*Mass* and *weight* are different measures. Since they are scalar multiples of each other, it is often easy to treat them as the same measure. In this section we effectively treat them as the same, as our technique for finding mass is the same as for finding weight. The density functions used will simply have different units.

# Section13.4Center of Mass¶ permalink

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 \(\iint_R\ dA\text{.}\) That is, summing up the areas of lots of little subregions of \(R\) gave us the total area. Informally, we think of \(\iint_R\ dA\) as meaning “sum up lots of little areas over \(R\text{.}\)”

To find the signed volume under a surface, we evaluated the double integral \(\iint_R f(x,y)\ dA\text{.}\) Recall that the “\(dA\)” is not just a “bookend” at the end of an integral; rather, it is multiplied by \(f(x,y)\text{.}\) We regard \(f(x,y)\) as giving a height, and \(dA\) still giving an area: \(f(x,y)\ dA\) gives a volume. Thus, informally, \(\iint_Rf(x,y)\ dA\) means “sum up lots of little volumes over \(R\text{.}\)”

We now extend these ideas to other contexts.

# Subsection13.4.1Mass and Weight

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 13.4.1(a), with the shape of some planar region \(R\text{,}\) as shown in part (b).

We can write a simple double integral that represents the mass of the lamina: \(\iint_R\ dm\text{,}\) where “\(dm\)” 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 \(R\text{.}\)”

To evaluate this double integral, partition \(R\) into \(n\) subregions as we have done in the past. The \(i^{\,\text{ th } }\) subregion has area \(\Delta A_i\text{.}\) A fundamental property of mass is that “mass=density×area.” If the lamina has a constant density \(\delta\text{,}\) then the mass of this \(i^{\,\text{ th } }\) subregion is \(\Delta m_i=\delta\Delta A_i\text{.}\) 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 \(\delta= \delta(x,y)\text{,}\) then we can approximate the mass of the \(i^{\,\text{ th } }\) subregion of \(R\) by multiplying \(\Delta A_i\) by \(\delta(x_i,y_i)\text{,}\) where \((x_i,y_i)\) is a point in that subregion. That is, for a small enough subregion of \(R\text{,}\) the density across that region is almost constant.

The total mass \(M\) of the lamina is approximately the sum of approximate masses of subregions: \begin{equation*} M \approx \sum_{i=1}^n \Delta m_i = \sum_{i=1}^n \delta(x_i,y_i)\Delta A_i. \end{equation*}

Taking the limit as the size of the subregions shrinks to 0 gives us the actual mass; that is, integrating \(\delta(x,y)\) over \(R\) gives the mass of the lamina.

##### Definition13.4.4Mass of a Lamina with Vairable Density

Let \(\delta(x,y)\) be a continuous density function of a lamina corresponding to a plane region \(R\text{.}\) The mass \(M\) of the lamina is \begin{equation*} \text{ mass } M = \iint_R\ dm = \iint_R \delta(x,y)\ dA. \end{equation*}

##### Example13.4.5Finding the mass of a lamina with constant density

Find the mass of a square lamina, with side length 1, with a density of \(\delta = 3\)gm/cm\(^2\text{.}\)

##### Example13.4.7Finding 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 13.4.6), with variable density \(\delta(x,y) = (x+y+2)\)gm/cm\(^2\text{.}\)

##### Example13.4.9Finding the weight of a lamina with variable density

Find the weight of the lamina represented by the circle with radius 2ft, centered at the origin, with density function \(\delta(x,y) = (x^2+y^2+1)\)lb/ft\(^2\text{.}\) Compare this to the weight of the same lamina with density \(\delta(x,y) = (2\sqrt{x^2+y^2}+1)\)lb/ft\(^2\text{.}\)

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 \(\iint_R f(x,y)\ dA\) as giving the volume under \(f\) over \(R\text{;}\) we can understand \(\iint_R\delta(x,y)\ dA\) in the same way. The “volume” under \(\delta\) over \(R\) is actually mass; by compressing the “volume” under \(\delta\) onto the \(x\)-\(y\) plane, we get “more mass” in some areas than others — i.e., areas of greater density.

Knowing the mass of a lamina is one of several important measures. Another is the *center of mass*, which we discuss next.

# Subsection13.4.2Center of Mass

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 3kg, the disk will behave physically as though it were a point-mass of 3kg 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:
\begin{equation*}
\frac{0.9+0.73+0.85}{3} \approx 0.837 = 83.7\text{ \% } .
\end{equation*}

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:
\begin{equation*}
(0.1)(0.9) + (0.6)(0.73)+(0.3)(0.85) = 0.783 = 78.3\text{ \% } .
\end{equation*}

Each grade is multiplied by a *weight*.

In general, given values \(x_1,x_2,\ldots,x_n\) and weights \(w_1,w_2,\ldots,w_n\text{,}\) the weighted average of the \(n\) values is \begin{equation*} \sum_{i=1}^n w_ix_i\Bigg/\sum_{i=1}^n w_i. \end{equation*}

In the grading example above, the sum of the weights 0.1, 0.6 and 0.3 is 1, so we don't see the division by the sum of weights in that instance.

How this relates to center of mass is given in the following theorem.

##### Theorem13.4.11Center of Mass of Discrete Linear System

Let point masses \(m_1,m_2,\ldots,m_n\) be distributed along the \(x\)-axis at locations \(x_1,x_2,\ldots,x_n\text{,}\) respectively. The center of mass \(\overline{x}\) of the system is located at \begin{equation*} \overline{x} = \sum_{i=1}^nm_ix_i\Bigg/\sum_{i=1}^n m_i. \end{equation*}

##### Example13.4.12Finding the center of mass of a discrete linear system

Point masses of 2gm are located at \(x=-1\text{,}\) \(x=2\) and \(x=3\) are connected by a thin rod of negligible weight. Find the center of mass of the system.

Point masses of 10gm, 2gm and 1gm are located at \(x=-1\text{,}\) \(x=2\) and \(x=3\text{,}\) 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 13.4.11, we are finding a weighted measure of distances from the \(y\)-axis, so we refer to this as *the moment about the \(y\)-axis*, represented by \(M_y\text{.}\) Letting \(M\) be the total mass of the system, we have \(\overline{x} = M_y/M\text{.}\)

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.

##### Definition13.4.16Moments about the \(x\)- and \(y\)- Axes.

Let point masses \(m_1\text{,}\) \(m_2,\ldots,m_n\) be located at points \((x_1,y_1)\text{,}\) \((x_2,y_2)\ldots,(x_n,y_n)\text{,}\) respectively, in the \(x\)-\(y\) plane.

The

*moment about the \(y\)-axis*, \(M_y\text{,}\) is \(\ds M_y = \sum_{i=1}^n m_ix_i\text{.}\)The

*moment about the \(x\)-axis*, \(M_x\text{,}\) is \(\ds M_x = \sum_{i=1}^n m_iy_i\text{.}\)

One can think that these definitions are “backwards” as \(M_y\) sums up “\(x\)” distances. But remember, “\(x\)” distances are measurements of distance from the \(y\)-axis, hence defining the moment about the \(y\)-axis.

We now define the center of mass of discrete points in the plane.

##### Theorem13.4.17Center of Mass of Discrete Planar System

Let point masses \(m_1\text{,}\) \(m_2,\ldots,m_n\) be located at points \((x_1,y_1)\text{,}\) \((x_2,y_2)\ldots,(x_n,y_n)\text{,}\) respectively, in the \(x\)-\(y\) plane, and let \(\ds M = \sum_{i=1}^n m_i\text{.}\)

The center of mass of the system is at \((\overline{x},\overline{y})\text{,}\) where \begin{equation*} \overline{x}= \frac{M_y}{M} \text{ and } \overline{y} = \frac{M_x}{M}. \end{equation*}

##### Example13.4.18Finding the center of mass of a discrete planar system

Let point masses of 1kg, 2kg and 5kg be located at points \((2,0)\text{,}\) \((1,1)\) and \((3,1)\text{,}\) 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 \(R\) in the \(x\)-\(y\) plane with density function \(\delta(x,y)\text{.}\) Partition \(R\) into \(n\) subdivisions, each with area \(\Delta A_i\text{.}\) As done before, we can approximate the mass of the \(i^{\,\text{ th } }\) subregion with \(\delta(x_i,y_i)\Delta A_i\text{,}\) where \((x_i,y_i)\) is a point inside the \(i^{\,\text{ th } }\) subregion. We can approximate the moment of this subregion about the \(y\)-axis with \(x_i\delta(x_i,y_i)\Delta A_i\) — that is, by multiplying the approximate mass of the region by its approximate distance from the \(y\)-axis. Similarly, we can approximate the moment about the \(x\)-axis with \(y_i\delta(x_i,y_i)\Delta A_i\text{.}\) By summing over all subregions, we have: \begin{align*} \text{ mass: } M \amp \approx \sum_{i=1}^n \delta(x_i,y_i)\Delta A_i \text{ (as seen before) }\\ \text{ moment about the \(x\)-axis: } M_x \amp \approx \sum_{i=1}^n y_i\delta(x_i,y_i)\Delta A_i\\ \text{ moment about the \(y\)-axis: } M_y \amp \approx \sum_{i=1}^n x_i\delta(x_i,y_i)\Delta A_i \end{align*}

By taking limits, where size of each subregion shrinks to 0 in both the \(x\) and \(y\) directions, we arrive at the double integrals given in the following theorem.

##### Theorem13.4.20Center of Mass of a Planar Lamina, Moments

Let a planar lamina be represented by a region \(R\) in the \(x\)-\(y\) plane with density function \(\delta(x,y)\text{.}\)

\(\ds \text{ mass: } M = \iint_R\delta(x,y)\ dA\)

\(\ds \text{ moment about the \(x\)-axis: } M_x = \iint_Ry\delta(x,y)\ dA\)

\(\ds \text{ moment about the \(y\)-axis: } M_y = \iint_Rx\delta(x,y)\ dA\)

The center of mass of the lamina is \begin{equation*} (\overline{x},\overline{y}) = \left(\frac{M_y}{M},\frac{M_x}M\right). \end{equation*}

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 \(M\text{,}\) \(M_x\) and \(M_y\) and leave the details of the integration to the reader.

##### Example13.4.21Finding the center of mass of a lamina

Find the center mass of a square lamina, with side length 1, with a density of \(\delta = 3\)gm/cm\(^2\text{.}\) (Note: this is the lamina from Example 13.4.5.)

##### Example13.4.23Finding 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 13.4.22), with variable density \(\delta(x,y) = (x+y+2)\)gm/cm\(^2\text{.}\) (Note: this is the lamina from Example 13.4.7.)

##### Example13.4.24Finding the center of mass of a lamina

Find the center of mass of the lamina represented by the circle with radius 2ft, centered at the origin, with density function \(\delta(x,y) = (x^2+y^2+1)\)lb/ft\(^2\text{.}\) (Note: this is one of the lamina used in Example 13.4.9.)

##### Example13.4.25Finding the center of mass of a lamina

Find the center of mass of the lamina represented by the region \(R\) shown in Figure 13.4.26, half an annulus with outer radius 6 and inner radius 5, with constant density 2lb/ft\(^{2}\text{.}\)

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.

# Subsection13.4.3Exercises

Terms and Concepts

In the following exercises, point masses are given along a line or in the plane. Find the center of mass \(\overline{x}\) or \((\overline{x},\overline{y})\text{,}\) as appropriate. (All masses are in grams and distances are in cm.)