Applications of Double Integrals

This section will be given different coverage by different instructors. We have tried to keep the essential elements of double integrals in the first three sections of this chapter and include many applications of double integrals in this section. The idea of double integrals being an accumulation of density functions is mentioned in the first three sections of this chapter but is explored fully in this section.

So far, we have interpreted the double integral of a function \(f\) over a region \(D\) in three main ways, as described in . However, our focus has been primarily on using double integrals to find the volume the surface defined by \(z=f(x,y)\) bounds above the \(xy\)-plane over \(D\) minus the volume the surface bounds below the \(xy\)-plane over \(D\). In this section, we investigate several other applications of double integrals, using the integration process as seen in Preview Activity: 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 the second interpretation of double integrals in : a double integral can be used to determine the mass of a thin plate with a density distribution, \(\delta(x,y)\). 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.

The simplest possible density function we could have considered in the Preview Activity would have been \(\delta(x,y)=1\), so we begin by considering what such a double integral represents. In fact, we can do this without needing to think of density at all. If the integrand in a double integral is \(1\), then our volume interpretation tells us that the double integral represents the volume of the solid underneath the plane \(z=1\) and above the region \(D\) of integration in the \(xy\)-plane. Since this solid is cylindrical and has height \(1\), its volume is also the area of the base region \(D\), as we can see in the activity below.

We now formally state the conclusion from our earlier discussion and Activity.

Given a closed, bounded region \(D\) in the plane, the area of \(D\), denoted \(A(D)\), is given by the double integral \[\begin{aligned}\end{aligned}\]

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, the following holds in general.

If \(\delta(x, y)\) describes the density of a lamina defined by a planar region \(D\), then the mass of \(D\) is given by the double integral \(\displaystyle{\iint_D \delta(x,y) \, dA}\).

Calculating probabilities is an 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. Suppose that a dart board is in the form of a disk \(D\) with radius 10 inches. If we assume that a player throws a dart at random without 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 \(\frac{1}{100 \pi}\), or the ratio of the area of the desired target to the total area of \(D\). (This makes the perhaps unreasonable assumption that the dart thrower always hits the board itself at some point.) Similarly, the probability that the dart strikes a point in a disk \(D_3\) of radius 3 inches inside the larger disk is given by the area of \(D_3\) divided by the area of \(D\). In other words, the probability that the dart strikes the disk \(D_3\) is \[\begin{aligned}\end{aligned}\]

The integrand, \(\frac{1}{100\pi}\), 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. However, we can envision situations where the distribution function varies. For example, if the player is fairly good and is aiming for the bullseye (the center of \(D\)), then the distribution function \(f\) could be skewed toward the center, perhaps \[\begin{aligned}\end{aligned}\] for some constant positive \(K\). If we assume that the player is consistent enough so that the dart always strikes the board, then the total probability that the dart strikes the board somewhere is 1, and the distribution function \(f\) must satisfyThis makes \(K = \frac{1}{\pi\left(1-e^{-100}\right)}\), which you can check. \[\begin{aligned}\end{aligned}\]

For such a function \(f\), the probability that the dart strikes in the disk \(D_1\) of radius 1 would be \[\begin{aligned}\end{aligned}\]

Indeed, the probability that the dart strikes in any region \(R\) that lies within \(D\) is given by \[\begin{aligned}\end{aligned}\]

Note that it is possible that \(D\) could be an infinite region and the limits on the integral in Equation could be infinite. When we have such a probability density function \(f=f(x,y)\), the probability that the point \((x,y)\) is in some region \(R\) contained in the domain \(D\) is determined by \[\begin{aligned}\end{aligned}\]

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

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 \(N\) distinct point-masses in the plane.

Let \(m_1\), \(m_2\), \(\ldots\), \(m_N\) be \(N\) masses located in the plane. Think of these masses as connected by rigid rods of negligible weight from some central point \((x,y)\). A picture with four masses is shown in Figure. Now imagine balancing this system by placing it on a thin pole at the point \((x,y)\) perpendicular to the plane containing the masses. Unless the masses are perfectly balanced, the system will fall off the pole. The point \((\overline{x}, \overline{y})\) 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 \(x=\overline{x}\) and \(y=\overline{y}\) 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 \((x_1,y_1)\) be the location of mass \(m_1\), \((x_2,y_2)\) the location of mass \(m_2\), etc. In order to balance perfectly, the moments in the \(x\) direction and in the \(y\) direction must be in equilibrium. We determine these moments and solve the resulting system to find the equilibrium point \((\overline{x}, \overline{y})\) at the center of mass.

The force that mass \(m_1\) exerts to tilt the system from the line \(y=\overline{y}\) is \[\begin{aligned}\end{aligned}\], where \(g\) is the gravitational constant. Similarly, the force mass \(m_2\) exerts to tilt the system from the line \(y= \overline{y}\) is \[\begin{aligned}\end{aligned}\].

In general, the force that mass \(m_k\) exerts to tilt the system from the line \(y= \overline{y}\) is \[\begin{aligned}\end{aligned}\]. For the system to balance, we need the forces to sum to 0, so that \[\begin{aligned}\end{aligned}\]. Solving for \(\overline{y}\), we find that \[\begin{aligned}\end{aligned}\].

A similar argument shows that \[\begin{aligned}\end{aligned}\].

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.