Center of Mass and Moments of Inertia in Three Dimensions

Learning Objectives

Use triple integrals to locate the center of mass of a three-dimensional object.

All the expressions of double integrals discussed so far can be modified to become triple integrals.

definition

If we have a solid object $Q$ with a density function ${\rho}{({x},{y},{z})}$ at any point $(x, y, z)$ in space, then its mass is

${m} = {\underset{Q}{\displaystyle\iiint}}{\rho}{({x},{y},{z})}{dV}.$

Its moments about the $xy$ -plane, the $xz$ -plane, and the $yz$ -plane are

$\begin{aligned} {{M}_{xy}} = & {\underset{Q}{\displaystyle\iiint}}{z}{\rho}{({x},{y},{z})}{dV}, {{M}_{xz}} = {\underset{Q}{\displaystyle\iiint}}{y}{\rho}{({x},{y},{z})}{dV}, \\ {{M}_{yz}} = & {\underset{Q}{\displaystyle\iiint}}{x}{\rho}{({x},{y},{z})}{dV}. \end{aligned}$

If the center of mass of the object is the point ${\left ( {\overline{x}},{\overline{y}},{\overline{z}} \right )},$ then

$\large{{\overline{x}} = {\dfrac{{M}_{yz}}{m}}, \ {\overline{y}} = {\dfrac{{M}_{xz}}{m}}, \ {\overline{z}} = {\dfrac{{M}_{xy}}{m}}}.$

Also, if the solid object is homogeneous (with constant density), then the center of mass becomes the centroid of the solid. Finally, the moments of inertia about the $yz$ -plane, the $xz$ -plane, and the $xy$ -plane are

$\begin{aligned} {{I}_{x}} = & {\underset{Q}{\displaystyle\iiint}}{({{y}^{2}} + {{z}^{2}})}{\rho}{({x},{y},{z})}{dV}, \\ {{I}_{y}} = &{\underset{Q}{\displaystyle\iiint}}{({{x}^{2}} + {{z}^{2}})}{\rho}{({x},{y},{z})}{dV}, \\ {{I}_{z}} = & {\underset{Q}{\displaystyle\iiint}}{({{x}^{2}} + {{y}^{2}})}{\rho}{({x},{y},{z})}{dV}. \end{aligned}$

Example: finding the mass of a solid

Suppose that $Q$ is a solid region bounded by $x+2y+3z=6$ and the coordinate planes and has density ${\rho}{({x},{y},{z})} = {{x}^{2}}{y}{z}$ . Find the total mass.

Show Solution

The region $Q$ is a tetrahedron (Figure 1) meeting the axes at the points $(6, 0, 0)$ , $(0, 3, 0)$ , and $(0, 0, 2)$ . To find the limits of integration, let $z=0$ in the slanted plane ${z} = {\frac{1}{3}}{({6} - {x} - {{2}{y}})}$ . Then for $x$ and $y$ find the projection of $Q$ onto the $xy$ -plane, which is bounded by the axes and the line $x+2y=6$ . Hence the mass is

m = \underset{Q}{∭} ρ (x, y, z) d V = \int_{x = 0}^{x = 6} \int_{y = 0}^{y = \frac{1}{2} (6 - x)} \int_{z = 0}^{z = \frac{1}{3} (6 - x - 2 y)} x^{2} y z d z d y d x = \frac{108}{35} \approx 3.086.

${m} = {\underset{Q}{\displaystyle\iiint}}{\rho}{({x},{y},{z})}{dV} = \displaystyle\int^{{x} = {6}}_{{x} = {0}} \ \displaystyle\int^{y = \frac{1}{2}({6} - x)}_{y = 0} \ \displaystyle\int^{z = \frac{1}{3}(6 - x - 2y)}_{z = 0} \ x^{2}yz \ dz \ dy \ dx = \frac{108}{35} \ \approx \ 3.086.$

In x y z space, the solid Q is shown with corners (0, 0, 0), (0, 0, 2), (0, 3, 0), and (6, 0, 0). Alternatively, you could consider the solid as being bounded by the x y, x z, and y z planes and the plane x + 2y + 3z = 6, forming an irregular tetrahedron.

Figure 1. Finding the mass of a three-dimensional solid $Q$ .

try it

Consider the same region $Q$ (Figure 1), and use the density function ${\rho}{({x},{y},{z})} = {{x}^{2}}{y}{z}$ . Find the mass.

Show Solution

$\frac{54}{35}=1.543$

Example: finding the center of Mass of a solid

Suppose $Q$ is a solid region bounded by the plane $x+2y+3z=6$ and the coordinate planes with density ${\rho}{({x},{y},{z})} = {{x}^{2}}{y}{z}$ (see Figure 1). Find the center of mass using decimal approximation. Use the mass found in Example “Finding the Mass of a Solid”.

Show Solution

We have used this tetrahedron before and know the limits of integration, so we can proceed to the computations right away. First, we need to find the moments about the $xy$ -plane, the $xz$ -plane, and the $yz$ -plane:

$\begin{aligned} {{M}_{xy}} = &{\underset{Q}{\displaystyle\iiint}}{z}{\rho}{({x},{y},{z})}{dV} = {\displaystyle\int^{{x} = {6}}_{{x} = {0}}} \ {\displaystyle\int^{{y} = {1/2}{({6} - {x})}}_{{y} = {0}}} \ {\displaystyle\int^{{z} = {1/3}{({6} - {x} - {2y})}}_{{z} = {0}}} \ {{x}^{2}}{{yz}^{2}}{dz} \ {dy} \ {dx} = {\frac{54}{35}} \ {\approx} \ {1.543}, \\ {{M}_{xz}} = & {\underset{Q}{\displaystyle\iiint}}{y}{\rho}{({x},{y},{z})}{dV} = {\displaystyle\int^{{x} = {6}}_{{x} = {0}}} \ {\displaystyle\int^{{y} = {1/2}{({6} - {x})}}_{{y} = {0}}} \ {\displaystyle\int^{{z} = {1/3}{({6} - {x} - {2y})}}_{{z} = {0}}} \ {{x}^{2}}{{y}^{2}}{z} \ {dz} \ {dy} \ {dx} = {\frac{81}{35}} \ {\approx} \ {2.314}, \\ {{M}_{yz}} = & {\underset{Q}{\displaystyle\iiint}}{x}{\rho}{({x},{y},{z})}{dV} = {\displaystyle\int^{{x} = {6}}_{{x} = {0}}} \ {\displaystyle\int^{{y} = {1/2}{({6} - {x})}}_{{y} = {0}}} \ {\displaystyle\int^{{z} = {1/3}{({6} - {x} - {2y})}}_{{z} = {0}}} \ {{x}^{3}}{yz} \ {dz} \ {dy} \ {dx} = {\frac{243}{35}} \ {\approx} \ {6.943}. \end{aligned}$

Hence the center of mass is

$\begin{aligned} {\overline{x}} = & {\frac{{M}_{yz}}{m}}, \ {\overline{y}} = {\frac{{M}_{xz}}{m}}, \ {\overline{z}} = {\frac{{M}_{xy}}{m}}, \\ {\overline{x}} = & {\frac{{M}_{yz}}{m}} = {\frac{243/35}{108/35}} = {\frac{243}{108}} = {2.25}, \\ {\overline{y}} = & {\frac{{M}_{xz}}{m}} = {\frac{81/35}{108/35}} = {\frac{81}{108}} = {0.75}, \\ {\overline{z}} = & {\frac{{M}_{xy}}{m}} = {\frac{54/35}{108/35}} = {\frac{54}{108}} = {0.5}. \end{aligned}$

The center of mass for the tetrahedron $Q$ is the point ${({2.25},{0.75},{0.5})}.$

Try It

Consider the same region $Q$ (Figure 1) and use the density function ${\rho}{({x},{y},{z})} = xy^{2}z$ . Find the center of mass.

Show Solution

$\left(\frac32,\frac98,\frac12\right)$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 5.41” here (opens in new window).

We conclude this section with an example of finding moments of inertia $I_x$ , $I_y$ , and $I_z$ .

example: finding the moments of inertia of a solid

Suppose that $Q$ is a solid region and is bounded by $x+2y+3z=6$ and the coordinate planes with density ${\rho}{({x},{y},{z})} = {{x}^{2}}{y}{z}$ (see Figure 1). Find the moments of inertia of the tetrahedron $Q$ about the $yz$ -plane, the $xz$ -plane, and the $xy$ -plane.

Show Solution

Once again, we can almost immediately write the limits of integration and hence we can quickly proceed to evaluating the moments of inertia. Using the formula stated before, the moments of inertia of the tetrahedron

Q

$Q$ about the

x y

$xy$ -plane, the

x z

$xz$ -plane, and the

y z

$yz$ -plane are

\begin{aligned} I_{x} = & \underset{Q}{∭} (y^{2} + z^{2}) ρ (x, y, z) d V, \\ I_{y} = & \underset{Q}{∭} (x^{2} + z^{2}) ρ (x, y, z) d V, \end{aligned}

and

I_{z} = \underset{Q}{∭} (x^{2} + y^{2}) ρ (x, y, z) d V with ρ (x, y, z) = x^{2} y z .

${{I}_{z}} = {\underset{Q}{\displaystyle\iiint}}{({{x}^{2}} + {{y}^{2}})}{\rho}{({x},{y},{z})}{dV} \ {\text{with}} \ {\rho}{({x},{y},{z})} = {{x}^{2}}{yz}.$

Proceeding with the computations, we have

[latex]\begin{align}

{{I}_{x}} &= {\underset{Q}{\displaystyle\iiint}}{({{y}^{2}} + {{z}^{2}})}x^2yz \ {dV}=\displaystyle\int_{x=0}^{x=6}\displaystyle\int_{y=0}^{y=\frac{1}{2}(6-x)}\displaystyle\int_{z=0}^{z=\frac13(6-x-2y)}(y^2+z^2)x^2yz \ dz \ dy \ dx = \frac{117}{35}\approx3.343, \\ {{I}_{y}} &= {\underset{Q}{\displaystyle\iiint}}{({{x}^{2}} + {{z}^{2}})}x^2yz \ {dV}=\displaystyle\int_{x=0}^{x=6}\displaystyle\int_{y=0}^{y=\frac{1}{2}(6-x)}\displaystyle\int_{z=0}^{z=\frac13(6-x-2y)}(x^2+z^2)x^2yz \ dz \ dy \ dx = \frac{684}{35}\approx19.543, \\ {{I}_{z}} &= {\underset{Q}{\displaystyle\iiint}}{({{x}^{2}} + {{y}^{2}})}x^2yz \ {dV}=\displaystyle\int_{x=0}^{x=6}\displaystyle\int_{y=0}^{y=\frac{1}{2}(6-x)}\displaystyle\int_{z=0}^{z=\frac13(6-x-2y)}(x^2+y^2)x^2yz \ dz \ dy \ dx = \frac{729}{35}\approx20.829. \end{align}[/latex]

Thus, the moments of inertia of the tetrahedron $Q$ about the $yz$ -plane, the $xz$ -plane, and the $xy$ -plane are ${117/35}, {684/35},$ and ${729/35}$ , respectively.

try it

Consider the same region $Q$ (Figure 1), and use the density function ${\rho}{({x},{y},{z})} = {x}{{y}^{2}}{z}$ . Find the moments of inertia about the three coordinate planes.

Show Solution

The moments of inertia of the tetrahedron $Q$ about the $yz$ -plane, the $xz$ -plane, and the $xy$ -plane are $99/35$ , $36/7$ , and $243/35$ , respectively.

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 5.42” here (opens in new window).

Module 5: Multiple Integration