y,z)}{dz}{dy}{dx}[/latex].
The order of integration here is first with respect to [latex]z[/latex], then [latex]y[/latex], and then [latex]x[/latex]. Express this integral by changing the order of integration to be first with respect to [latex]x[/latex], then [latex]z[/latex], and then [latex]y[/latex]. Verify that the value of the integral is the same if we let [latex]f(x, y, z)=xyz[/latex].
try it
Write five different iterated integrals equal to the given integral [latex]{\displaystyle\int^{z=4}_{z=0}} \ {\displaystyle\int^{y={4-z}}_{y=0}} \ {\displaystyle\int^{x={\sqrt{y}}}_{x=0}}{f}{(x,y,z)}{dx}{dy}{dz}[/latex].
Example: changing integration order and coordinate systems
Evaluate the triple integral [latex]\underset{E}{\displaystyle\iiint}{\sqrt{{x^2}+{z^2}}}{dV}[/latex], where [latex]E[/latex] is the region bounded by the paraboloid [latex]y=x^{2}+z^{2}[/latex] (Figure 9) and the plane [latex]y=4[/latex].