APEX Calculus: for University of Lethbridge

This no longer looks like a “double integral,” but more like a “triple integral.” Just as our first introduction to double integrals was in the context of finding the area of a plane region, our introduction into triple integrals will be in the context of finding the volume of a space region.

To formally find the volume of a closed, bounded region \(D\) in space, such as the one shown in Figure 15.6.4.(a), we start with an approximation. Break \(D\) into \(n\) rectangular solids; the solids near the boundary of \(D\) may possibly not include portions of \(D\) and/or include extra space. In Figure 15.6.4.(b), we zoom in on a portion of the boundary of \(D\) to show a rectangular solid that contains space not in \(D\text{;}\) as this is an approximation of the volume, this is acceptable and this error will be reduced as we shrink the size of our solids.

Let \(\norm{\Delta D}\) represent the length of the longest diagonal of rectangular solids in the subdivision of \(D\text{.}\) As \(\norm{\Delta D}\to 0\text{,}\) the volume of each solid goes to 0, as do each of \(\dx_i\text{,}\) \(\dy_i\) and \(\ddz_i\text{,}\) for all \(i\text{.}\) Our calculus experience tells us that taking a limit as \(\norm{\Delta D}\to 0\) turns our approximation of \(V\) into an exact calculation of \(V\text{.}\) Before we state this result in a theorem, we use a definition to define some terms.

Our informal understanding of the notation \(\iiint_D\, dV\) is “sum up lots of little volumes over \(D\text{,}\)” analogous to our understanding of \(\iint_R\, dA\) and \(\iint_R\, dm\text{.}\)

We now state the major theorem of this section.

Examples will help us understand triple integration, including integrating with various orders of integration.

In the previous example, we collapsed the surface into the \(x\)-\(y\text{,}\) \(x\)-\(z\text{,}\) and \(yz\)-planes as we determined the “curve to curve, point to point” bounds of integration. Since the surface was a triangular portion of a plane, this collapsing, or projecting, was simple: the projection of a straight line in space onto a coordinate plane is a line.

The following example shows us how to do this when dealing with more complicated surfaces and curves.

The following theorem states two things that should make “common sense” to us. First, using the triple integral to find volume of a region \(D\) should always return a positive number; we are computing volume here, not signed volume. Secondly, to compute the volume of a “complicated” region, we could break it up into subregions and compute the volumes of each subregion separately, summing them later to find the total volume.

We use this latter property in the next example.

We give one more example of finding the volume of a space region.

If all one wanted to do in Example 15.6.23 was find the volume of the region \(D\text{,}\) one would have likely stopped at the first integration setup (with order \(dz\, dy\, dx\)) and computed the volume from there. However, we included the other two methods 1) to show that it could be done, “messy” or not, and 2) because sometimes we “have” to use a less desirable order of integration in order to actually integrate.

There are uses for triple integration beyond merely finding volume, just as there are uses for integration beyond “area under the curve.” These uses start with understanding how to integrate functions of three variables, which is effectively no different than integrating functions of two variables. This leads us to a definition, followed by an example.

We now know how to evaluate a triple integral of a function of three variables; we do not yet understand what it means. We build up this understanding in a way very similar to how we have understood integration and double integration.

Let \(h(x,y,z)\) be a continuous function of three variables, defined over some space region \(D\text{.}\) We can partition \(D\) into \(n\) rectangular-solid subregions, each with dimensions \(\dx_i\times\dy_i\times\ddz_i\text{.}\) Let \((x_i,y_i,z_i)\) be some point in the \(i\)th subregion, and consider the product \(h(x_i,y_i,z_i)\dx_i\dy_i\ddz_i\text{.}\) It is the product of a function value (that’s the \(h(x_i,y_i,z_i)\) part) and a small volume \(\Delta V_i\) (that’s the \(\dx_i\dy_i\ddz_i\) part). One of the simplest understanding of this type of product is when \(h\) describes the density of an object, for then \(h\times\text{ volume } =\text{ mass }\text{.}\)

While this limit has lots of interpretations depending on the function \(h\text{,}\) in the case where \(h\) describes density, \(S\) is the total mass of the object described by the region \(D\text{.}\)

We now use the above limit to define the triple integral, give a theorem that relates triple integrals to iterated iteration, followed by the application of triple integrals to find the centers of mass of solid objects.

The following theorem assures us that the above limit exists for continuous functions \(h\) and gives us a method of evaluating the limit.

Here we fix a \(k\) value, which establishes the \(z\)-height of the rectangular solids on one “level” of all the rectangular solids in the space region \(D\text{.}\) The inner double summation adds up all the volumes of the rectangular solids on this level, while the outer summation adds up the volumes of each level.

This triple summation understanding leads to the \(\iiint_D\) notation of the triple integral, as well as the method of evaluation shown in Theorem 15.6.29.

We now apply triple integration to find the centers of mass of solid objects.

One may wish to review Section 15.4 for a reminder of the relevant terms and concepts.

As stated before, there are many uses for triple integration beyond finding volume. When \(h(x,y,z)\) describes a rate of change function over some space region \(D\text{,}\) then \(\ds \iiint_D h(x,y,z)\, dV\) gives the total change over \(D\text{.}\) Our one specific example of this was computing mass; a density function is simply a “rate of mass change per volume” function. Integrating density gives total mass.

While knowing how to integrate is important, it is arguably much more important to know how to set up integrals. It takes skill to create a formula that describes a desired quantity; modern technology is very useful in evaluating these formulas quickly and accurately.

In Section 15.7, we learn about two new coordinate systems (each related to polar coordinates) that allow us to integrate over closed regions in space more easily than when using rectangular coordinates.