5.6.1Use double integrals to locate the center of mass of a two-dimensional object.
5.6.2Use double integrals to find the moment of inertia of a two-dimensional object.
5.6.3Use triple integrals to locate the center of mass of a three-dimensional object.
We have already discussed a few applications of multiple integrals, such as finding areas, volumes, and the average value of a function over a bounded region. In this section we develop computational techniques for finding the center of mass and moments of inertia of several types of physical objects, using double integrals for a lamina (flat plate) and triple integrals for a three-dimensional object with variable density. The density is usually considered to be a constant number when the lamina or the object is homogeneous; that is, the object has uniform density.
Center of Mass in Two Dimensions
The center of mass is also known as the center of gravity if the object is in a uniform gravitational field. If the object has uniform density, the center of mass is the geometric center of the object, which is called the centroid. Figure 5.64 shows a point as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass.
Figure 5.64A lamina is perfectly balanced on a spindle if the lamina’s center of mass sits on the spindle.
To find the coordinates of the center of mass of a lamina, we need to find the moment of the lamina about the and the moment about the We also need to find the mass of the lamina. Then
Refer to Moments and Centers of Mass for the definitions and the methods of single integration to find the center of mass of a one-dimensional object (for example, a thin rod). We are going to use a similar idea here except that the object is a two-dimensional lamina and we use a double integral.
If we allow a constant density function, then give the centroid of the lamina.
Suppose that the lamina occupies a region in the and let be its density (in units of mass per unit area) at any point Hence, where and are the mass and area of a small rectangle containing the point and the limit is taken as the dimensions of the rectangle go to (see the following figure).
Figure 5.65The density of a lamina at a point is the limit of its mass per area in a small rectangle about the point as the area goes to zero.
Just as before, we divide the region into tiny rectangles with area and choose as sample points. Then the mass of each is equal to (Figure 5.66). Let and be the number of subintervals in and respectively. Also, note that the shape might not always be rectangular but the limit works anyway, as seen in previous sections.
Figure 5.66Subdividing the lamina into tiny rectangles each containing a sample point
Hence, the mass of the lamina is
Let’s see an example now of finding the total mass of a triangular lamina.
Finding the Total Mass of a Lamina
Consider a triangular lamina with vertices and with density Find the total mass.
A sketch of the region is always helpful, as shown in the following figure.
Figure 5.67A lamina in the with density
Using the expression developed for mass, we see that
The computation is straightforward, giving the answer
Consider the same region as in the previous example, and use the density function Find the total mass. Hint: Use trigonometric substitution and then use the power reducing formulas for trigonometric functions.
Now that we have established the expression for mass, we have the tools we need for calculating moments and centers of mass. The moment about the for is the limit of the sums of moments of the regions about the Hence
Similarly, the moment about the for is the limit of the sums of moments of the regions about the Hence
Consider the same triangular lamina with vertices and with density Find the moments and
Use double integrals for each moment and compute their values:
The computation is quite straightforward.
Consider the same lamina as above, and use the density function Find the moments and
Finally we are ready to restate the expressions for the center of mass in terms of integrals. We denote the x-coordinate of the center of mass by and the y-coordinate by Specifically,
Finding the Center of Mass
Again consider the same triangular region with vertices and with density function Find the center of mass.
Using the formulas we developed, we have
Therefore, the center of mass is the point
If we choose the density instead to be uniform throughout the region (i.e., constant), such as the value 1 (any constant will do), then we can compute the centroid,
Notice that the center of mass is not exactly the same as the centroid of the triangular region. This is due to the variable density of If the density is constant, then we just use (constant). This value cancels out from the formulas, so for a constant density, the center of mass coincides with the centroid of the lamina.
Again use the same region as above and the density function Find the center of mass.
Once again, based on the comments at the end of Example 5.57, we have expressions for the centroid of a region on the plane:
We should use these formulas and verify the centroid of the triangular region referred to in the last three examples.
Finding Mass, Moments, and Center of Mass
Find the mass, moments, and the center of mass of the lamina of density occupying the region under the curve in the interval (see the following figure).
Figure 5.68Locating the center of mass of a lamina with density
First we compute the mass We need to describe the region between the graph of and the vertical lines and
Now compute the moments and
Finally, evaluate the center of mass,
Hence the center of mass is
Calculate the mass, moments, and the center of mass of the region between the curves and with the density function in the interval
Finding a Centroid
Find the centroid of the region under the curve over the interval (see the following figure).
Figure 5.69Finding a centroid of a region below the curve
To compute the centroid, we assume that the density function is constant and hence it cancels out:
Thus the centroid of the region is
Calculate the centroid of the region between the curves and with uniform density in the interval
Moments of Inertia
For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition of moments and centers of mass in Section 6.6 of Volume 1. The moment of inertia of a particle of mass about an axis is where is the distance of the particle from the axis. We can see from Figure 5.66 that the moment of inertia of the subrectangle about the is Similarly, the moment of inertia of the subrectangle about the is The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis.
The moment of inertia about the for the region is the limit of the sum of moments of inertia of the regions about the Hence
Similarly, the moment of inertia about the for is the limit of the sum of moments of inertia of the regions about the Hence
Sometimes, we need to find the moment of inertia of an object about the origin, which is known as the polar moment of inertia. We denote this by and obtain it by adding the moments of inertia and Hence
All these expressions can be written in polar coordinates by substituting and For example,
Finding Moments of Inertia for a Triangular Lamina
Use the triangular region with vertices and and with density as in previous examples. Find the moments of inertia.
Using the expressions established above for the moments of inertia, we have
Again use the same region as above and the density function Find the moments of inertia.
As mentioned earlier, the moment of inertia of a particle of mass about an axis is where is the distance of the particle from the axis, also known as the radius of gyration.
Hence the radii of gyration with respect to the the and the origin are
respectively. In each case, the radius of gyration tells us how far (perpendicular distance) from the axis of rotation the entire mass of an object might be concentrated. The moments of an object are useful for finding information on the balance and torque of the object about an axis, but radii of gyration are used to describe the distribution of mass around its centroidal axis. There are many applications in engineering and physics. Sometimes it is necessary to find the radius of gyration, as in the next example.
Finding the Radius of Gyration for a Triangular Lamina
Consider the same triangular lamina with vertices and and with density as in previous examples. Find the radii of gyration with respect to the the and the origin.
If we compute the mass of this region we find that We found the moments of inertia of this lamina in Example 5.58. From these data, the radii of gyration with respect to the and the origin are, respectively,
Use the same region from Example 5.61 and the density function Find the radii of gyration with respect to the the and the origin.
Center of Mass and Moments of Inertia in Three Dimensions
All the expressions of double integrals discussed so far can be modified to become triple integrals.
If we have a solid object with a density function at any point in space, then its mass is
Its moments about the the and the are
If the center of mass of the object is the point then
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 the and the are
Finding the Mass of a Solid
Suppose that is a solid region bounded by and the coordinate planes and has density Find the total mass.
The region is a tetrahedron (Figure 5.70) meeting the axes at the points and To find the limits of integration, let in the slanted plane Then for and find the projection of onto the which is bounded by the axes and the line Hence the mass is
Figure 5.70Finding the mass of a three-dimensional solid
Consider the same region (Figure 5.70), and use the density function Find the mass.
Finding the Center of Mass of a Solid
Suppose is a solid region bounded by the plane and the coordinate planes with density (see Figure 5.70). Find the center of mass using decimal approximation. Use the mass found in Example 5.62
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 the and the
Hence the center of mass is
The center of mass for the tetrahedron is the point
Consider the same region (Figure 5.70) and use the density function Find the center of mass.
We conclude this section with an example of finding moments of inertia and
Finding the Moments of Inertia of a Solid
Suppose that is a solid region and is bounded by and the coordinate planes with density (see Figure 5.70). Find the moments of inertia of the tetrahedron about the the and the
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 about the the and the are
Proceeding with the computations, we have
Thus, the moments of inertia of the tetrahedron about the the and the are respectively.
Consider the same region (Figure 5.70), and use the density function Find the moments of inertia about the three coordinate planes.
Section 5.6 Exercises
In the following exercises, the region occupied by a lamina is shown in a graph. Find the mass of with the density function
In the following exercises, consider a lamina occupying the region and having the density function given in the preceding group of exercises. Use a computer algebra system (CAS) to answer the following questions.
Find the moments and about the and respectively.
Calculate and plot the center of mass of the lamina.
[T] Use a CAS to locate the center of mass on the graph of
Let be the solid bounded by the the cylinder and the plane where is a real number. Find the moment of the solid about the if its density given in cylindrical coordinates is where is a differentiable function with the first and second derivatives continuous and differentiable on
A solid has a volume given by where is the projection of the solid onto the and are real numbers, and its density does not depend on the variable Show that its center of mass lies in the plane
Consider the solid enclosed by the cylinder and the planes and where and are real numbers. The density of is given by where is a differential function whose derivative is continuous on Show that if then the moment of inertia about the of is null.
[T] The average density of a solid is defined as where and are the volume and the mass of respectively. If the density of the unit ball centered at the origin is use a CAS to find its average density. Round your answer to three decimal places.