So far, we have interpreted the double integral of a function over a domain in two different ways. First, tells us a difference of volumes β the volume the surface defined by bounds above the -plane on minus the volume the surface bounds below the -plane on . In addition, determines the average value of on . In this section, we investigate several other applications of double integrals, using the integration process as seen in Preview Activity 11.4.1: we partition into small regions, approximate the desired quantity on each small region, then use the integral to sum these values exactly in the limit.
The following preview activity explores how a double integral can be used to determine the density of a thin plate with a mass density distribution. Recall that in single-variable calculus, we considered a similar problem and computed the mass of a one-dimensional rod with a mass-density distribution. There, as here, the key idea is that if density is constant, mass is the product of density and volume.
Suppose that we have a flat, thin object (called a lamina) whose density varies across the object. We can think of the density on a lamina as a measure of mass per unit area. As an example, consider a circular plate of radius 1 cm centered at the origin whose density varies depending on the distance from its center so that the density in grams per square centimeter at point is
Suppose that we partition the plate into subrectangles , where and , of equal area , and select a point in for each and . What is the meaning of the quantity ?
Determine an iterated integral which, if evaluated, would give the exact mass of the plate. Do not actually evaluate the integral. (This integral is considerably easier to evaluate in polar coordinates, which we will learn more about in Section 11.5.)
Density is a measure of some quantity per unit area or volume. For example, we can measure the human population density of some region as the number of humans in that region divided by the area of that region. In physics, the mass density of an object is the mass of the object per unit area or volume. As suggested by Preview Activity 11.4.1, the following holds in general.
Let be a half-disk lamina of radius 3 in quadrants IV and I, centered at the origin as shown in Figure 11.4.1. Assume the density at point is given by . Find the exact mass of the lamina.
If we consider the situation where the mass-density distribution is constant, we can also see how a double integral may be used to determine the area of a region. Assuming that over a closed bounded region , where the units of are βmass per unit of area,β it follows that is the mass of the lamina. But since the density is constant, the numerical value of the integral is simply the area.
The volume of a solid with constant height is given by the area of the base times the height. Hence, we may interpret the area of the region as the volume of a solid with base and of uniform height 1. That is, the area of is given by . Write an iterated integral whose value is . (Hint: Which order of integration might be more efficient? Why?)
The center of mass of an object is a point at which the object will balance perfectly. For example, the center of mass of a circular disk of uniform density is located at its center. For any object, if we throw it through the air, it will spin around its center of mass and behave as if all the mass is located at the center of mass.
In order to understand the role that integrals play in determining the center of mass of an object with a nonuniform mass distribution, we start by finding the center of mass of a collection of distinct point-masses in the plane.
Let ,,, be masses located in the plane. Think of these masses as connected by rigid rods of negligible weight from some central point . A picture with four masses is shown in Figure 11.4.3. Now imagine balancing this system by placing it on a thin pole at the point perpendicular to the plane containing the masses. Unless the masses are perfectly balanced, the system will fall off the pole. The point at which the system will balance perfectly is called the center of mass of the system. Our goal is to determine the center of mass of a system of discrete masses, then extend this to a continuous lamina.
Each mass exerts a force (called a moment) around the lines and that causes the system to tilt in the direction of the mass. These moments are dependent on the mass and the distance from the given line. Let be the location of mass , the location of mass , etc. In order to balance perfectly, the moments in the direction and in the direction must be in equilibrium. We determine these moments and solve the resulting system to find the equilibrium point at the center of mass.
The value is called the total moment with respect to the -axis; is the total moment with respect to the -axis. Hence, the respective quotients of the moments to the total mass, , determines the center of mass of a point-mass system:
Now, suppose that rather than a point-mass system, we have a continuous lamina with a variable mass-density . We may estimate its center of mass by partitioning the lamina into subrectangles of equal area , and treating the resulting partitioned lamina as a point-mass system. In particular, we select a point in the th subrectangle, and observe that the quanity
is density times area, so approximates the mass of the small portion of the lamina determined by the subrectangle .
The center of mass of a lamina can then be thought of as a weighted average of all of the points in the lamina with the weights given by the density at each point. The centroid of a lamina is the just the average of all of the points in the lamina, or the center of mass if the density at each point is 1.
The numerators of and are called the respective moments of the lamina about the coordinate axes. Thus, the moment of a lamina with density about the -axis is
In this activity we determine integrals that represent the center of mass of a lamina described by the triangular region bounded by the -axis and the lines and in the first quadrant if the density at point is . A picture of the lamina is shown in Figure 11.4.4.
Calculating probabilities is a very important application of integration in the physical, social, and life sciences. To understand the basics, consider the game of darts in which a player throws a dart at a board and tries to hit a particular target. Let us suppose that a dart board is in the form of a disk with radius 10 inches. If we assume that a player throws a dart at random, and is not aiming at any particular point, then it is equally probable that the dart will strike any single point on the board. For instance, the probability that the dart will strike a particular 1 square inch region is , or the ratio of the area of the desired target to the total area of (assuming that the dart thrower always hits the board itself at some point). Similarly, the probability that the dart strikes a point in the disk of radius 3 inches is given by the area of divided by the area of . In other words, the probability that the dart strikes the disk is
The integrand, , may be thought of as a distribution function, describing how the dart strikes are distributed across the board. In this case the distribution function is constant since we are assuming a uniform distribution, but we can easily envision situations where the distribution function varies. For example, if the player is fairly good and is aiming for the bulls eye (the center of ), then the distribution function could be skewed toward the center, say
for some constant positive . If we assume that the player is consistent enough so that the dart always strikes the board, then the probability that the dart strikes the board somewhere is 1, and the distribution function will have to satisfyβ1β
The preceding discussion highlights the general idea behind calculating probabilities. We assume we have a joint probability density function, a function of two independent variables and defined on a domain that satisfies the conditions
Note that it is possible that could be an infinite region and the limits on the integral in Equation (11.4.1) could be infinite. When we have such a probability density function , the probability that the point is in some region contained in the domain (the notation we use here is ββ) is determined by
A firm manufactures smoke detectors. Two components for the detectors come from different suppliers β one in Michigan and one in Ohio. The company studies these components for their reliability and their data suggests that if is the life span (in years) of a randomly chosen component from the Michigan supplier and the life span (in years) of a randomly chosen component from the Ohio supplier, then the joint probability density function might be given by
Theoretically, the components might last forever, so the domain of the function is the set of all such that and . To show that is a probability density function on we need to demonstrate that
or that
Use your knowledge of improper integrals to verify that is indeed a probability density function.
Assume that the smoke detector fails only if both of the supplied components fail. To determine the probability that a randomly selected detector will fail within one year, we will need to determine the probability that the life span of each component is between 0 and 1 years. Set up an appropriate iterated integral, and evaluate the integral to determine the probability.
Suppose that the manufacturer determines that one of the components is more likely to fail than the other, and hence conjectures that the probability density function is instead What is the value of ?
Given a joint probability density function is a function of two independent variables and defined on a domain , if is some subregion of , then the probability that is in is given by
A lamina occupies the region inside the circle but outside the circle . The density at each point is inversely proportional to its distance from the orgin.
A triangular plate is bounded by the graphs of the equations ,, and . The plateβs density at is given by , measured in grams per square centimeter (and and are measured in centimeters).
Set up an iterated integral whose value is the mass of the plate. Include a labeled sketch of the region of integration. Why did you choose the order of integration you did?
Let denote the time (in minutes) that a person spends waiting in a checkout line at a grocery store and the time (in minutes) that it takes to check out. Suppose the joint probability density for and is
What is the exact probability that a person spends between 0 to 5 minutes waiting in line, and then 0 to 5 minutes waiting to check out?
Set up, but do not evaluate, an iterated integral whose value determines the exact probability that a person spends at most 10 minutes total both waiting in line and checking out at this grocery store.
Set up, but do not evaluate, an iterated integral expression whose value determines the exact probability that a person spends at least 10 minutes total both waiting in line and checking out, but not more than 20 minutes.