We have used iterated integrals to find areas of plane regions and signed volumes under surfaces. A brief recap of these uses will be useful in this section as we apply iterated integrals to compute the mass and center of mass of planar regions.
To find the area of a planar region, we evaluated the double integral . That is, summing up the areas of lots of little subregions of gave us the total area. Informally, we think of as meaning “sum up lots of little areas over .”
To find the signed volume under a surface, we evaluated the double integral . Recall that the “” is not just a “bookend” at the end of an integral; rather, it is multiplied by . We regard as giving a height, and still giving an area: gives a volume. Thus, informally, means “sum up lots of little volumes over .”
Consider a thin sheet of material with constant thickness and finite area. Mathematicians (and physicists and engineers) call such a sheet a lamina. So consider a lamina, as shown in Figure 14.4.1.(a), with the shape of some planar region , as shown in Figure 14.4.1.(b).
We can write a simple double integral that represents the mass of the lamina: , where “” means “a little mass.” That is, the double integral states the total mass of the lamina can be found by “summing up lots of little masses over .”
To evaluate this double integral, partition into subregions as we have done in the past. The th subregion has area . A fundamental property of mass is that “mass=density×area.” If the lamina has a constant density , then the mass of this th subregion is . That is, we can compute a small amount of mass by multiplying a small amount of area by the density.
If density is variable, with density function , then we can approximate the mass of the th subregion of by multiplying by , where is a point in that subregion. That is, for a small enough subregion of , the density across that region is almost constant.
We represent the lamina with a square region in the plane as shown in Figure 14.4.4. As the density is constant, it does not matter where we place the square.
Figure14.4.4.A region representing a lamina in Example 14.4.3
Example14.4.5.Finding the mass of a lamina with variable density.
Find the mass of a square lamina, represented by the unit square with lower lefthand corner at the origin (see Figure 14.4.4), with variable density g/cm.
The variable density , in this example, is very uniform, giving a density of 3 in the center of the square and changing linearly. A graph of can be seen in Figure 14.4.6; notice how “same amount” of density is above as below. We’ll comment on the significance of this momentarily.
The mass is found by integrating over . The order of integration is not important; we choose arbitrarily. Thus:
g.
It turns out that since the density of the lamina is so uniformly distributed “above and below” that the mass of the lamina is the same as if it had a constant density of 3. The density functions in Example 14.4.3 and Example 14.4.5 are graphed in Figure 14.4.6, which illustrates this concept.
Example14.4.7.Finding the weight of a lamina with variable density.
Find the weight of the lamina represented by the disk with radius 2 ft, centered at the origin, with density function lb/ft. Compare this to the weight of the lamina with the same shape and density lb/ft.
A direct application of Definition 14.4.2 states that the weight of the lamina is . Since our lamina is in the shape of a circle, it makes sense to approach the double integral using polar coordinates.
The density function becomes
.
The circle is bounded by and . Thus the weight is:
lb .
Now compare this with the density function . Converting this to polar coordinates gives
.
Thus the weight is:
lb .
One would expect different density functions to return different weights, as we have here. The density functions were chosen, though, to be similar: each gives a density of 1 at the origin and a density of 5 at the outside edge of the circle, as seen in Figure 14.4.8.
Plotting the density functions can be useful as our understanding of mass can be related to our understanding of “volume under a surface.” We interpreted as giving the volume under over ; we can understand in the same way. The “volume” under over is actually mass; by compressing the “volume” under onto the -plane, we get “more mass” in some areas than others — i.e., areas of greater density.
Consider a disk of radius 1 with uniform density. It is common knowledge that the disk will balance on a point if the point is placed at the center of the disk. What if the disk does not have a uniform density? Through trial-and-error, we should still be able to find a spot on the disk at which the disk will balance on a point. This balance point is referred to as the center of mass, or center of gravity. It is though all the mass is “centered” there. In fact, if the disk has a mass of 3 kg, the disk will behave physically as though it were a point-mass of 3 kg located at its center of mass. For instance, the disk will naturally spin with an axis through its center of mass (which is why it is important to “balance” the tires of your car: if they are “out of balance”, their center of mass will be outside of the axle and it will shake terribly).
We find the center of mass based on the principle of a weighted average. Consider a college class in which your homework average is 90%, your test average is 73%, and your final exam grade is an 85%. Experience tells us that our final grade is not the average of these three grades: that is, it is not:
That is, you are probably not pulling a B in the course. Rather, your grades are weighted. Let’s say the homework is worth 10% of the grade, tests are 60% and the exam is 30%. Then your final grade is:
Example14.4.10.Finding the center of mass of a discrete linear system.
Point masses of 2 g are located at , and are connected by a thin rod of negligible weight. Find the center of mass of the system.
Point masses of 10 g, 2 g and 1 g are located at , and , respectively, are connected by a thin rod of negligible weight. Find the center of mass of the system.
In a discrete system (i.e., mass is located at individual points, not along a continuum) we find the center of mass by dividing the mass into a moment of the system. In general, a moment is a weighted measure of distance from a particular point or line. In the case described by Theorem 14.4.9, we are finding a weighted measure of distances from the -axis, so we refer to this as the moment about the -axis, represented by . Letting be the total mass of the system, we have .
We can extend the concept of the center of mass of discrete points along a line to the center of mass of discrete points in the plane rather easily. To do so, we define some terms then give a theorem.
One can think that these definitions are “backwards” as sums up “” distances. But remember, “” distances are measurements of distance from the -axis, hence defining the moment about the -axis.
Example14.4.14.Finding the center of mass of a discrete planar system.
Let point masses of 1 kg, 2 kg and 5 kg be located at points , and , respectively, and are connected by thin rods of negligible weight. Find the center of mass of the system.
We finally arrive at our true goal of this section: finding the center of mass of a lamina with variable density. While the above measurement of center of mass is interesting, it does not directly answer more realistic situations where we need to find the center of mass of a contiguous region. However, understanding the discrete case allows us to approximate the center of mass of a planar lamina; using calculus, we can refine the approximation to an exact value.
We begin by representing a planar lamina with a region in the -plane with density function . Partition into subdivisions, each with area . As done before, we can approximate the mass of the th subregion with , where is a point inside the th subregion. We can approximate the moment of this subregion about the -axis with — that is, by multiplying the approximate mass of the region by its approximate distance from the -axis. Similarly, we can approximate the moment about the -axis with . By summing over all subregions, we have:
By taking limits, where size of each subregion shrinks to 0 in both the and directions, we arrive at the double integrals given in the following theorem.
We start our practice of finding centers of mass by revisiting some of the lamina used previously in this section when finding mass. We will just set up the integrals needed to compute , and and leave the details of the integration to the reader.
Example14.4.19.Finding the center of mass of a lamina.
Find the center of mass of a square lamina, represented by the unit square with lower lefthand corner at the origin (see Figure 14.4.18), with variable density g/cm. (Note: this is the lamina from Example 14.4.5.)
While the mass of this lamina is the same as the lamina in the previous example, the greater density found with greater and values pulls the center of mass from the center slightly towards the upper righthand corner.
Example14.4.20.Finding the center of mass of a lamina.
Find the center of mass of the lamina represented by the circle with radius 2 ft, centered at the origin, with density function lb/ft. (Note: this is one of the lamina used in Example 14.4.7.)
As done in Example 14.4.7, it is best to describe using polar coordinates. Thus when we compute , we will integrate not , but rather . We compute , and :
lb.
Since and the density of are both symmetric about the and axes, it should come as no big surprise that the moments about each axis is 0. Thus the center of mass is .
Example14.4.21.Finding the center of mass of a lamina.
Find the center of mass of the lamina represented by the region shown in Figure 14.4.22, half an annulus with outer radius 6 ft and inner radius 5 ft, with constant density 2 lb⁄ft2.
Once again it will be useful to represent in polar coordinates. Using the description of and/or the illustration, we see that is bounded by and . As the lamina is symmetric about the -axis, we should expect . We compute , and :
lb.
Thus the center of mass is . The center of mass is indicated in Figure 14.4.22; note how it lies outside of !
This section has shown us another use for iterated integrals beyond finding area or signed volume under the curve. While there are many uses for iterated integrals, we give one more application in the following section: computing surface area.
In the following exercises, point masses are given along a line or in the plane. Find the center of mass or , as appropriate. (All masses are in grams and distances are in cm.)
The moment of inertia is a measure of the tendency of a lamina to resist rotating about an axis or continue to rotate about an axis. is the moment of inertia about the -axis, is the moment of inertia about the -axis, and is the moment of inertia about the origin. These are computed as follows: