5.4.1 Recognize when a function of three variables is integrable over a rectangular box.
5.4.2 Evaluate a triple integral by expressing it as an iterated integral.
5.4.3 Recognize when a function of three variables is integrable over a closed and bounded region.
5.4.4 Simplify a calculation by changing the order of integration of a triple integral.
5.4.5 Calculate the average value of a function of three variables.
In Double Integrals over Rectangular Regions, we discussed the double integral of a function of two variables over a rectangular region in the plane. In this section we define the triple integral of a function of three variables over a rectangular solid box in space, Later in this section we extend the definition to more general regions in
Integrable Functions of Three Variables
We can define a rectangular box in as We follow a similar procedure to what we did in Double Integrals over Rectangular Regions. We divide the interval into subintervals of equal length divide the interval into subintervals of equal length and divide the interval into subintervals of equal length Then the rectangular box is subdivided into subboxes as shown in Figure 5.40.
Figure 5.40A rectangular box in divided into subboxes by planes parallel to the coordinate planes.
For each consider a sample point in each sub-box We see that its volume is Form the triple Riemann sum
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 over a rectangular box is defined as
5.10
if this limit exists.
When the triple integral exists on the function is said to be integrable on Also, the triple integral exists if is continuous on Therefore, we will use continuous functions for our examples. However, continuity is sufficient but not necessary; in other words, is bounded on and continuous except possibly on the boundary of The sample point can be any point in the rectangular sub-box 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 5.9
Fubini’s Theorem for Triple Integrals
If is continuous on a rectangular box then
This integral is also equal to any of the other five possible orderings for the iterated triple integral.
For and 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).
Example 5.36
Evaluating a Triple Integral
Evaluate the triple integral
Solution
The order of integration is specified in the problem, so integrate with respect to first, then y, and then
Example 5.37
Evaluating a Triple Integral
Evaluate the triple integral where as shown in the following figure.
Figure 5.41Evaluating a triple integral over a given rectangular box.
Solution
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 y first, then x, and then z.
Now try to integrate in a different order just to see that we get the same answer. Choose to integrate with respect to first, then and then
Checkpoint 5.23
Evaluate the triple integral where
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 in The general bounded regions we will consider are of three types. First, let be the bounded region that is a projection of onto the -plane. Suppose the region in has the form
For two functions and such that for all in as shown in the following figure.
Figure 5.42We can describe region as the space between and above the projection of onto the -plane.
Theorem 5.10
Triple Integral over a General Region
The triple integral of a continuous function over a general three-dimensional region
in where is the projection of onto the -plane, is
Similarly, we can consider a general bounded region in the -plane and two functions and such that for all in Then we can describe the solid region in as
where is the projection of onto the -plane and the triple integral is
Finally, if is a general bounded region in the -plane and we have two functions and such that for all in then the solid region in can be described as
where is the projection of onto the -plane and the triple integral is
Figure 5.43A box where the projection in the -plane is of Type I.
Then the triple integral becomes
If in the -plane is of Type II (Figure 5.44), then
Figure 5.44A box where the projection in the -plane is of Type II.
Then the triple integral becomes
Example 5.38
Evaluating a Triple Integral over a General Bounded Region
Evaluate the triple integral of the function over the solid tetrahedron bounded by the planes and
Solution
Figure 5.45 shows the solid tetrahedron and its projection on the -plane.
Figure 5.45The solid has a projection on the -plane of Type I.
We can describe the solid region tetrahedron as
Hence, the triple integral is
To simplify the calculation, first evaluate the integral We have
Now evaluate the integral obtaining
Finally, evaluate
Putting it all together, we have
Just as we used the double integral to find the area of a general bounded region we can use to find the volume of a general solid bounded region The next example illustrates the method.
Example 5.39
Finding a Volume by Evaluating a Triple Integral
Find the volume of a right pyramid that has the square base in the -plane and vertex at the point as shown in the following figure.
Figure 5.46Finding the volume of a pyramid with a square base.
Solution
In this pyramid the value of changes from and at each height the cross section of the pyramid for any value of is the square Hence, the volume of the pyramid is where
Thus, we have
Hence, the volume of the pyramid is cubic units.
Checkpoint 5.24
Consider the solid sphere Write the triple integral for an arbitrary function as an iterated integral. Then evaluate this triple integral with Notice that this gives the volume of a sphere using a triple integral.
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 5.40
Changing the Order of Integration
Consider the iterated integral
The order of integration here is first with respect to z, then y, and then x. Express this integral by changing the order of integration to be first with respect to x, then z, and then Verify that the value of the integral is the same if we let
Solution
The best way to do this is to sketch the region and its projections onto each of the three coordinate planes. Thus, let
and
We need to express this triple integral as
Knowing the region we can draw the following projections (Figure 5.47):
on the -plane is
on the -plane is and
on the -plane is
Figure 5.47The three cross sections of on the three coordinate planes.
Now we can describe the same region as and consequently, the triple integral becomes
Now assume that in each of the integrals. Then we have
The answers match.
Checkpoint 5.25
Write five different iterated integrals equal to the given integral
Example 5.41
Changing Integration Order and Coordinate Systems
Evaluate the triple integral where is the region bounded by the paraboloid (Figure 5.48) and the plane
Figure 5.48Integrating a triple integral over a paraboloid.
Solution
The projection of the solid region onto the -plane is the region bounded above by and below by the parabola as shown.
Figure 5.49Cross section in the -plane of the paraboloid in Figure 5.48.
Thus, we have
The triple integral becomes
This expression is difficult to compute, so consider the projection of onto the -plane. This is a circular disc So we obtain
Here the order of integration changes from being first with respect to then and then to being first with respect to then to and then to It will soon be clear how this change can be beneficial for computation. We have
Now use the polar substitution and in the -plane. This is essentially the same thing as when we used polar coordinates in the -plane, except we are replacing by Consequently the limits of integration change and we have, by using
Average Value of a Function of Three Variables
Recall that we found the average value of a function of two variables by evaluating the double integral over a region on the plane and then dividing by the area of the region. Similarly, we can find the average value of a function in three variables by evaluating the triple integral over a solid region and then dividing by the volume of the solid.
Theorem 5.11
Average Value of a Function of Three Variables
If is integrable over a solid bounded region with positive volume then the average value of the function is
Note that the volume is
Example 5.42
Finding an Average Temperature
The temperature at a point of a solid bounded by the coordinate planes and the plane is Find the average temperature over the solid.
Solution
Use the theorem given above and the triple integral to find the numerator and the denominator. Then do the division. Notice that the plane has intercepts and The region looks like
Hence the triple integral of the temperature is
The volume evaluation is
Hence the average value is degrees Celsius.
Checkpoint 5.26
Find the average value of the function over the cube with sides of length units in the first octant with one vertex at the origin and edges parallel to the coordinate axes.
Section 5.4 Exercises
In the following exercises, evaluate the triple integrals over the rectangular solid box
[T] The volume of a solid is given by the integral Use a computer algebra system (CAS) to graph and find its volume. Round your answer to two decimal places.
214.
[T] The volume of a solid is given by the integral Use a CAS to graph and find its volume Round your answer to two decimal places.
In the following exercises, use two circular permutations of the variables to write new integrals whose values equal the value of the original integral. A circular permutation of is the arrangement of the numbers in one of the following orders:
The midpoint rule for the triple integral over the rectangular solid box is a generalization of the midpoint rule for double integrals. The region is divided into subboxes of equal sizes and the integral is approximated by the triple Riemann sum where is the center of the box and is the volume of each subbox. Apply the midpoint rule to approximate over the solid by using a partition of eight cubes of equal size. Round your answer to three decimal places.
234.
[T]
Apply the midpoint rule to approximate over the solid by using a partition of eight cubes of equal size. Round your answer to three decimal places.
Use a CAS to improve the above integral approximation in the case of a partition of cubes of equal size, where
Suppose that the temperature in degrees Celsius at a point of a solid bounded by the coordinate planes and is Find the average temperature over the solid.
236.
Suppose that the temperature in degrees Fahrenheit at a point of a solid bounded by the coordinate planes and is Find the average temperature over the solid.
237.
Show that the volume of a right square pyramid of height and side length is by using triple integrals.
238.
Show that the volume of a regular right hexagonal prism of edge length is by using triple integrals.
239.
Show that the volume of a regular right hexagonal pyramid of edge length is by using triple integrals.
240.
If the charge density at an arbitrary point of a solid is given by the function then the total charge inside the solid is defined as the triple integral Assume that the charge density of the solid enclosed by the paraboloids and is equal to the distance from an arbitrary point of to the origin. Set up the integral that gives the total charge inside the solid