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Section 6.3 Density, Mass, and Center of Mass

Studying the units on the integrand and variable of integration helps us understand the meaning of a definite integral. For instance, if \(v(t)\) is the velocity of an object moving along an axis, measured in feet per second, and \(t\) measures time in seconds, then both the definite integral and its Riemann sum approximation,

\begin{equation*} \int_a^b v(t) \, dt \approx \sum_{i=1}^n v(t_i) \Delta t\text{,} \end{equation*}

have units given by the product of the units of \(v(t)\) and \(t\text{:}\)

\begin{equation*} \text{(feet/sec)} \cdot \text{(sec)} = \text{feet}\text{.} \end{equation*}

Thus, \(\int_a^b v(t) \, dt\) measures the total change in position of the moving object in feet.

Unit analysis will be particularly helpful to us in what follows.

Preview Activity 6.3.1.

In each of the following scenarios, we consider the distribution of a quantity along an axis.

  1. Suppose that the function \(c(x) = 200 + 100 e^{-0.1x}\) models the density of traffic on a straight road, measured in cars per mile, where \(x\) is number of miles east of a major interchange, and consider the definite integral \(\int_0^2 (200 + 100 e^{-0.1x}) \, dx\text{.}\)

    1. What are the units on the product \(c(x) \cdot \Delta x\text{?}\)

    2. What are the units on the definite integral and its Riemann sum approximation given by

      \begin{equation*} \int_0^2 c(x) \, dx \approx \sum_{i=1}^n c(x_i) \Delta x? \end{equation*}

    3. Evaluate the definite integral \(\int_0^2 c(x) \, dx = \int_0^2 \left(200 + 100 e^{-0.1x}\right) \, dx\) and write one sentence to explain the meaning of the value you find.

  2. On a 6 foot long shelf filled with books, the function \(B\) models the distribution of the weight of the books, in pounds per inch, where \(x\) is the number of inches from the left end of the bookshelf. Let \(B(x)\) be given by the rule \(B(x) = 0.5 + \frac{1}{(x+1)^2}\text{.}\)

    1. What are the units on the product \(B(x) \cdot \Delta x\text{?}\)

    2. What are the units on the definite integral and its Riemann sum approximation given by

      \begin{equation*} \int_{12}^{36} B(x) \, dx \approx \sum_{i=1}^n B(x_i) \Delta x? \end{equation*}

    3. Evaluate the definite integral \(\int_{0}^{72} B(x) \, dx = \int_0^{72} \left(0.5 + \frac{1}{(x+1)^2}\right) \, dx\) and write one sentence to explain the meaning of the value you find.

Subsection 6.3.1 Density

The mass of a quantity, typically measured in metric units such as grams or kilograms, is a measure of the amount of the quantity. In a corresponding way, the density of an object measures the distribution of mass per unit volume. For instance, if a brick has mass 3 kg and volume 0.002 m\(^3\text{,}\) then the density of the brick is

\begin{equation*} \frac{3 \mbox{kg} }{0.002 \mbox{m} ^3} = 1500 \frac{\mbox{kg} }{\mbox{m} ^3}\text{.} \end{equation*}

As another example, the mass density of water is 1000 kg/m\(^3\text{.}\) Each of these relationships demonstrate the following general principle.

For an object of constant density \(d\text{,}\) with mass \(m\) and volume \(V\text{,}\)

\begin{equation*} d = \frac{m}{V}, \ \text{or} \ m = d \cdot V\text{.} \end{equation*}

But what happens when the density is not constant?

The formula \(m = d \cdot V\) is reminiscent of two other equations that we have used in our work: for a body moving in a fixed direction, distance = rate \(\cdot\) time, and, for a rectangle, its area is given by \(A = l \cdot w\text{.}\) These formulas hold when the principal quantities involved, such as the rate the body moves and the height of the rectangle, are constant. When these quantities are not constant, we have turned to the definite integral for assistance. By working with small slices on which the quantity of interest (such as velocity) is approximately constant, we can use a definite integral to add up the values on the pieces.

For example, if we have a nonnegative velocity function that is not constant, over a short time interval \(\Delta t\) we know that the distance traveled is approximately \(v(t) \Delta t\text{,}\) since \(v(t)\) is almost constant on a small interval. Similarly, if we are thinking about the area under a nonnegative function \(f\) whose value is changing, on a short interval \(\Delta x\) the area under the curve is approximately the area of the rectangle whose height is \(f(x)\) and whose width is \(\Delta x\text{:}\) \(f(x) \Delta x\text{.}\) Both of these principles are represented visually in Figure 6.3.1.

Figure 6.3.1. At left, estimating a small amount of distance traveled, \(v(t) \Delta t\text{,}\) and at right, a small amount of area under the curve, \(f(x) \Delta x\text{.}\)

In a similar way, if the density of some object is not constant, we can use a definite integral to compute the overall mass of the object. We will focus on problems where the density varies in only one dimension, say along a single axis.

Let's consider a thin bar of length \(b\) whose left end is at the origin, where \(x = 0\text{,}\) and assume that the bar has constant cross-sectional area of 1 cm\(^2\text{.}\) We let \(\rho(x)\) represent the mass density function of the bar, measured in grams per cubic centimeter. That is, given a location \(x\text{,}\) \(\rho(x)\) tells us approximately how much mass will be found in a one-centimeter wide slice of the bar at \(x\text{.}\)

Figure 6.3.2. A thin bar of constant cross-sectional area 1 cm\(^2\) with density function \(\rho(x)\) g/cm\(^3\text{.}\)

The volume of a thin slice of the bar of width \(\Delta x\text{,}\) as pictured in Figure 6.3.2, is the cross-sectional area times \(\Delta x\text{.}\) Since the cross-sections each have constant area 1 cm\(^2\text{,}\) it follows that the volume of the slice is \(1 \Delta x\) cm\(^3\text{.}\) And because mass is the product of density and volume, we see that the mass of this slice is approximately

\begin{equation*} \text{mass} _{\text{slice} } \approx \rho(x) \ \frac{\mbox{g} }{\mbox{cm} ^3} \cdot 1 \Delta x \ \mbox{cm} ^3 = \rho(x) \cdot \Delta x \ \mbox{g}\text{.} \end{equation*}

The corresponding Riemann sum (and the integral that it approximates),

\begin{equation*} \sum_{i=1}^n \rho(x_i) \Delta x \approx \int_0^b \rho(x) \, dx\text{,} \end{equation*}

therefore measure the mass of the bar between \(0\) and \(b\text{.}\) (The Riemann sum is an approximation, while the integral will be the exact mass.)

For objects whose cross-sectional area is constant and whose mass is distributed relative to horizontal location, \(x\text{,}\) it makes sense to think of the density function \(\rho(x)\) with units “mass per unit length,” such as g/cm. Thus, when we compute \(\rho(x) \cdot \Delta x\) on a small slice \(\Delta x\text{,}\) the resulting units are g/cm \(\cdot\) cm = g, which thus measures the mass of the slice. The general principle follows.

For an object of constant cross-sectional area whose mass is distributed along a single axis according to the function \(\rho(x)\) (whose units are units of mass per unit of length), the total mass, \(M\text{,}\) of the object between \(x = a\) and \(x = b\) is given by

\begin{equation*} M = \int_a^b \rho(x) \, dx\text{.} \end{equation*}
Activity 6.3.2.

Consider the following situations in which mass is distributed in a non-constant manner.

  1. Suppose that a thin rod with constant cross-sectional area of 1 cm\(^2\) has its mass distributed according to the density function \(\rho(x) = 2e^{-0.2x}\text{,}\) where \(x\) is the distance in cm from the left end of the rod, and the units on \(\rho(x)\) are g/cm. If the rod is 10 cm long, determine the exact mass of the rod.

  2. Consider the cone that has a base of radius 4 m and a height of 5 m. Picture the cone lying horizontally with the center of its base at the origin and think of the cone as a solid of revolution.

    1. Write and evaluate a definite integral whose value is the volume of the cone.

    2. Next, suppose that the cone has uniform density of 800 kg/m\(^3\text{.}\) What is the mass of the solid cone?

    3. Now suppose that the cone's density is not uniform, but rather that the cone is most dense at its base. In particular, assume that the density of the cone is uniform across cross sections parallel to its base, but that in each such cross section that is a distance \(x\) units from the origin, the density of the cross section is given by the function \(\rho(x) = 400 + \frac{200}{1+x^2}\text{,}\) measured in kg/m\(^3\text{.}\) Determine and evaluate a definite integral whose value is the mass of this cone of non-uniform density. Do so by first thinking about the mass of a given slice of the cone \(x\) units away from the base; remember that in such a slice, the density will be essentially constant.

  3. Let a thin rod of constant cross-sectional area 1 cm\(^2\) and length 12 cm have its mass be distributed according to the density function \(\rho(x) = \frac{1}{25}(x-15)^2\text{,}\) measured in g/cm. Find the exact location \(z\) at which to cut the bar so that the two pieces will each have identical mass.

Hint

  1. Remember that \(M = \int_a^b \rho(x) \, dx\text{.}\)

  2. Think about slices: what's the volume of a slice? what's the mass of a slice?

  3. Consider \(\int_0^b \rho(x) \, dx\text{.}\)

Answer

  1. \(M = 0.1 - 0.1e^{-4} \approx 0.0981684\) grams.

    1. \(V = \int_{0}^{5} \pi (4 - \frac{4}{5}x)^2 \, dx = \frac{80\pi}{3} \approx 83.7758 \mbox{m}^3\text{.}\)

    2. \(M = \int_{0}^{5} (400 + \frac{200}{1+x^2}) \cdot \pi (4-\frac{4}{5}x)^2 \, dx = \frac{32}{5}\pi(25015 - 5 \ln(17576) + 72\tan^{-1}(5) \approx 33597.4664 \mbox{kg}\)

  3. \(b \approx 3.0652\text{.}\)

Solution

  1. Since the mass, \(M\text{,}\) is given by \(M = \int_a^b \rho(x) \, dx\text{,}\) it follows that

    \begin{equation*} M = \int_{0}^{10} 2e^{-0.2x} = \left. \frac{2}{-0.2}e^{-0.2x} \right|_{0}^{20} = -0.1e^{-4}+0.1e^{0}\text{,} \end{equation*}

    and hence \(M = 0.1 - 0.1e^{-4} \approx 0.0981684\) grams.

  2. Consider the cone that has a base of radius 4 m and a height of 5 m.

    1. With the cone having its base centered at the origin and being a solid of revolution about the \(x\)-axis, the cone is generated by the line \(y = 4 - \frac{4}{5}x\text{.}\) Using the disk method, it follows that the cone's volume is

      \begin{equation*} V = \int_{0}^{5} \pi (4 - \frac{4}{5}x)^2 \, dx = \frac{80\pi}{3} \approx 83.7758 \mbox{m}^3\text{.} \end{equation*}

    2. If the cone has uniform density 800 kg/m\(^3\text{,}\) we know that the cone's mass is the product of its density and volume, and thus

      \begin{equation*} M = \rho \cdot V = 800 \cdot \frac{80\pi}{3} = \frac{6400\pi}{3} \approx 67020.6433 \mbox{kg}\text{.} \end{equation*}

    3. If the cone has non-uniform density, but is uniform across cross sections parallel to its base such that in each cross section a distance \(x\) units from the origin has its density given by \(\rho(x) = 400 + \frac{200}{1+x^2}\) kg/m\(^3\text{,}\) we naturally consider slices of the cone that are perpendicular to the \(x\)-axis. For each such slice, the radius is \(r(x) = 4 - \frac{4}{5}x\text{,}\) which makes the volume of the slice

      \begin{equation*} V_{\text{slice}} \approx \pi (4-\frac{4}{5}x)^2 \Delta x\text{.} \end{equation*}

      Since the density is constant on the slice with value \(\rho(x) = 400 + \frac{200}{1+x^2}\text{,}\) it follows that the mass of the slice is

      \begin{equation*} M_{\text{slice}} \approx (400 + \frac{200}{1+x^2}) \cdot \pi (4-\frac{4}{5}x)^2 \Delta x\text{.} \end{equation*}

      Using a definite integral to let \(\Delta x \to 0\) and add up the values of all of the slices, we thus find the mass of the cone with this density distribution to be

      \begin{align*} M \amp = \int_{0}^{5} (400 + \frac{200}{1+x^2}) \cdot \pi (4-\frac{4}{5}x)^2 \, dx\\ \amp = \frac{32}{5}\pi(25015 - 5 \ln(17576) + 72\tan^{-1}(5) \approx 33597.4664 \mbox{kg}\\ \amp \approx 33597.4664 \mbox{kg}\text{.} \end{align*}

  3. The total mass of the bar is given by \(M = \int_{0}^{12} \frac{1}{25}(x-15)^2 \, dx = \frac{1116}{25}\text{.}\) To find where to cut the bar in two so that the two pieces have equal mass, we have to determine the value of \(b\) such that

    \begin{equation*} \int_{0}^{b} \frac{1}{25}(x-15)^2 \, dx = \frac{1}{2} \cdot \frac{1116}{25}\text{.} \end{equation*}

    Since \(\int_{0}^{b} \frac{1}{25}(x-15)^2 \, dx = \frac{1}{75}( (b-15)^3 - (-15)^3 )\text{,}\) we need to have \(b\) satisfy the equation

    \begin{equation*} \frac{1}{75}( (b-15)^3 - (-15)^3 ) = \frac{1}{2} \cdot \frac{1116}{25}\text{.} \end{equation*}

    Thus, \((b-15)^3 - (-15)^3 = 75 \cdot \frac{558}{25} = 3 \cdot 558 = 1674\text{,}\) so \((b-15)^3 = 1675 - 3375 = -1700\text{.}\) It follows that \(b-15 = \sqrt[3]{-1700} \approx -11.9348\text{,}\) so \(b \approx 3.0652\text{.}\)

Subsection 6.3.2 Weighted Averages

The concept of an average is a natural one, and one that we have used repeatedly as part of our understanding of the meaning of the definite integral. If we have \(n\) values \(a_1\text{,}\) \(a_2\text{,}\) \(\ldots\text{,}\) \(a_n\text{,}\) we know that their average is given by

\begin{equation*} \frac{a_1 + a_2 + \cdots + a_n}{n}\text{,} \end{equation*}

and for a quantity being measured by a function \(f\) on an interval \([a,b]\text{,}\) the average value of the quantity on \([a,b]\) is

\begin{equation*} \frac{1}{b-a} \int_a^b f(x) \, dx\text{.} \end{equation*}

As we continue to think about problems involving the distribution of mass, it is natural to consider the idea of a weighted average, where certain quantities involved are counted more in the average.

A common use of weighted averages is in the computation of a student's GPA, where grades are weighted according to credit hours. Let's consider the scenario in Table 6.3.3.

class grade grade points credits
chemistry B+ 3.3 5
calculus A- 3.7 4
history B- 2.7 3
psychology B- 2.7 3
Table 6.3.3. A college student's semester grades.

If all of the classes were of the same weight (i.e., the same number of credits), the student's GPA would simply be calculated by taking the average

\begin{equation*} \frac{3.3 + 3.7 + 2.7 + 2.7}{4} = 3.1\text{.} \end{equation*}

But since the chemistry and calculus courses have higher weights (of 5 and 4 credits respectively), we actually compute the GPA according to the weighted average

\begin{equation*} \frac{3.3 \cdot 5 + 3.7 \cdot 4 + 2.7 \cdot 3 + 2.7 \cdot 3}{5 + 4 + 3 + 3} = 3.1\overline{6}\text{.} \end{equation*}

The weighted average reflects the fact that chemistry and calculus, as courses with higher credits, have a greater impact on the students' grade point average. Note particularly that in the weighted average, each grade gets multiplied by its weight, and we divide by the sum of the weights.

In the following activity, we explore further how weighted averages can be used to find the balancing point of a physical system.

Activity 6.3.3.

For quantities of equal weight, such as two children on a teeter-totter, the balancing point is found by taking the average of their locations. When the weights of the quantities differ, we use a weighted average of their respective locations to find the balancing point.

  1. Suppose that a shelf is 6 feet long, with its left end situated at \(x = 0\text{.}\) If one book of weight 1 lb is placed at \(x_1 = 0\text{,}\) and another book of weight 1 lb is placed at \(x_2 = 6\text{,}\) what is the location of \(\overline{x}\text{,}\) the point at which the shelf would (theoretically) balance on a fulcrum?

  2. Now, say that we place four books on the shelf, each weighing 1 lb: at \(x_1 = 0\text{,}\) at \(x_2 = 2\text{,}\) at \(x_3 = 4\text{,}\) and at \(x_4 = 6\text{.}\) Find \(\overline{x}\text{,}\) the balancing point of the shelf.

  3. How does \(\overline{x}\) change if we change the location of the third book? Say the locations of the 1-lb books are \(x_1 = 0\text{,}\) \(x_2 = 2\text{,}\) \(x_3 = 3\text{,}\) and \(x_4 = 6\text{.}\)

  4. Next, suppose that we place four books on the shelf, but of varying weights: at \(x_1 = 0\) a 2-lb book, at \(x_2 = 2\) a 3-lb book, at \(x_3 = 4\) a 1-lb book, and at \(x_4 = 6\) a 1-lb book. Use a weighted average of the locations to find \(\overline{x}\text{,}\) the balancing point of the shelf. How does the balancing point in this scenario compare to that found in (b)?

  5. What happens if we change the location of one of the books? Say that we keep everything the same in (d), except that \(x_3 = 5\text{.}\) How does \(\overline{x}\) change?

  6. What happens if we change the weight of one of the books? Say that we keep everything the same in (d), except that the book at \(x_3 = 4\) now weighs 2 lbs. How does \(\overline{x}\) change?

  7. Experiment with a couple of different scenarios of your choosing where you move one of the books to the left, or you decrease the weight of one of the books.

  8. Write a couple of sentences to explain how adjusting the location of one of the books or the weight of one of the books affects the location of the balancing point of the shelf. Think carefully here about how your changes should be considered relative to the location of the balancing point \(\overline{x}\) of the current scenario.

Hint

  1. Find the average location.

  2. Note that you are averaging 4 locations.

  3. Compare (b).

  4. If the book at location \(x_1\) weighs 2 pounds, it's like there are two books at \(x_1\text{,}\) so \(2x_1\) must play a role in the average.

  5. Use a weighted average.

  6. Compare (e).

  7. Experiment.

  8. Compare the effects of moving locations left and adding weight to the left of the balancing point.

Answer

  1. \(\overline{x} = \frac{x_1 + x_2}{2} = 3\text{.}\)

  2. \(\overline{x} = \frac{x_1 + x_2 + x_3 + x_4}{4} = 3\text{.}\)

  3. \(\overline{x} = \frac{x_1 + x_2 + x_3 + x_4}{4} = 2.75\text{.}\)

  4. \(\overline{x} = \frac{2x_1 + 3x_2 + 1x_3 + 1x_4}{7} = \frac{16}{7}\text{.}\)

  5. \(\overline{x} = \frac{2x_1 + 3x_2 + 1x_3 + 1x_4}{7} = \frac{17}{7}\text{.}\)

  6. \(\overline{x} = \frac{2x_1 + 3x_2 + 1x_3 + 2x_4}{7} = \frac{22}{7}\text{.}\)

  7. Answers will vary.

  8. If we have an existing arrangement and balancing point, moving one of the locations to the left will move the balancing point to the left; similarly, moving one of the locations to the right will move the balancing point to the right. If instead we add weight to an existing location, if that location is left of the balancing point, the balancing point will move left; the behavior is similar if on the right.

Solution

  1. The shelf will balance at \(\overline{x} = \frac{x_1 + x_2}{2} = \frac{0 + 6}{2} = 3\text{.}\)

  2. The shelf will balance at the average of the locations: \(\overline{x} = \frac{x_1 + x_2 + x_3 + x_4}{4} = \frac{0 + 2 + 4 + 6}{4} = 3\text{.}\)

  3. Changing the location of \(x_3\) to the left will move the average to the left, so the new location for \(\overline{x}\) is \(\overline{x} = \frac{x_1 + x_2 + x_3 + x_4}{4} = \frac{0 + 2 + 3 + 6}{4} = \frac{11}{4} = 2.75\text{.}\)

  4. We can think of the weighted books almost as if there were that number of 1-lb books at each location. The balancing point is given by the weighted average

    \begin{equation*} \overline{x} = \frac{2x_1 + 3x_2 + 1x_3 + 1x_4}{7} = \frac{2 \cdot 0 + 3 \cdot 2 + 1 \cdot 4 + 1 \cdot 6}{7} = \frac{16}{7}\text{.} \end{equation*}

    Note particularly that we divide by the total weight of the books, not the total number.

  5. Here we find

    \begin{equation*} \overline{x} = \frac{2x_1 + 3x_2 + 1x_3 + 1x_4}{7} = \frac{2 \cdot 0 + 3 \cdot 2 + 1 \cdot 5 + 1 \cdot 6}{7} = \frac{17}{7}\text{,} \end{equation*}

    which moves the balancing point slightly to the right.

  6. In this situation, we find the balancing point moves significantly to the right:

    \begin{equation*} \overline{x} = \frac{2x_1 + 3x_2 + 1x_3 + 2x_4}{7} = \frac{2 \cdot 0 + 3 \cdot 2 + 1 \cdot 4 + 2 \cdot 6}{7} = \frac{22}{7}\text{.} \end{equation*}

  7. Try, for instance, having \(x_2 = 1\) and \(x_3 = 2\) and experiment with different weights.

  8. If we have an existing arrangement and balancing point, moving one of the locations to the left will move the balancing point to the left; similarly, moving one of the locations to the right will move the balancing point to the right. If instead we add weight to an existing location, if that location is left of the balancing point, the balancing point will move left; the behavior is similar if on the right.

Subsection 6.3.3 Center of Mass

In Activity 6.3.3, we saw that the balancing point of a system of point-masses 1 (such as books on a shelf) is found by taking a weighted average of their respective locations. In the activity, we were computing the center of mass of a system of masses distributed along an axis, which is the balancing point of the axis on which the masses rest.

In the activity, we actually used weight rather than mass. Since weight is proportional to mass, the computations for the balancing point result in the same location regardless of whether we use weight or mass. The gravitational constant is present in both the numerator and denominator of the weighted average.
Center of Mass (point-masses).

For a collection of \(n\) masses \(m_1\text{,}\) \(\ldots\text{,}\) \(m_n\) that are distributed along a single axis at the locations \(x_1\text{,}\) \(\ldots\text{,}\) \(x_n\text{,}\) the center of mass is given by

\begin{equation*} \overline{x} = \frac{x_1 m_1 + x_2 m_2 + \cdots + x_n m_n}{m_1 + m_2 + \cdots + m_n}\text{.} \end{equation*}

Now consider a thin bar over which density is distributed continuously. If the density is constant, it is obvious that the balancing point of the bar is its midpoint. But if density is not constant, we must compute a weighted average. Let's say that the function \(\rho(x)\) tells us the density distribution along the bar, measured in g/cm. If we slice the bar into small sections, we can think of the bar as holding a collection of adjacent point-masses. The mass \(m_i\) of a slice of thickness \(\Delta x\) at location \(x_i\text{,}\) is \(m_i \approx \rho(x_i) \Delta x\text{.}\)

If we slice the bar into \(n\) pieces, we can approximate its center of mass by

\begin{equation*} \overline{x} \approx \frac{x_1 \cdot \rho(x_1) \Delta x + x_2 \cdot \rho(x_2) \Delta x + \cdots + x_n \cdot \rho(x_n) \Delta x }{\rho(x_1) \Delta x + \rho(x_2) \Delta x + \cdots + \rho(x_n) \Delta x}\text{.} \end{equation*}

Rewriting the sums in sigma notation, we have

\begin{equation} \overline{x} \approx \frac{\sum_{i = 1}^{n} x_i \cdot \rho(x_i) \Delta x}{\sum_{i = 1}^{n} \rho(x_i) \Delta x}\text{.}\label{npa}\tag{6.3.1} \end{equation}

The greater the number of slices, the more accurate our estimate of the balancing point will be. The sums in Equation (6.3.1) can be viewed as Riemann sums, so in the limit as \(n \to \infty\text{,}\) we find that the center of mass is given by the quotient of two integrals.

Center of Mass (continuous mass distribution).

For a thin rod of density \(\rho(x)\) distributed along an axis from \(x = a\) to \(x = b\text{,}\) the center of mass of the rod is given by

\begin{equation*} \overline{x} = \frac{\int_a^b x \rho(x) \, dx}{\int_a^b \rho(x) \, dx}\text{.} \end{equation*}

Note that the denominator of \(\overline{x}\) is the mass of the bar, and that this quotient of integrals is simply the continuous version of the weighted average of locations, \(x\text{,}\) along the bar.

Activity 6.3.4.

Consider a thin bar of length 20 cm whose density is distributed according to the function \(\rho(x) = 4 + 0.1x\text{,}\) where \(x = 0\) represents the left end of the bar. Assume that \(\rho\) is measured in g/cm and \(x\) is measured in cm.

  1. Find the total mass, \(M\text{,}\) of the bar.

  2. Without doing any calculations, do you expect the center of mass of the bar to be equal to 10, less than 10, or greater than 10? Why?

  3. Compute \(\overline{x}\text{,}\) the exact center of mass of the bar.

  4. What is the average density of the bar?

  5. Now consider a different density function, given by \(p(x) = 4e^{0.020732x}\text{,}\) also for a bar of length 20 cm whose left end is at \(x = 0\text{.}\) Plot both \(\rho(x)\) and \(p(x)\) on the same axes. Without doing any calculations, which bar do you expect to have the greater center of mass? Why?

  6. Compute the exact center of mass of the bar described in (e) whose density function is \(p(x) = 4e^{0.020732x}\text{.}\) Check the result against the prediction you made in (e).

Hint

  1. Recall \(M = \int_a^b \rho(x) \, dx\text{.}\)

  2. Think about whether there's more mass to the left of 10 or to the right of 10.

  3. Use the formula just developed in the preceding subsection of the text.

  4. Recall that when density, \(\rho\) is constant, \(M = \rho \cdot V\text{.}\)

  5. Think about how the two density functions distribute mass. Note that they appear to distribute the same total mass, as the areas under the two curves appear to be equal.

  6. Use \(p(x)\) in place of \(\rho(x)\) in standard calculations.

Answer

  1. \(M = \int_{0}^{20} 4 + 0.1x \, dx = 100\) g.

  2. Greater than 10.

  3. \(\overline{x} = \frac{\int_{0}^{20} x (4 + 0.1x)) \, dx}{\int_{0}^{20} 4 + 0.1x \, dx} = \frac{32}{3}\text{.}\)

  4. 5 g/cm.

  5. Slightly to the right of the center of mass for \(\rho(x)\text{.}\)

  6. \(\overline{x} = \frac{\int_{0}^{20} x 4e^{0.020732x} \, dx}{\int_{0}^{20} 4e^{0.020732x} \, dx} \approx 10.6891\text{,}\)

Solution

  1. \(M = \int_{0}^{20} 4 + 0.1x \, dx = 100\) g.

  2. The center of mass will be greater than 10 since the mass is more concentrated as \(x\) increases.

  3. Using the formula we developed, \(\overline{x} = \frac{\int_{0}^{20} x \rho(x) \, dx}{\int_{0}^{20} \rho(x) \, dx}\text{,}\) so

    \begin{equation*} \overline{x} = \frac{\int_{0}^{20} x (4 + 0.1x)) \, dx}{\int_{0}^{20} 4 + 0.1x \, dx} = \frac{\frac{3200}{3}}{100} = \frac{32}{3}\text{,} \end{equation*}

    which is indeed just greater than 10.

  4. Recall that average density is given by mass divided by length, so since the bar is 20 cm long and its mass is 100 g, the average density is 100/20 = 5 g/cm. This can also be computed using the formula for average value of a function:

    \begin{equation*} \rho_{\text{AVG} [0,20]} = \frac{1}{20-0} \int_{0}^{20} \rho(x) \,dx = \frac{100}{20}\text{.} \end{equation*}

  5. Since \(p(x) = 4e^{0.020732x}\) lies below \(\rho(x) = 4 + 0.1x\) up until where they intersect at about \(17.53\text{,}\) after which \(p(x) \gt \rho(x)\text{,}\) we see that \(p(x)\) distributes mass more to the right. It also appears as if the two curves may bound the same total area, and thus the generate the same total mass. If that is the case, we expect \(\overline{x}\) for \(p(x)\) to lie slightly to the right of the center of mass for \(\rho(x)\text{.}\)

  6. We see that

    \begin{equation*} \overline{x} = \frac{\int_{0}^{20} x 4e^{0.020732x} \, dx}{\int_{0}^{20} 4e^{0.020732x} \, dx} \approx 10.6891\text{,} \end{equation*}

    which is indeed slightly right of \(\overline{x}\) for \(\rho(x)\text{.}\)

Subsection 6.3.4 Summary

  • For an object of constant density \(D\text{,}\) with volume \(V\) and mass \(m\text{,}\) we know that \(m = D \cdot V\text{.}\)

  • If an object with constant cross-sectional area (such as a thin bar) has its density distributed along an axis according to the function \(\rho(x)\text{,}\) then we can find the mass of the object between \(x = a\) and \(x = b\) by

    \begin{equation*} m = \int_a^b \rho(x) \, dx\text{.} \end{equation*}

  • For a system of point-masses distributed along an axis, say \(m_1, \ldots, m_n\) at locations \(x_1, \ldots, x_n\text{,}\) the center of mass, \(\overline{x}\text{,}\) is given by the weighted average

    \begin{equation*} \overline{x} = \frac{\sum_{i=1}^n x_i m_i}{\sum_{i=1}^n m_i}\text{.} \end{equation*}

    If instead we have mass continuously distributed along an axis, such as by a density function \(\rho(x)\) for a thin bar of constant cross-sectional area, the center of mass of the portion of the bar between \(x = a\) and \(x = b\) is given by

    \begin{equation*} \overline{x} = \frac{\int_a^b x \rho(x) \, dx}{\int_a^b \rho(x) \, dx}\text{.} \end{equation*}

    In each situation, \(\overline{x}\) represents the balancing point of the system of masses or of the portion of the bar.

Exercises 6.3.5 Exercises

1. Center of mass for a linear density function.
2. Center of mass for a nonlinear density function.
3. Interpreting the density of cars on a road.
4. Center of mass in a point-mass system.
5.

Let a thin rod of length \(a\) have density distribution function \(\rho(x) = 10e^{-0.1x}\text{,}\) where \(x\) is measured in cm and \(\rho\) in grams per centimeter.

  1. If the mass of the rod is 30 g, what is the value of \(a\text{?}\)

  2. For the 30g rod, will the center of mass lie at its midpoint, to the left of the midpoint, or to the right of the midpoint? Why?

  3. For the 30g rod, find the center of mass, and compare your prediction in (b).

  4. At what value of \(x\) should the 30g rod be cut in order to form two pieces of equal mass?

Answer

  1. \(a = -10 \ln(0.7) \approx 3.567 \text{ cm}\text{.}\)

  2. Left of the midpoint.

  3. \(\overline{x} \approx \frac{50.3338}{30} \approx 1.687\text{.}\)

  4. \(q = -10\ln(0.85) \approx 1.625\) cm.

Solution

  1. To find mass we slice the rod horizontally into uniform slices of length \(\Delta x\) and approximate the density on the \(i\)th slice with the constant density \(\rho(x_i)\text{.}\) Then the approximate mass of the \(i\)th slice is \(\rho(x_i) \Delta x\text{.}\) Adding the approximations and taking the limit as the number of slices goes to infinity gives us the definite integral

    \begin{equation*} \int_0^a \rho(x) \ dx \end{equation*}

    to represent the mass of the rod. Using the substitution \(u = -0.1x\) and \(du = -0.1 \ dx\) we have

    \begin{align*} \int_0^a \rho(x) \ dx &= \int_0^a 10e^{-0.1x} \ dx\\ &= \int_0^{-0.1a} -100 e^u \ du\\ &= -100e^u \biggm|_0^{-0.1a}\\ &= -100\left(e^{-0.1a} - 1\right)\\ &= 100\left(1-e^{-0.1a}\right)\text{.} \end{align*}

    So if the mass is \(30\) g, then \(100\left(1-e^{-0.1a}\right) = 30\text{,}\) so \(1-e^{-0.1a} = 0.3\text{,}\) and \(e^{-0.1a} = 0.7\text{.}\) It follows that \(-0.1a = \ln(0.7)\text{,}\) and thus, \(a = -10 \ln(0.7) \approx 3.567 \text{ cm}\text{.}\)

  2. Since the density is a decreasing function, we should expect that the center of mass of the rod will lie to the left of the midpoint.

  3. To find the center of mass, we slice the rod horizontally into uniform slices of length \(\Delta x\) and approximate the mass on the \(i\)th slice with the constant density \(\rho(x_i)\) to be \(\rho(x_i) \Delta x\) as in (a). The position of the \(i\)th slice is approximately \(x_i/\text{.}\) Treating the rod as a collection of \(n\) discrete masses (the slices), the center of mass \(\overline{x}\) is approximately

    \begin{equation*} \overline{x} \approx \frac{\sum_{i=1}^n x_i \rho(x_i) \Delta x}{\sum_{i=1}^n \rho(x_i) \Delta x}\text{.} \end{equation*}

    Letting the number of slices go to infinity gives us the exact location of the center of mass of the rod,

    \begin{equation*} \overline{x} = \frac{\int_0^a x \rho(x) \ dx}{\int_0^a \rho(x) \ dx}\text{.} \end{equation*}

    The mass of the rod is \(30\) g, so now we just need to evaluate \(\int_0^a x \rho(x) \ dx = \int_0^a 10xe^{-0.1x} \ dx\text{.}\) Integrating by parts with \(u = x\text{,}\) \(du = dx\text{,}\) \(dv = e^{-0.1x} \ dx\text{,}\) and \(v = -10e^{-0.1x}\) gives us

    \begin{align*} \int_0^a 10xe^{-0.1x} \ dx &= 10\left(-10xe^{-0.1x}\biggm|_0^a + \int_0^a -10e^{-0.1x} \ dx \right)\\ &= 10\left(-10ae^{-0.1a} + 100(1-e^{-0.1a})\right)\\ &\approx 10\left(-10(3.567)ae^{-0.1(3.567)} + 100(1-e^{-0.1(3.567)})\right)\\ &\approx 50.3338\text{.} \end{align*}

    So the center of mass is \(\overline{x} \approx \frac{50.3338}{30} \approx 1.687\text{,}\) which is less than midway across the \(3.567\) cm rod as expected.

  4. Each piece would have mass \(15\) grams. Let \(q\) be a value of \(x\) between 0 and \(a\) so the the mass of the rod on the interval \([0, q]\) is 15. Using the substitution \(u = -0.1x\) and \(du = -0.1 \ dx\) we have that the mass of the rod on the interval \([0, q]\) is

    \begin{align*} \int_0^q \rho(x) \ dx &= \int_0^q 10e^{-0.1x} \ dx\\ &= \int_0^{-0.1q} -100e^{u} \ du\\ &= -100e^u\biggm|_0^{-0.1q}\\ &= 100\left(1-e^{-0.1q}\right) \end{align*}

    We want this mass to be \(15\) grams, so we find \(q\) such that \(100\left(1-e^{-0.1q}\right) = 15\text{.}\) Solving for \(q\) in the usual way, we find \(q = -10\ln(0.85) \approx 1.625\) cm. Note that this is not the same as the center of mass. The center of mass takes into account the torque (determined by distance from the center of mass), which is not incorporated into the calculation of the point to split the object into two pieces of equal mass.

6.

Consider two thin bars of constant cross-sectional area, each of length 10 cm, with respective mass density functions \(\rho(x) = \frac{1}{1+x^2}\) and \(p(x) = e^{-0.1x}\text{.}\)

  1. Find the mass of each bar.

  2. Find the center of mass of each bar.

  3. Now consider a new 10 cm bar whose mass density function is \(f(x) = \rho(x) + p(x)\text{.}\)

    1. Explain how you can easily find the mass of this new bar with little to no additional work.

    2. Similarly, compute \(\int_0^{10} xf(x) \, dx\) as simply as possible, in light of earlier computations.

    3. True or false: the center of mass of this new bar is the average of the centers of mass of the two earlier bars. Write at least one sentence to say why your conclusion makes sense.

Answer

  1. \(M_1 = \arctan(10) \approx 1.47113\text{;}\) \(M_2 = 10 - 10e^{-1} \approx 6.32121\text{.}\)

  2. \(\overline{x_1} \approx 1.56857 \text{;}\) \(\overline{x_2} \approx 4.18023 \text{.}\)

    1. \begin{equation*} M = \int_0^{10} \rho(x) \, dx + \int_0^{10} p(x) \, dx \approx 1.47113 + 6.32121 = 7.79234\text{.} \end{equation*}
    2. \begin{equation*} \int_0^{10} x(\rho(x) + p(x))) \, dx = 28.73167\text{.} \end{equation*}

    3. False.

Solution

  1. To find the mass of the first bar, we use the fact that \(M = \int_0^a \rho(x) \, dx\text{,}\) with \(a = 10\) and \(\rho(x) = \frac{1}{1+x^2}\text{.}\) Thus, the mass of the first bar is

    \begin{equation*} M_1 = \int_0^{10} \frac{1}{1+x^2} \, dx = \left. \arctan(x) \right|_0^{10}\text{,} \end{equation*}

    so \(M_1 = \arctan(10) \approx 1.47113\text{.}\)

    Similar computations for the second bar shows that its mass is

    \begin{equation*} M_2 = \int_0^{10} e^{-0.1x} \, dx = \left. -10e^{-0.1x} \right|_0^{10}\text{,} \end{equation*}

    so \(M_2 = 10 - 10e^{-1} \approx 6.32121\text{.}\)

  2. To find the center of mass of each bar, we use the standard formula:

    \begin{equation*} \overline{x} = \frac{\int_0^{10} x \rho(x) \, dx}{\int_0^{10} \rho(x) \, dx}\text{.} \end{equation*}

    Having already evaluated the denominator for each mass distribution, we simply need to compute the numerator. It is straightforward to show that

    \begin{equation*} \int_0^{10} x \frac{1}{1+x^2} \, dx = \frac{1}{2} \ln(101) \approx 2.30756\text{,} \end{equation*}

    so the first bar's center of mass is

    \begin{equation*} \overline{x_1} = \frac{\arctan(10)}{\frac{1}{2}\ln(101)} \approx \frac{2.30756}{1.47113} \approx 1.56857\text{.} \end{equation*}

    For the second bar,

    \begin{equation*} \int_0^{10} x e^{-0.1x} \, dx = 100-200e^{-1} \approx 26.42411\text{,} \end{equation*}

    so its center of mass is

    \begin{equation*} \overline{x_2} = \frac{10 - 10e^{-1}}{100-200e^{-1}} \approx \frac{26.42411}{6.32121} \approx 4.18023\text{.} \end{equation*}

  3. Consider a new 10 cm bar whose mass density function is \(f(x) = \rho(x) + p(x)\text{.}\)

    1. The mass of this new bar is \(M = \int_0^{10} (\rho(x) + p(x)) \, dx\text{.}\) Using the additive property of the definite integral,

      \begin{equation*} M = \int_0^{10} \rho(x) \, dx + \int_0^{10} p(x) \, dx \approx 1.47113 + 6.32121 = 7.79234\text{.} \end{equation*}

    2. Applying the same idea to \(\int_0^{10} xf(x) \, dx\text{,}\) we have

      \begin{equation*} \int_0^{10} x(\rho(x) + p(x))) \, dx = \int_0^{10} x \rho(x) \, dx + \int_0^{10} x p(x) \, dx \approx 2.30756 + 26.42411 = 28.73167\text{.} \end{equation*}

    3. False. The center of mass of the new bar is

      \begin{equation*} \overline{x} = \frac{\int_0^{10} x(\rho(x) + p(x)) \, dx}{\int_0^{10} (\rho(x) + p(x)) \, dx} \approx \frac{2.30756 + 26.42411}{1.47113 + 6.32121} = 3.68717 \end{equation*}

      while the average of the two bars' centers of mass is

      \begin{equation*} \frac{\overline{x_1} + \overline{x_2}}{2} \approx \frac{\frac{2.30756}{1.47113} + \frac{26.42411}{6.32121}}{2} = 2.8744\text{.} \end{equation*}

      Although the computations prove the conclusion, the result is also intuitive: the average treats the two bars as equal contributors and doesn't take into account that one bar has much more mass than the other. Indeed, a weighted average is needed to correctly compute the center of mass, which is what our earlier computations in (i) and (ii) accomplish.

7.

Consider the curve given by \(y = f(x) = 2xe^{-1.25x} + (30-x) e^{-0.25(30-x)}\text{.}\)

  1. Plot this curve in the window \(x = 0 \ldots 30\text{,}\) \(y = 0 \ldots 3\) (with constrained scaling so the units on the \(x\) and \(y\) axis are equal), and use it to generate a solid of revolution about the \(x\)-axis. Explain why this curve could generate a reasonable model of a baseball bat.

  2. Let \(x\) and \(y\) be measured in inches. Find the total volume of the baseball bat generated by revolving the given curve about the \(x\)-axis. Include units on your answer.

  3. Suppose that the baseball bat has constant weight density, and that the weight density is \(0.6\) ounces per cubic inch. Find the total weight of the bat whose volume you found in (b).

  4. Because the baseball bat does not have constant cross-sectional area, we see that the amount of weight concentrated at a location \(x\) along the bat is determined by the volume of a slice at location \(x\text{.}\) Explain why we can think about the function \(\rho(x) = 0.6 \pi f(x)^2\) (where \(f\) is the function given at the start of the problem) as being the weight density function for how the weight of the baseball bat is distributed from \(x = 0\) to \(x = 30\text{.}\)

  5. Compute the center of mass of the baseball bat.

Answer

  2. \(V = \int_0^{30} \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \, dx \approx 52.0666\) cubic inches.

  3. \(W \approx 0.6 \cdot 52.0666 = 31.23996\) ounces.

  4. At a given \(x\)-location, the amount of weight concentrated there is approximately the weight density (\(0.6\) ounces per cubic inch) times the volume of the slice, which is \(V_{text{slice}} \approx \pi f(x)^2\text{.}\)

  5. \(\overline{x} \approx 23.21415\)

Solution

  1. The plot shown above (with the curve reflected across the \(x\)-axis to generate the solid of revolution) shows why this solid of revolution is a reasonable approximation to a baseball bat (if perhaps a bit thin between \(x = 3\) and \(x = 9\text{.}\)

  2. We use the disk method to find the volume of the solid of revolution. In the usual way, the volume of a typical slice is

    \begin{equation*} V_{\text{slice}} \approx \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \triangle x \end{equation*}

    and thus using a definite integral to sum the volumes of the slices, we get

    \begin{equation*} V = \int_0^{30} \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \, dx\text{.} \end{equation*}

    Using technology to evaluate the integral, we find that \(V \approx 52.0666\) cubic inches.

  3. Given that the bat has (constant) weight density \(0.6\) ounces per cubic inch, it follows that the total weight of the bat is \(W \approx 0.6 \cdot 52.0666 = 31.23996\) ounces.

  4. If we slice the baseball bat perpendicular to the horizontal axes, we get thin cylindrical disks as cross-sections. At a given \(x\)-location, the amount of weight concentrated there is approximately the weight density (\(0.6\) ounces per cubic inch) times the volume of the slice, which is \(V_{text{slice}} \approx \pi f(x)^2\text{.}\) Thus, we see that the amount of weight concentrated at a location \(x\) along the bat is given by \(\rho(x) = 0.6 \pi f(x)^2\text{,}\) which we can therefore view as being the weight density function for how the weight of the baseball bat is distributed from \(x = 0\) to \(x = 30\text{.}\)

  5. Even though we are dealing with ``weight'' instead of ``mass'', we can compute the center of mass in the usual way since weight differs from mass only by the gravitational constant, and this constant is present in both the numerator and denominator of the center of mass. Thus, using the standard formula,

    \begin{equation*} \overline{x} = \frac{\int_0^{30} x \rho(x) \ dx}{\int_0^{30} \rho(x) \ dx} \end{equation*}

    so

    \begin{align*} \overline{x} &= \frac{\int_0^{30} x \cdot 0.6 \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2) \ dx}{\int_0^{30} 0.6 \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \ dx}\\ &= \frac{\int_0^{30} x \cdot (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2) \ dx}{\int_0^{30} (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \ dx}\\ &\approx \frac{384.735}{16.5733} = 23.21415\text{,} \end{align*}

    so the center of mass is just over 2/3 along the length of the bat.

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