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Moments and Centers of Mass

Find the center of mass of objects distributed along a line.

Center of Mass and Moments

Let’s begin by looking at the center of mass in a one-dimensional context. Consider a long, thin wire or rod of negligible mass resting on a fulcrum, as shown in (a). Now suppose we place objects having masses \({m}_{1}\) and \({m}_{2}\) at distances \({d}_{1}\) and \({d}_{2}\) from the fulcrum, respectively, as shown in (b).

The most common real-life example of a system like this is a playground seesaw, or teeter-totter, with children of different weights sitting at different distances from the center. On a seesaw, if one child sits at each end, the heavier child sinks down and the lighter child is lifted into the air. If the heavier child slides in toward the center, though, the seesaw balances. Applying this concept to the masses on the rod, we note that the masses balance each other if and only if \({m}_{1}{d}_{1}={m}_{2}{d}_{2}.\)

In the seesaw example, we balanced the system by moving the masses (children) with respect to the fulcrum. However, we are really interested in systems in which the masses are not allowed to move, and instead we balance the system by moving the fulcrum. Suppose we have two point masses, \({m}_{1}\) and \({m}_{2},\) located on a number line at points \({x}_{1}\) and \({x}_{2},\) respectively (). The center of mass, \(\overset{-}{x},\) is the point where the fulcrum should be placed to make the system balance.

Thus, we have

\[\begin{array}{lll}{m}_{1}|\overset{-}{x}-{x}_{1}| & = & {m}_{2}|{x}_{2}-\overset{-}{x}| \\ {m}_{1}(\overset{-}{x}-{x}_{1}) & = & {m}_{2}({x}_{2}-\overset{-}{x}) \\ {m}_{1}\overset{-}{x}-{m}_{1}{x}_{1} & = & {m}_{2}{x}_{2}-{m}_{2}\overset{-}{x} \\ \overset{-}{x}({m}_{1}+{m}_{2}) & = & {m}_{1}{x}_{1}+{m}_{2}{x}_{2} \\ \overset{-}{x} & = & \frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}}{{m}_{1}+{m}_{2}}.\end{array}\]

The expression in the numerator, \({m}_{1}{x}_{1}+{m}_{2}{x}_{2},\) is called the first moment of the system with respect to the origin. If the context is clear, we often drop the word first and just refer to this expression as the moment of the system. The expression in the denominator, \({m}_{1}+{m}_{2},\) is the total mass of the system. Thus, the center of mass of the system is the point at which the total mass of the system could be concentrated without changing the moment.

This idea is not limited just to two point masses. In general, if n masses, \({m}_{1},{m}_{2}\text{,\ldots },{m}_{n},\) are placed on a number line at points \({x}_{1},{x}_{2}\text{,\ldots },{x}_{n},\) respectively, then the center of mass of the system is given by

\[\overset{-}{x}=\frac{\sum _{i=1}^{n}{m}_{i}{x}_{i}}{\sum _{i=1}^{n}{m}_{i}}.\]

We apply this theorem in the following example.

Condensed — the full section is in OpenStax Calculus Volume 2.

Center of Mass of Thin Plates

So far we have looked at systems of point masses on a line and in a plane. Now, instead of having the mass of a system concentrated at discrete points, we want to look at systems in which the mass of the system is distributed continuously across a thin sheet of material. For our purposes, we assume the sheet is thin enough that it can be treated as if it is two-dimensional. Such a sheet is called a lamina. Next we develop techniques to find the center of mass of a lamina. In this section, we also assume the density of the lamina is constant.

Laminas are often represented by a two-dimensional region in a plane. The geometric center of such a region is called its centroid. Since we have assumed the density of the lamina is constant, the center of mass of the lamina depends only on the shape of the corresponding region in the plane; it does not depend on the density. In this case, the center of mass of the lamina corresponds to the centroid of the delineated region in the plane. As with systems of point masses, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the x- and y-axes.

We first consider a lamina in the shape of a rectangle. Recall that the center of mass of a lamina is the point where the lamina balances. For a rectangle, that point is both the horizontal and vertical center of the rectangle. Based on this understanding, it is clear that the center of mass of a rectangular lamina is the point where the diagonals intersect, which is a result of the symmetry principle, and it is stated here without proof.

Let’s turn to more general laminas. Suppose we have a lamina bounded above by the graph of a continuous function \(f(x),\) below by the x-axis, and on the left and right by the lines \(x=a\) and \(x=b,\) respectively, as shown in the following figure.

As with systems of point masses, to find the center of mass of the lamina, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the x- and y-axes. As we have done many times before, we approximate these quantities by partitioning the interval \([a,b]\) and constructing rectangles.

\[m\approx \sum _{i=1}^{n}\rho f({x}_{i}^{*})\text{\Delta }x.\]\[m=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\rho f({x}_{i}^{*})\text{\Delta }x=\rho {\int }_{a}^{b}f(x)dx.\]\[{M}_{x}=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\rho \frac{{[f({x}_{i}^{*})]}^{2}}{2}\text{\Delta }x=\rho {\int }_{a}^{b}\frac{{[f(x)]}^{2}}{2}dx.\]

Condensed — the full section is in OpenStax Calculus Volume 2.

The Symmetry Principle

We stated the symmetry principle earlier, when we were looking at the centroid of a rectangle. The symmetry principle can be a great help when finding centroids of regions that are symmetric. Consider the following example.

Example

Try it.

Let R be the region bounded above by the graph of the function \(f(x)=4-{x}^{2}\) and below by the x-axis. Find the centroid of the region.

Solution

The region is depicted in the following figure.

The region is symmetric with respect to the y-axis. Therefore, the x-coordinate of the centroid is zero. We need only calculate \(\overset{-}{y}.\) Once again, for the sake of convenience, assume \(\rho =1.\)

First, we calculate the total mass:

\[\begin{array}{ll}m & =\rho {\int }_{a}^{b}f(x)dx \\ & ={\int }_{-2}^{2}(4-{x}^{2})dx \\ & ={[4x-\frac{{x}^{3}}{3}]\ |}_{-2}^{2}=\frac{32}{3}.\end{array}\]

Next, we calculate the moments. We only need \({M}_{x}\text{:}\)

\[\begin{array}{ll}{M}_{x} & =\rho {\int }_{a}^{b}\frac{{[f(x)]}^{2}}{2}dx \\ & =\frac{1}{2}{\int }_{-2}^{2}{[4-{x}^{2}]}^{2}dx=\frac{1}{2}{\int }_{-2}^{2}(16-8{x}^{2}+{x}^{4})dx \\ & =\frac{1}{2}{[\frac{{x}^{5}}{5}-\frac{8{x}^{3}}{3}+16x]\ |}_{-2}^{2}=\frac{256}{15}.\end{array}\]

Then we have

\[\overset{-}{y}=\frac{{M}_{x}}{m}=\frac{256}{15}\cdot \frac{3}{32}=\frac{8}{5}.\]

The centroid of the region is \((0,8\text{/}5).\)

Condensed — the full section is in OpenStax Calculus Volume 2.

Theorem of Pappus

This section ends with a discussion of the theorem of Pappus for volume, which allows us to find the volume of particular kinds of solids by using the centroid. (There is also a theorem of Pappus for surface area, but it is much less useful than the theorem for volume.)

We can prove the case when the region is bounded above by the graph of a function \(f(x)\) and below by the graph of a function \(g(x)\) over an interval \([a,b],\) and for which the axis of revolution is the y-axis. In this case, the area of the region is \(A={\int }_{a}^{b}[f(x)-g(x)]dx.\) Since the axis of rotation is the y-axis, the distance traveled by the centroid of the region depends only on the x-coordinate of the centroid, \(\overset{-}{x},\) which is

\[\overset{-}{x}=\frac{{M}_{y}}{m},\]

where

\[m=\rho {\int }_{a}^{b}[f(x)-g(x)]dx\ \text{and}\ {M}_{y}=\rho {\int }_{a}^{b}x[f(x)-g(x)]dx.\]

Then,

\[d=2\pi \frac{\rho {\int }_{a}^{b}x[f(x)-g(x)]dx}{\rho {\int }_{a}^{b}[f(x)-g(x)]dx}\]

and thus

\[d\cdot A=2\pi {\int }_{a}^{b}x[f(x)-g(x)]dx.\]

However, using the method of cylindrical shells, we have

\[V=2\pi {\int }_{a}^{b}x[f(x)-g(x)]dx.\]

So,

\[V=d\cdot A\]

and the proof is complete.

□

Example

Try it.

Let R be a circle of radius 2 centered at \((4,0).\) Use the theorem of Pappus for volume to find the volume of the torus generated by revolving R around the y-axis.

Solution

The region and torus are depicted in the following figure.

The region R is a circle of radius 2, so the area of R is \(A=4\pi\) units2. By the symmetry principle, the centroid of R is the center of the circle. The centroid travels around the y-axis in a circular path of radius 4, so the centroid travels \(d=8\pi\) units. Then, the volume of the torus is \(A\cdot d=32{\pi }^{2}\) units3.

Key Concepts

  • Mathematically, the center of mass of a system is the point at which the total mass of the system could be concentrated without changing the moment. Loosely speaking, the center of mass can be thought of as the balancing point of the system.
  • For point masses distributed along a number line, the moment of the system with respect to the origin is \(M=\sum _{i=1}^{n}{m}_{i}{x}_{i}.\) For point masses distributed in a plane, the moments of the system with respect to the x- and y-axes, respectively, are \({M}_{x}=\sum _{i=1}^{n}{m}_{i}{y}_{i}\) and \({M}_{y}=\sum _{i=1}^{n}{m}_{i}{x}_{i},\) respectively.
  • For a lamina bounded above by a function \(f(x),\) the moments of the system with respect to the x- and y-axes, respectively, are \({M}_{x}=\rho {\int }_{a}^{b}\frac{{[f(x)]}^{2}}{2}dx\) and \({M}_{y}=\rho {\int }_{a}^{b}xf(x)dx.\)
  • The x- and y-coordinates of the center of mass can be found by dividing the moments around the y-axis and around the x-axis, respectively, by the total mass. The symmetry principle says that if a region is symmetric with respect to a line, then the centroid of the region lies on the line.
  • The theorem of Pappus for volume says that if a region is revolved around an external axis, the volume of the resulting solid is equal to the area of the region multiplied by the distance traveled by the centroid of the region.

Key Equations

Mass of a lamina\(m=\rho {\int }_{a}^{b}f(x)dx\)
Moments of a lamina\({M}_{x}=\rho {\int }_{a}^{b}\frac{{[f(x)]}^{2}}{2}dx\ \text{and}\ {M}_{y}=\rho {\int }_{a}^{b}xf(x)dx\)
Center of mass of a lamina\(\overset{-}{x}=\frac{{M}_{y}}{m}\ \text{and}\ \overset{-}{y}=\frac{{M}_{x}}{m}\)

Moments and Centers of Mass

For the following exercises, calculate the center of mass for the collection of masses given.

For the following exercises, compute the center of mass \(\overset{-}{x}.\)

For the following exercises, compute the center of mass \((\overset{-}{x},\overset{-}{y}).\) Use symmetry to help locate the center of mass whenever possible.

For the following exercises, use a calculator to draw the region, then compute the center of mass \((\overset{-}{x},\overset{-}{y}).\) Use symmetry to help locate the center of mass whenever possible.

For the following exercises, use the theorem of Pappus to determine the volume of the shape.

For the following exercises, use a calculator to draw the region enclosed by the curve. Find the area \(M\) and the centroid \((\overset{-}{x},\overset{-}{y})\) for the given shapes. Use symmetry to help locate the center of mass whenever possible.

Center of Mass and Moments

Let’s begin by looking at the center of mass in a one-dimensional context. Consider a long, thin wire or rod of negligible mass resting on a fulcrum, as shown in (a). Now suppose we place objects having masses \({m}_{1}\) and \({m}_{2}\) at distances \({d}_{1}\) and \({d}_{2}\) from the fulcrum, respectively, as shown in (b).

The most common real-life example of a system like this is a playground seesaw, or teeter-totter, with children of different weights sitting at different distances from the center. On a seesaw, if one child sits at each end, the heavier child sinks down and the lighter child is lifted into the air. If the heavier child slides in toward the center, though, the seesaw balances. Applying this concept to the masses on the rod, we note that the masses balance each other if and only if \({m}_{1}{d}_{1}={m}_{2}{d}_{2}.\)

In the seesaw example, we balanced the system by moving the masses (children) with respect to the fulcrum. However, we are really interested in systems in which the masses are not allowed to move, and instead we balance the system by moving the fulcrum. Suppose we have two point masses, \({m}_{1}\) and \({m}_{2},\) located on a number line at points \({x}_{1}\) and \({x}_{2},\) respectively (). The center of mass, \(\overset{-}{x},\) is the point where the fulcrum should be placed to make the system balance.

Thus, we have

\[\begin{array}{lll}{m}_{1}|\overset{-}{x}-{x}_{1}| & = & {m}_{2}|{x}_{2}-\overset{-}{x}| \\ {m}_{1}(\overset{-}{x}-{x}_{1}) & = & {m}_{2}({x}_{2}-\overset{-}{x}) \\ {m}_{1}\overset{-}{x}-{m}_{1}{x}_{1} & = & {m}_{2}{x}_{2}-{m}_{2}\overset{-}{x} \\ \overset{-}{x}({m}_{1}+{m}_{2}) & = & {m}_{1}{x}_{1}+{m}_{2}{x}_{2} \\ \overset{-}{x} & = & \frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}}{{m}_{1}+{m}_{2}}.\end{array}\]

The expression in the numerator, \({m}_{1}{x}_{1}+{m}_{2}{x}_{2},\) is called the first moment of the system with respect to the origin. If the context is clear, we often drop the word first and just refer to this expression as the moment of the system. The expression in the denominator, \({m}_{1}+{m}_{2},\) is the total mass of the system. Thus, the center of mass of the system is the point at which the total mass of the system could be concentrated without changing the moment.

This idea is not limited just to two point masses. In general, if n masses, \({m}_{1},{m}_{2}\text{,\ldots },{m}_{n},\) are placed on a number line at points \({x}_{1},{x}_{2}\text{,\ldots },{x}_{n},\) respectively, then the center of mass of the system is given by

\[\overset{-}{x}=\frac{\sum _{i=1}^{n}{m}_{i}{x}_{i}}{\sum _{i=1}^{n}{m}_{i}}.\]

We apply this theorem in the following example.

Condensed — the full section is in OpenStax Calculus Volume 1.

Center of Mass of Thin Plates

So far we have looked at systems of point masses on a line and in a plane. Now, instead of having the mass of a system concentrated at discrete points, we want to look at systems in which the mass of the system is distributed continuously across a thin sheet of material. For our purposes, we assume the sheet is thin enough that it can be treated as if it is two-dimensional. Such a sheet is called a lamina. Next we develop techniques to find the center of mass of a lamina. In this section, we also assume the density of the lamina is constant.

Laminas are often represented by a two-dimensional region in a plane. The geometric center of such a region is called its centroid. Since we have assumed the density of the lamina is constant, the center of mass of the lamina depends only on the shape of the corresponding region in the plane; it does not depend on the density. In this case, the center of mass of the lamina corresponds to the centroid of the delineated region in the plane. As with systems of point masses, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the x- and y-axes.

We first consider a lamina in the shape of a rectangle. Recall that the center of mass of a lamina is the point where the lamina balances. For a rectangle, that point is both the horizontal and vertical center of the rectangle. Based on this understanding, it is clear that the center of mass of a rectangular lamina is the point where the diagonals intersect, which is a result of the symmetry principle, and it is stated here without proof.

Let’s turn to more general laminas. Suppose we have a lamina bounded above by the graph of a continuous function \(f(x),\) below by the x-axis, and on the left and right by the lines \(x=a\) and \(x=b,\) respectively, as shown in the following figure.

As with systems of point masses, to find the center of mass of the lamina, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the x- and y-axes. As we have done many times before, we approximate these quantities by partitioning the interval \([a,b]\) and constructing rectangles.

\[m\approx \sum _{i=1}^{n}\rho f({x}_{i}^{*})\text{\Delta }x.\]\[m=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\rho f({x}_{i}^{*})\text{\Delta }x=\rho {\int }_{a}^{b}f(x)dx.\]\[{M}_{x}=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\rho \frac{{[f({x}_{i}^{*})]}^{2}}{2}\text{\Delta }x=\rho {\int }_{a}^{b}\frac{{[f(x)]}^{2}}{2}dx.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

The Symmetry Principle

We stated the symmetry principle earlier, when we were looking at the centroid of a rectangle. The symmetry principle can be a great help when finding centroids of regions that are symmetric. Consider the following example.

Example

Try it.

Let R be the region bounded above by the graph of the function \(f(x)=4-{x}^{2}\) and below by the x-axis. Find the centroid of the region.

Solution

The region is depicted in the following figure.

The region is symmetric with respect to the y-axis. Therefore, the x-coordinate of the centroid is zero. We need only calculate \(\overset{-}{y}.\) Once again, for the sake of convenience, assume \(\rho =1.\)

First, we calculate the total mass:

\[\begin{array}{ll}m & =\rho {\int }_{a}^{b}f(x)dx \\ & ={\int }_{-2}^{2}(4-{x}^{2})dx \\ & ={[4x-\frac{{x}^{3}}{3}]\ |}_{-2}^{2}=\frac{32}{3}.\end{array}\]

Next, we calculate the moments. We only need \({M}_{x}\text{:}\)

\[\begin{array}{ll}{M}_{x} & =\rho {\int }_{a}^{b}\frac{{[f(x)]}^{2}}{2}dx \\ & =\frac{1}{2}{\int }_{-2}^{2}{[4-{x}^{2}]}^{2}dx=\frac{1}{2}{\int }_{-2}^{2}(16-8{x}^{2}+{x}^{4})dx \\ & =\frac{1}{2}{[\frac{{x}^{5}}{5}-\frac{8{x}^{3}}{3}+16x]\ |}_{-2}^{2}=\frac{256}{15}.\end{array}\]

Then we have

\[\overset{-}{y}=\frac{{M}_{x}}{m}=\frac{256}{15}\cdot \frac{3}{32}=\frac{8}{5}.\]

The centroid of the region is \((0,8\text{/}5).\)

Condensed — the full section is in OpenStax Calculus Volume 1.

Theorem of Pappus

This section ends with a discussion of the theorem of Pappus for volume, which allows us to find the volume of particular kinds of solids by using the centroid. (There is also a theorem of Pappus for surface area, but it is much less useful than the theorem for volume.)

We can prove the case when the region is bounded above by the graph of a function \(f(x)\) and below by the graph of a function \(g(x)\) over an interval \([a,b],\) and for which the axis of revolution is the y-axis. In this case, the area of the region is \(A={\int }_{a}^{b}[f(x)-g(x)]dx.\) Since the axis of rotation is the y-axis, the distance traveled by the centroid of the region depends only on the x-coordinate of the centroid, \(\overset{-}{x},\) which is

\[\overset{-}{x}=\frac{{M}_{y}}{m},\]

where

\[m=\rho {\int }_{a}^{b}[f(x)-g(x)]dx\ \text{and}\ {M}_{y}=\rho {\int }_{a}^{b}x[f(x)-g(x)]dx.\]

Then,

\[d=2\pi \frac{\rho {\int }_{a}^{b}x[f(x)-g(x)]dx}{\rho {\int }_{a}^{b}[f(x)-g(x)]dx}\]

and thus

\[d\cdot A=2\pi {\int }_{a}^{b}x[f(x)-g(x)]dx.\]

However, using the method of cylindrical shells, we have

\[V=2\pi {\int }_{a}^{b}x[f(x)-g(x)]dx.\]

So,

\[V=d\cdot A\]

and the proof is complete.

□

Example

Try it.

Let R be a circle of radius 2 centered at \((4,0).\) Use the theorem of Pappus for volume to find the volume of the torus generated by revolving R around the y-axis.

Solution

The region and torus are depicted in the following figure.

The region R is a circle of radius 2, so the area of R is \(A=4\pi\) units2. By the symmetry principle, the centroid of R is the center of the circle. The centroid travels around the y-axis in a circular path of radius 4, so the centroid travels \(d=8\pi\) units. Then, the volume of the torus is \(A\cdot d=32{\pi }^{2}\) units3.

Key Concepts

  • Mathematically, the center of mass of a system is the point at which the total mass of the system could be concentrated without changing the moment. Loosely speaking, the center of mass can be thought of as the balancing point of the system.
  • For point masses distributed along a number line, the moment of the system with respect to the origin is \(M=\sum _{i=1}^{n}{m}_{i}{x}_{i}.\) For point masses distributed in a plane, the moments of the system with respect to the x- and y-axes, respectively, are \({M}_{x}=\sum _{i=1}^{n}{m}_{i}{y}_{i}\) and \({M}_{y}=\sum _{i=1}^{n}{m}_{i}{x}_{i},\) respectively.
  • For a lamina bounded above by a function \(f(x),\) the moments of the system with respect to the x- and y-axes, respectively, are \({M}_{x}=\rho {\int }_{a}^{b}\frac{{[f(x)]}^{2}}{2}dx\) and \({M}_{y}=\rho {\int }_{a}^{b}xf(x)dx.\)
  • The x- and y-coordinates of the center of mass can be found by dividing the moments around the y-axis and around the x-axis, respectively, by the total mass. The symmetry principle says that if a region is symmetric with respect to a line, then the centroid of the region lies on the line.
  • The theorem of Pappus for volume says that if a region is revolved around an external axis, the volume of the resulting solid is equal to the area of the region multiplied by the distance traveled by the centroid of the region.

Key Equations

Mass of a lamina\(m=\rho {\int }_{a}^{b}f(x)dx\)
Moments of a lamina\({M}_{x}=\rho {\int }_{a}^{b}\frac{{[f(x)]}^{2}}{2}dx\ \text{and}\ {M}_{y}=\rho {\int }_{a}^{b}xf(x)dx\)
Center of mass of a lamina\(\overset{-}{x}=\frac{{M}_{y}}{m}\ \text{and}\ \overset{-}{y}=\frac{{M}_{x}}{m}\)

Moments and Centers of Mass

For the following exercises, calculate the center of mass for the collection of masses given.

For the following exercises, compute the center of mass \(\overset{-}{x}.\)

For the following exercises, compute the center of mass \((\overset{-}{x},\overset{-}{y}).\) Use symmetry to help locate the center of mass whenever possible.

For the following exercises, use a calculator to draw the region, then compute the center of mass \((\overset{-}{x},\overset{-}{y}).\) Use symmetry to help locate the center of mass whenever possible.

For the following exercises, use the theorem of Pappus to determine the volume of the shape.

For the following exercises, use a calculator to draw the region enclosed by the curve. Find the area \(M\) and the centroid \((\overset{-}{x},\overset{-}{y})\) for the given shapes. Use symmetry to help locate the center of mass whenever possible.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Suppose four point masses are placed on a number line as follows:

    \[\begin{array}{llll}{m}_{1}=30\ \text{kg,}\ \text{placed at}\ {x}_{1}=-2\ \text{m} & & & {m}_{2}=5\ \text{kg,}\ \text{placed at}\ {x}_{2}=3\ \text{m} \\ {m}_{3}=10\ \text{kg,}\ \text{placed at}\ {x}_{3}=6\ \text{m} & & & {m}_{4}=15\ \text{kg,}\ \text{placed at}\ {x}_{4}=-3\ \text{m}.\end{array}\]

    Find the moment of the system with respect to the origin and find the center of mass of the system.

    发送答案

    First, we need to calculate the moment of the system:

    \[\begin{array}{ll}M & =\sum _{i=1}^{4}{m}_{i}{x}_{i} \\ & =-60+15+60-45=-30.\end{array}\]

    Now, to find the center of mass, we need the total mass of the system:

    \[\begin{array}{ll}m & =\sum _{i=1}^{4}{m}_{i} \\ & =30+5+10+15=60\ \text{kg}\text{.}\end{array}\]

    Then we have

    \[\overset{-}{x}=\frac{M}{m}=\frac{-30}{60}=-\frac{1}{2}.\]

    The center of mass is located 1/2 m to the left of the origin.

  2. Suppose four point masses are placed on a number line as follows:

    \[\begin{array}{llll}{m}_{1}=12\ \text{kg,}\ \text{placed at}\ {x}_{1}=-4\ \text{m} & & & {m}_{2}=12\ \text{kg,}\ \text{placed at}\ {x}_{2}=4\ \text{m} \\ {m}_{3}=30\ \text{kg,}\ \text{placed at}\ {x}_{3}=2\ \text{m} & & & {m}_{4}=6\ \text{kg,}\ \text{placed at}\ {x}_{4}=-6\ \text{m}.\end{array}\]

    Find the moment of the system with respect to the origin and find the center of mass of the system.

    发送答案

    \(M=24,\overset{-}{x}=\frac{2}{5}\ \text{m}\)

  3. Suppose three point masses are placed in the xy-plane as follows (assume coordinates are given in meters):

    \[\begin{array}{l}{m}_{1}=2\ \text{kg, placed at}\ (-1,3), \\ {m}_{2}=6\ \text{kg, placed at}\ (1,1), \\ {m}_{3}=4\ \text{kg, placed at}\ (2,-2).\end{array}\]

    Find the center of mass of the system.

    发送答案

    First we calculate the total mass of the system:

    \[m=\sum _{i=1}^{3}{m}_{i}=2+6+4=12\ \text{kg}\text{.}\]

    Next we find the moments with respect to the x- and y-axes:

    \[\begin{array}{l} \\ \\ {M}_{y}=\sum _{i=1}^{3}{m}_{i}{x}_{i}=-2+6+8=12, \\ {M}_{x}=\sum _{i=1}^{3}{m}_{i}{y}_{i}=6+6-8=4.\end{array}\]

    Then we have

    \[\overset{-}{x}=\frac{{M}_{y}}{m}=\frac{12}{12}=1\ \text{and}\ \overset{-}{y}=\frac{{M}_{x}}{m}=\frac{4}{12}=\frac{1}{3}.\]

    The center of mass of the system is \((1,1\text{/}3),\) in meters.

  4. Suppose three point masses are placed on a number line as follows (assume coordinates are given in meters):

    \[\begin{array}{l}{m}_{1}=5\ \text{kg, placed at}\ (-2,-3), \\ {m}_{2}=3\ \text{kg, placed at}\ (2,3), \\ {m}_{3}=2\ \text{kg, placed at}\ (-3,-2).\end{array}\]

    Find the center of mass of the system.

    发送答案

    \((-1,-1)\) m

  5. Let R be the region bounded above by the graph of the function \(f(x)=\sqrt{x}\) and below by the x-axis over the interval \([0,4].\) Find the centroid of the region.

    发送答案

    The region is depicted in the following figure.

    Since we are only asked for the centroid of the region, rather than the mass or moments of the associated lamina, we know the density constant \(\rho\) cancels out of the calculations eventually. Therefore, for the sake of convenience, let’s assume \(\rho =1.\)

    First, we need to calculate the total mass:

    \[\begin{array}{ll}m & =\rho {\int }_{a}^{b}f(x)dx={\int }_{0}^{4}\sqrt{x}\ dx \\ & ={\frac{2}{3}{x}^{3\text{/}2}|}_{0}^{4}=\frac{2}{3}[8-0]=\frac{16}{3}.\end{array}\]

    Next, we compute the moments:

    \[\begin{array}{ll}{M}_{x} & =\rho {\int }_{a}^{b}\frac{{[f(x)]}^{2}}{2}dx \\ & ={\int }_{0}^{4}\frac{x}{2}dx={\frac{1}{4}{x}^{2}|}_{0}^{4}=4\end{array}\]

    and

    \[\begin{array}{ll}{M}_{y} & =\rho {\int }_{a}^{b}xf(x)dx \\ & ={\int }_{0}^{4}x\sqrt{x}dx={\int }_{0}^{4}{x}^{3\text{/}2}dx \\ & ={\frac{2}{5}{x}^{5\text{/}2}|}_{0}^{4}=\frac{2}{5}[32-0]=\frac{64}{5}.\end{array}\]

    Thus, we have

    \[\overset{-}{x}=\frac{{M}_{y}}{m}=\frac{64\text{/}5}{16\text{/}3}=\frac{64}{5}\cdot \frac{3}{16}=\frac{12}{5}\ \text{and}\ \overset{-}{y}=\frac{{M}_{x}}{m}=\frac{4}{16\text{/}3}=4\cdot \frac{3}{16}=\frac{3}{4}.\]

    The centroid of the region is \((12\text{/}5,3\text{/}4).\)

  6. Let R be the region bounded above by the graph of the function \(f(x)={x}^{2}\) and below by the x-axis over the interval \([0,2].\) Find the centroid of the region.

    发送答案

    The centroid of the region is \((3\text{/}2,6\text{/}5).\)

  7. Let R be the region bounded above by the graph of the function \(f(x)=1-{x}^{2}\) and below by the graph of the function \(g(x)=x-1.\) Find the centroid of the region.

    发送答案

    The region is depicted in the following figure.

    The graphs of the functions intersect at \((-2,-3)\) and \((1,0),\) so we integrate from −2 to 1. Once again, for the sake of convenience, assume \(\rho =1.\)

    First, we need to calculate the total mass:

    \[\begin{array}{ll}m & =\rho {\int }_{a}^{b}[f(x)-g(x)]dx \\ & ={\int }_{-2}^{1}[1-{x}^{2}-(x-1)]dx={\int }_{-2}^{1}(2-{x}^{2}-x)dx \\ & ={[2x-\frac{1}{3}{x}^{3}-\frac{1}{2}{x}^{2}]\ |}_{-2}^{1}=[2-\frac{1}{3}-\frac{1}{2}]-[-4+\frac{8}{3}-2]=\frac{9}{2}.\end{array}\]

    Next, we compute the moments:

    \[\begin{array}{ll}{M}_{x} & =\rho {\int }_{a}^{b}\frac{1}{2}({[f(x)]}^{2}-{[g(x)]}^{2})dx \\ & =\frac{1}{2}{\int }_{-2}^{1}({(1-{x}^{2})}^{2}-{(x-1)}^{2})dx=\frac{1}{2}{\int }_{-2}^{1}({x}^{4}-3{x}^{2}+2x)dx \\ & =\frac{1}{2}{[\frac{{x}^{5}}{5}-{x}^{3}+{x}^{2}]\ |}_{-2}^{1}=-\frac{27}{10}\end{array}\]

    and

    \[\begin{array}{ll}{M}_{y} & =\rho {\int }_{a}^{b}x[f(x)-g(x)]dx \\ & ={\int }_{-2}^{1}x[(1-{x}^{2})-(x-1)]dx={\int }_{-2}^{1}x[2-{x}^{2}-x]dx \\ ={\int }_{-2}^{1}(2x-{x}^{3}-{x}^{2})dx & \ \\ & ={[{x}^{2}-\frac{{x}^{4}}{4}-\frac{{x}^{3}}{3}]\ |}_{-2}^{1}=-\frac{9}{4}.\end{array}\]

    Therefore, we have

    \[\overset{-}{x}=\frac{{M}_{y}}{m}=-\frac{9}{4}\cdot \frac{2}{9}=-\frac{1}{2}\ \text{and}\ \overset{-}{y}=\frac{{M}_{x}}{m}=-\frac{27}{10}\cdot \frac{2}{9}=-\frac{3}{5}.\]

    The centroid of the region is \((\text{-}(1\text{/}2),\text{-}(3\text{/}5)).\)

  8. Let R be the region bounded above by the graph of the function \(f(x)=6-{x}^{2}\) and below by the graph of the function \(g(x)=3-2x.\) Find the centroid of the region.

    发送答案

    The centroid of the region is \((1,13\text{/}5).\)

  9. Let R be the region bounded above by the graph of the function \(f(x)=4-{x}^{2}\) and below by the x-axis. Find the centroid of the region.

    发送答案

    The region is depicted in the following figure.

    The region is symmetric with respect to the y-axis. Therefore, the x-coordinate of the centroid is zero. We need only calculate \(\overset{-}{y}.\) Once again, for the sake of convenience, assume \(\rho =1.\)

    First, we calculate the total mass:

    \[\begin{array}{ll}m & =\rho {\int }_{a}^{b}f(x)dx \\ & ={\int }_{-2}^{2}(4-{x}^{2})dx \\ & ={[4x-\frac{{x}^{3}}{3}]\ |}_{-2}^{2}=\frac{32}{3}.\end{array}\]

    Next, we calculate the moments. We only need \({M}_{x}\text{:}\)

    \[\begin{array}{ll}{M}_{x} & =\rho {\int }_{a}^{b}\frac{{[f(x)]}^{2}}{2}dx \\ & =\frac{1}{2}{\int }_{-2}^{2}{[4-{x}^{2}]}^{2}dx=\frac{1}{2}{\int }_{-2}^{2}(16-8{x}^{2}+{x}^{4})dx \\ & =\frac{1}{2}{[\frac{{x}^{5}}{5}-\frac{8{x}^{3}}{3}+16x]\ |}_{-2}^{2}=\frac{256}{15}.\end{array}\]

    Then we have

    \[\overset{-}{y}=\frac{{M}_{x}}{m}=\frac{256}{15}\cdot \frac{3}{32}=\frac{8}{5}.\]

    The centroid of the region is \((0,8\text{/}5).\)

  10. Let R be the region bounded above by the graph of the function \(f(x)=1-{x}^{2}\) and below by x-axis. Find the centroid of the region.

    发送答案

    The centroid of the region is \((0,2\text{/}5).\)

  11. Let R be a circle of radius 2 centered at \((4,0).\) Use the theorem of Pappus for volume to find the volume of the torus generated by revolving R around the y-axis.

    发送答案

    The region and torus are depicted in the following figure.

    The region R is a circle of radius 2, so the area of R is \(A=4\pi\) units2. By the symmetry principle, the centroid of R is the center of the circle. The centroid travels around the y-axis in a circular path of radius 4, so the centroid travels \(d=8\pi\) units. Then, the volume of the torus is \(A\cdot d=32{\pi }^{2}\) units3.

  12. Let R be a circle of radius 1 centered at \((3,0).\) Use the theorem of Pappus for volume to find the volume of the torus generated by revolving R around the y-axis.

    发送答案

    \(6{\pi }^{2}\) units3

  13. \({m}_{1}=2\) at \({x}_{1}=1\) and \({m}_{2}=4\) at \({x}_{2}=2\)

  14. \({m}_{1}=1\) at \({x}_{1}=-1\) and \({m}_{2}=3\) at \({x}_{2}=2\)

    发送答案

    \(\frac{5}{4}\)

  15. \(m=3\) at \(x=0,1,2,6\)

  16. Unit masses at \((x,y)=(1,0),(0,1),(1,1)\)

    发送答案

    \((\frac{2}{3},\frac{2}{3})\)

  17. \({m}_{1}=1\) at \((1,0)\) and \({m}_{2}=4\) at \((0,1)\)

  18. \({m}_{1}=1\) at \((1,0)\) and \({m}_{2}=3\) at \((2,2)\)

    发送答案

    \((\frac{7}{4},\frac{3}{2})\)

  19. \(\rho =1\) for \(x\in (-1,3)\)

  20. \(\rho ={x}^{2}\) for \(x\in (0,L)\)

    发送答案

    \(\frac{3L}{4}\)

  21. \(\rho =1\) for \(x\in (0,1)\) and \(\rho =2\) for \(x\in (1,2)\)

  22. \(\rho =\text{sin}\ x\) for \(x\in (0,\pi )\)

    发送答案

    \(\frac{\pi }{2}\)

  23. \(\rho =\text{cos}\ x\) for \(x\in (0,\frac{\pi }{2})\)

  24. \(\rho ={e}^{x}\) for \(x\in (0,2)\)

    发送答案

    \(\frac{{e}^{2}+1}{{e}^{2}-1}\)

  25. \(\rho ={x}^{3}+x{e}^{\text{-}x}\) for \(x\in (0,1)\)

  26. \(\rho =x\ \text{sin}\ x\) for \(x\in (0,\pi )\)

    发送答案

    \(\frac{{\pi }^{2}-4}{\pi }\)

  27. \(\rho =\sqrt{x}\) for \(x\in (1,4)\)

  28. \(\rho =\text{ln}\ x\) for \(x\in (1,e)\)

    发送答案

    \(\frac{1}{4}(1+{e}^{2})\)

  29. \(\rho =7\) in the square \(0\le x\le 1,\) \(0\le y\le 1\)

  30. \(\rho =3\) in the triangle with vertices \((0,0),\) \((a,0),\) and \((0,b)\)

    发送答案

    \((\frac{a}{3},\frac{b}{3})\)

  31. \(\rho =2\) for the region bounded by \(y=\text{cos}(x),\) \(y=\text{-}\text{cos}(x),\) \(x=-\frac{\pi }{2},\) and \(x=\frac{\pi }{2}\)

  32. [T] The region bounded by \(y=\text{cos}(2x),\) \(x=-\frac{\pi }{4},\) and \(x=\frac{\pi }{4}\)

    发送答案

    \((0,\frac{\pi }{8})\)

  33. [T] The region between \(y=2{x}^{2},\) \(y=0,\) \(x=0,\) and \(x=1\)

  34. [T] The region between \(y=\frac{5}{4}{x}^{2}\) and \(y=5\)

    发送答案

    \((0,3)\)

  35. [T] Region between \(y=\sqrt{x},\) \(y=\text{ln}(x),\) \(x=1,\) and \(x=4\)

  36. [T] The region bounded by \(y=0,\) \(\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1\)

    发送答案

    \((0,\frac{4}{\pi })\)

  37. [T] The region bounded by \(y=0,\) \(x=0,\) and \(\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1\)

  38. [T] The region bounded by \(y={x}^{2}\) and \(y={x}^{4}\) in the first quadrant

    发送答案

    \((\frac{5}{8},\frac{1}{3})\)

  39. Rotating \(y=mx\) around the \(x\)-axis between \(x=0\) and \(x=1\)

  40. Rotating \(y=mx\) around the \(y\)-axis between \(x=0\) and \(x=1\)

    发送答案

    \(\frac{2m\pi }{3}\)

Symbols used here

\sum_{k=1}^{n} a_k
summation
Add a_k for k = 1 up to n.
\int f(x)\,dx,\ \int_a^b
integral
Antiderivative (indefinite) or signed area from a to b (definite).
\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
\infty
infinity
Not a number: "grows without bound" in limits and intervals.
\pi
pi
Ratio of a circle's circumference to its diameter, 3.14159…
x \in A,\ A \subseteq B
element of, subset
x belongs to A; every element of A is in B.
i
imaginary unit
i² = −1.
\approx
approximately equal
Equal to the precision shown, not exactly.
\sup,\ \inf
supremum, infimum
Least upper bound, greatest lower bound.
e
Euler's number
2.71828…, the base whose exponential is its own derivative.
\sin,\ \cos,\ \tan
sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\lim_{x \to a} f(x)
limit
The value f(x) approaches as x approaches a.
f'(x),\ \frac{dy}{dx}
derivative
Instantaneous rate of change; slope of the graph.
y',\ y''
first and second derivative of y
Prime notation for derivatives with respect to x (or t).
C,\ C_1,\ C_2
arbitrary constants
Constants of integration fixed by initial conditions.

How to: Moments and Centers of Mass

  1. Find the center of mass of objects distributed along a line.
  2. Locate the center of mass of a thin plate.
  3. Use symmetry to help locate the centroid of a thin plate.
  4. Apply the theorem of Pappus for volume.
  5. The mass of the lamina is
  6. The moments
  7. The coordinates of the center of mass
  8. The mass of the lamina is

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.

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Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests