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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.
Introduction
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, \[\begin{aligned}\end{aligned}\], have units given by the product of the units of \(v(t)\) and \(t\): \[\begin{aligned}\end{aligned}\]. 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.
Exploration
Exploration
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\(^3\), then the density of the brick is \[\begin{aligned}\end{aligned}\].
As another example, the mass density of water is 1000 kg/m\(^3\). Each of these relationships demonstrate the following general principle.
For an object of constant density \(d\), with mass \(m\) and volume \(V\), \[\begin{aligned}\end{aligned}\].
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 \(A = l \cdot w\). 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 \(v(t) \Delta t\), 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 \(\Delta x\): \(f(x) \Delta x\). Both of these principles are represented visually in Figure.
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.
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, \(M\), of the object between \(x = a\) and \(x = b\) is given by \[\begin{aligned}\end{aligned}\].
Condensed — the full section is in Boelkins, Active Calculus.
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 \(a_1\), \(a_2\), \(\ldots\), \(a_n\), we know that their average is given by \[\begin{aligned}\end{aligned}\], and for a quantity being measured by a function \(f\) on an interval \([a,b]\), the average value of the quantity on \([a,b]\) is \[\begin{aligned}\end{aligned}\].
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.
| 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 |
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 \[\begin{aligned}\end{aligned}\].
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 \[\begin{aligned}\end{aligned}\].
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.
Center of Mass
In Activity, we saw that the balancing point of a system of point-masses 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. (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. Here we formally summarize the results of the activity, which hold for any finite collection of point-masses. This forms an important foundation for when mass varies continuously.
For a collection of \(n\) masses \(m_1\), \(\ldots\), \(m_n\) that are distributed along a single axis at the locations \(x_1\), \(\ldots\), \(x_n\), the center of mass is given by \[\begin{aligned}\end{aligned}\].
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 used the idea of a weighted average to find its balancing point. 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 \(x_i\), is \(m_i \approx \rho(x_i) \Delta x\).
If we slice the bar into \(n\) pieces, we can approximate its center of mass by \[\begin{aligned}\end{aligned}\]. Rewriting the sums in sigma notation, we have \[\begin{aligned}\end{aligned}\].
The greater the number of slices, the more accurate our estimate of the balancing point will be. The sums in Equation can be viewed as Riemann sums, so in the limit as \(n \to \infty\), we find that the center of mass is given by the quotient of two integrals.
For a thin rod of density \(\rho(x)\) distributed along an axis from \(x = a\) to \(x = b\), the center of mass of the rod is given by \[\begin{aligned}\end{aligned}\].
Condensed — the full section is in Boelkins, Active Calculus.
Summary
For an object of constant density \(D\), with volume \(V\) and mass \(m\), we know that \[\begin{aligned}\end{aligned}\].
If an object with constant cross-sectional area (such as a thin bar) has its density distributed along an axis according to the function \(\rho(x)\), then we can find the mass of the object between \(x = a\) and \(x = b\) by \[\begin{aligned}\end{aligned}\].
For a system of point-masses distributed along an axis, say \(m_1, \ldots, m_n\) at locations \(x_1, \ldots, x_n\), the center of mass, \(\overline{x}\), is given by the weighted average \[\begin{aligned}\end{aligned}\]. 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 \[\begin{aligned}\end{aligned}\]. In each situation, \(\overline{x}\) represents the balancing point of the system of masses or of the portion of the bar.
Practice (4)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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Let a thin rod of length \(a\) have density distribution function \(\rho(x) = 10e^{-0.1x}\), where \(x\) is measured in cm and \(\rho\) in grams per centimeter.
If the mass of the rod is 30 g, what is the value of \(a\)?
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?
For the 30g rod, find the center of mass, and compare your prediction in (b).
At what value of \(x\) should the 30g rod be cut in order to form two pieces of equal mass?
Zbulo përgjigjen
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 \(\rho(x_i)\). Then the approximate mass of the \(i\)th slice is \(\rho(x_i) \Delta x\). Adding the approximations and taking the limit as the number of slices goes to infinity gives us the definite integral \[\begin{aligned}\end{aligned}\] to represent the mass of the rod. Using the substitution \(u = -0.1x\) and \(du = -0.1 \ dx\) we have \[\begin{aligned}\int_0^a \rho(x) \ dx &= \int_0^a 10e^{-0.1x} \ dx \\ &= \int_0^{-0.1a} -100 e^u \ du \\ &= -100e^u \biggm|_0^{-0.1a} \\ &= -100\left(e^{-0.1a} - 1\right) \\ &= 100\left(1-e^{-0.1a}\right)\end{aligned}\]. So if the mass is \(30\) g, then \(100\left(1-e^{-0.1a}\right) = 30\), so \(1-e^{-0.1a} = 0.3\), and \(e^{-0.1a} = 0.7\). It follows that \(-0.1a = \ln(0.7)\), and thus, \(a = -10 \ln(0.7) \approx 3.567 \text{ cm}\).
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.
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 \(x_i/\). Treating the rod as a collection of \(n\) discrete masses (the slices), the center of mass \(\overline{x}\) is approximately \[\begin{aligned}\end{aligned}\]. Letting the number of slices go to infinity gives us the exact location of the center of mass of the rod, \[\begin{aligned}\end{aligned}\]. The mass of the rod is \(30\) g, so now we just need to evaluate \(\int_0^a x \rho(x) \ dx = \int_0^a 10xe^{-0.1x} \ dx\). Integrating by parts with \(u = x\), \(du = dx\), \(dv = e^{-0.1x} \ dx\), and \(v = -10e^{-0.1x}\) gives us \[\begin{aligned}\int_0^a 10xe^{-0.1x} \ dx &= 10\left(-10xe^{-0.1x}\biggm|_0^a + \int_0^a -10e^{-0.1x} \ dx \right) \\ &= 10\left(-10ae^{-0.1a} + 100(1-e^{-0.1a})\right) \\ &\approx 10\left(-10(3.567)ae^{-0.1(3.567)} + 100(1-e^{-0.1(3.567)})\right) \\ &\approx 50.3338\end{aligned}\]. So the center of mass is \(\overline{x} \approx \frac{50.3338}{30} \approx 1.687\), which is less than midway across the \(3.567\) cm rod as expected.
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 \[\begin{aligned}\int_0^q \rho(x) \ dx &= \int_0^q 10e^{-0.1x} \ dx \\ &= \int_0^{-0.1q} -100e^{u} \ du \\ &= -100e^u\biggm|_0^{-0.1q} \\ &= 100\left(1-e^{-0.1q}\right)\end{aligned}\] We want this mass to be \(15\) grams, so we find \(q\) such that \(100\left(1-e^{-0.1q}\right) = 15\). 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.
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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 \(p(x) = e^{-0.1x}\).
Find the mass of each bar.
Find the center of mass of each bar.
Now consider a new 10 cm bar whose mass density function is \(f(x) = \rho(x) + p(x)\).
Explain how you can easily find the mass of this new bar with little to no additional work.
Similarly, compute \(\int_0^{10} xf(x) \, dx\) as simply as possible, in light of earlier computations.
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.
Zbulo përgjigjen
To find the mass of the first bar, we use the fact that \(M = \int_0^a \rho(x) \, dx\), with \(a = 10\) and \(\rho(x) = \frac{1}{1+x^2}\). Thus, the mass of the first bar is \[\begin{aligned}\end{aligned}\], so \(M_1 = \arctan(10) \approx 1.47113\).
Similar computations for the second bar shows that its mass is \[\begin{aligned}\end{aligned}\], so \(M_2 = 10 - 10e^{-1} \approx 6.32121\).
To find the center of mass of each bar, we use the standard formula: \[\begin{aligned}\end{aligned}\]. Having already evaluated the denominator for each mass distribution, we simply need to compute the numerator. It is straightforward to show that \[\begin{aligned}\end{aligned}\], so the first bar's center of mass is \[\begin{aligned}\end{aligned}\]. For the second bar, \[\begin{aligned}\end{aligned}\], so its center of mass is \[\begin{aligned}\end{aligned}\].
Consider a new 10 cm bar whose mass density function is \(f(x) = \rho(x) + p(x)\).
The mass of this new bar is \(M = \int_0^{10} (\rho(x) + p(x)) \, dx\). Using the additive property of the definite integral, \[\begin{aligned}\end{aligned}\].
Applying the same idea to \(\int_0^{10} xf(x) \, dx\), we have \[\begin{aligned}\end{aligned}\].
False. The center of mass of the new bar is \[\begin{aligned}\end{aligned}\] while the average of the two bars' centers of mass is \[\begin{aligned}\end{aligned}\]. 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.
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Let a thin rod of length \(a\) have density distribution function \(\rho(x) = 10e^{-0.1x}\), where \(x\) is measured in cm and \(\rho\) in grams per centimeter.
If the mass of the rod is 30 g, what is the value of \(a\)?
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?
For the 30g rod, find the center of mass, and compare your prediction in (b).
At what value of \(x\) should the 30g rod be cut in order to form two pieces of equal mass?
Zbulo përgjigjen
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 \(\rho(x_i)\). Then the approximate mass of the \(i\)th slice is \(\rho(x_i) \Delta x\). Adding the approximations and taking the limit as the number of slices goes to infinity gives us the definite integral \[\begin{aligned}\end{aligned}\] to represent the mass of the rod. Using the substitution \(u = -0.1x\) and \(du = -0.1 \ dx\) we have \[\begin{aligned}\int_0^a \rho(x) \ dx &= \int_0^a 10e^{-0.1x} \ dx \\ &= \int_0^{-0.1a} -100 e^u \ du \\ &= -100e^u \biggm|_0^{-0.1a} \\ &= -100\left(e^{-0.1a} - 1\right) \\ &= 100\left(1-e^{-0.1a}\right)\end{aligned}\]. So if the mass is \(30\) g, then \(100\left(1-e^{-0.1a}\right) = 30\), so \(1-e^{-0.1a} = 0.3\), and \(e^{-0.1a} = 0.7\). It follows that \(-0.1a = \ln(0.7)\), and thus, \(a = -10 \ln(0.7) \approx 3.567 \text{ cm}\).
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.
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 \(x_i/\). Treating the rod as a collection of \(n\) discrete masses (the slices), the center of mass \(\overline{x}\) is approximately \[\begin{aligned}\end{aligned}\]. Letting the number of slices go to infinity gives us the exact location of the center of mass of the rod, \[\begin{aligned}\end{aligned}\]. The mass of the rod is \(30\) g, so now we just need to evaluate \(\int_0^a x \rho(x) \ dx = \int_0^a 10xe^{-0.1x} \ dx\). Integrating by parts with \(u = x\), \(du = dx\), \(dv = e^{-0.1x} \ dx\), and \(v = -10e^{-0.1x}\) gives us \[\begin{aligned}\int_0^a 10xe^{-0.1x} \ dx &= 10\left(-10xe^{-0.1x}\biggm|_0^a + \int_0^a -10e^{-0.1x} \ dx \right) \\ &= 10\left(-10ae^{-0.1a} + 100(1-e^{-0.1a})\right) \\ &\approx 10\left(-10(3.567)ae^{-0.1(3.567)} + 100(1-e^{-0.1(3.567)})\right) \\ &\approx 50.3338\end{aligned}\]. So the center of mass is \(\overline{x} \approx \frac{50.3338}{30} \approx 1.687\), which is less than midway across the \(3.567\) cm rod as expected.
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 \[\begin{aligned}\int_0^q \rho(x) \ dx &= \int_0^q 10e^{-0.1x} \ dx \\ &= \int_0^{-0.1q} -100e^{u} \ du \\ &= -100e^u\biggm|_0^{-0.1q} \\ &= 100\left(1-e^{-0.1q}\right)\end{aligned}\] We want this mass to be \(15\) grams, so we find \(q\) such that \(100\left(1-e^{-0.1q}\right) = 15\). 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.
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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 \(p(x) = e^{-0.1x}\).
Find the mass of each bar.
Find the center of mass of each bar.
Now consider a new 10 cm bar whose mass density function is \(f(x) = \rho(x) + p(x)\).
Explain how you can easily find the mass of this new bar with little to no additional work.
Similarly, compute \(\int_0^{10} xf(x) \, dx\) as simply as possible, in light of earlier computations.
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.
Zbulo përgjigjen
To find the mass of the first bar, we use the fact that \(M = \int_0^a \rho(x) \, dx\), with \(a = 10\) and \(\rho(x) = \frac{1}{1+x^2}\). Thus, the mass of the first bar is \[\begin{aligned}\end{aligned}\], so \(M_1 = \arctan(10) \approx 1.47113\).
Similar computations for the second bar shows that its mass is \[\begin{aligned}\end{aligned}\], so \(M_2 = 10 - 10e^{-1} \approx 6.32121\).
To find the center of mass of each bar, we use the standard formula: \[\begin{aligned}\end{aligned}\]. Having already evaluated the denominator for each mass distribution, we simply need to compute the numerator. It is straightforward to show that \[\begin{aligned}\end{aligned}\], so the first bar's center of mass is \[\begin{aligned}\end{aligned}\]. For the second bar, \[\begin{aligned}\end{aligned}\], so its center of mass is \[\begin{aligned}\end{aligned}\].
Consider a new 10 cm bar whose mass density function is \(f(x) = \rho(x) + p(x)\).
The mass of this new bar is \(M = \int_0^{10} (\rho(x) + p(x)) \, dx\). Using the additive property of the definite integral, \[\begin{aligned}\end{aligned}\].
Applying the same idea to \(\int_0^{10} xf(x) \, dx\), we have \[\begin{aligned}\end{aligned}\].
False. The center of mass of the new bar is \[\begin{aligned}\end{aligned}\] while the average of the two bars' centers of mass is \[\begin{aligned}\end{aligned}\]. 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.
Symbols used here
Antiderivative (indefinite) or signed area from a to b (definite).
Not a number: "grows without bound" in limits and intervals.
2.71828…, the base whose exponential is its own derivative.
i² = −1.
Equal to the precision shown, not exactly.
Least upper bound, greatest lower bound.
Ratio of a circle's circumference to its diameter, 3.14159…
Ratios of sides in a right triangle; coordinates on the unit circle.
The exponent b must be raised to for x; ln uses base e.
Add a_k for k = 1 up to n.
The value f(x) approaches as x approaches a.
Instantaneous rate of change; slope of the graph.
Prime notation for derivatives with respect to x (or t).
Constants of integration fixed by initial conditions.
How to: Density, mass, and center of mass
- How are mass, density, and volume related?
- How is the mass of an object with varying density computed?
- What is the center of mass of an object, and how are definite integrals used to compute it?
Questions people ask
What is a derivative in one sentence?
The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.
What is an integral in one sentence?
The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.
Why are derivatives and integrals opposites?
That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.
When do I use substitution and when integration by parts?
Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.
Provo timen.
Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.
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