We’ve already used multiple integrals to find areas, volumes, and average values of functions over bounded regions. Now we’ll extend these techniques to find the center of mass and moments of inertia for physical objects. You’ll use double integrals for laminas (flat plates) and triple integrals for three-dimensional objects with variable density. When an object has uniform density throughout, we call it homogeneous. For homogeneous objects, the density is constant.
The center of mass represents the balance point of an object. In a uniform gravitational field, this is also called the center of gravity. For objects with uniform density, the center of mass is simply the geometric center, which we call the centroid.
Figure 1 shows a lamina perfectly balanced at its center of mass [latex]P[/latex].
Figure 1. A 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 [latex]P(\overline{x}, \overline{y})[/latex] of a lamina, you need three quantities:
The moment [latex]M_x[/latex] about the [latex]x[/latex]-axis
The moment [latex]M_y[/latex] about the [latex]y[/latex]-axis
For a one-dimensional object like a thin rod with density [latex]\rho(x)[/latex] along the [latex]x[/latex]-axis from [latex]x = a[/latex] to [latex]x = b[/latex]:
Center of mass: [latex]\overline{x} = \frac{1}{m}\displaystyle\int_a^b x\rho(x)dx[/latex]
We’ll extend this same idea to two dimensions using double integrals.
When the density function is constant, these formulas give the centroid of the lamina—the geometric center independent of mass distribution.
Setting Up the Mass Calculation
Suppose that the lamina occupies a region [latex]R[/latex] in the [latex]xy[/latex]-plane, with density [latex]{\rho}{({{x},{y}})}[/latex] (measured in mass per unit area) at any point [latex](x, y)[/latex].
where [latex]{\Delta}{m}[/latex] and [latex]{\Delta}{A}[/latex] are the mass and area of a small rectangle containing the point [latex](x, y)[/latex], and the limit is taken as the rectangle’s dimensions approach zero (Figure 2).
Figure 2. The 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.
To find the total mass, we partition region [latex]R[/latex] into tiny rectangles [latex]R_{ij}[/latex], each with area [latex]\Delta A[/latex]. We choose sample points [latex]{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}[/latex] in each rectangle. The mass [latex]m_{ij}[/latex] of rectangle [latex]R_{ij}[/latex] is approximately:
Let [latex]k[/latex] and [latex]l[/latex] be the number of subintervals in [latex]x[/latex] and [latex]y[/latex] respectively (Figure 3). Note that while we use rectangles for convenience, this limiting process works for any shape, as you’ve seen in previous sections.
Figure 3. Subdividing the lamina into tiny rectangles [latex]R_{ij}[/latex], each containing a sample point [latex](x_{ij}^{\ast},y_{ij}^{\ast})[/latex].
Taking the limit as the partition becomes infinitely fine gives us the total mass:
For a lamina occupying region [latex]R[/latex] with density function [latex]\rho(x,y)[/latex], the total mass is:
[latex]m = \iint_R \rho(x,y) dA[/latex]
Let’s see how this works with a concrete example.
Consider a triangular lamina [latex]R[/latex] with vertices [latex](0, 0)[/latex], [latex](0, 3)[/latex], [latex](3, 0)[/latex] and with density [latex]{\rho}{({x},{y})} = {{x}{y}} \ {\text{kg/m}^2}[/latex]. Find the total mass.
A sketch of the region [latex]R[/latex] is always helpful, as shown in the following figure.
Figure 4. A lamina in the [latex]xy[/latex]-plane with density [latex]\rho(x,y)=xy[/latex].
Using the expression developed for mass, we see that
The computation is straightforward, giving the answer [latex]{m} = {\frac{27}{8}}[/latex] kg.
Calculating Moments
Now that we have the mass formula, we can develop expressions for moments. The moment [latex]M_x[/latex] about the [latex]x[/latex]-axis for region [latex]R[/latex] is the limit of the sums of moments of the rectangles [latex]R_{ij}[/latex] about the [latex]x[/latex]-axis:
For a lamina occupying region [latex]R[/latex] with density function [latex]\rho(x,y)[/latex]:
Moment about the [latex]x[/latex]-axis: [latex]M_x = \iint_R y\rho(x,y) dA[/latex]
Moment about the [latex]y[/latex]-axis: [latex]M_y = \iint_R x\rho(x,y) dA[/latex]
Consider the same triangular lamina [latex]R[/latex] with vertices [latex](0, 0)[/latex], [latex](0, 3)[/latex], [latex](3, 0)[/latex] and with density [latex]{\rho}{({x},{y})} = {xy}[/latex]. Find the moments [latex]M_x[/latex] and [latex]M_y[/latex].
Use double integrals for each moment and compute their values:
We can now express the center of mass coordinates in terms of integrals. We denote the [latex]x[/latex]-coordinate by [latex]\overline{x}[/latex] and the [latex]y[/latex]-coordinate by [latex]\overline{y}[/latex]:
For a lamina occupying region [latex]R[/latex] with density [latex]\rho(x,y)[/latex], the center of mass is at [latex](\overline{x}, \overline{y})[/latex] where:
Again consider the same triangular region [latex]R[/latex] with vertices [latex](0, 0)[/latex], [latex](0, 3)[/latex], [latex](3, 0)[/latex] and with density function [latex]{\rho}{({x},{y})} = {xy}[/latex]. Find the center of mass.
Therefore, the center of mass is the point [latex]\left(\frac65,\frac65\right)[/latex].
Analysis
If we choose the density [latex]{\rho}{({x},{y})}[/latex] 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 [latex]{\left ({\frac{6}{5}},{\frac{6}{5}} \right )}[/latex] is not exactly the same as the centroid [latex](1, 1)[/latex] of the triangular region. This is due to the variable density of [latex]R[/latex]. If the density is constant, then we just use [latex]{\rho}{({x},{y})} = {c}[/latex] (constant). This value cancels out from the formulas, so for a constant density, the center of mass coincides with the centroid of the lamina.
Centroid Formulas
Based on the observation in the previous example, when density is constant (or we’re simply finding the geometric center), we have simplified formulas for the centroid of a region:
For a region [latex]R[/latex] in the plane, the centroid (geometric center) is at [latex](x_c, y_c)[/latex] where:
[latex]x_c = \dfrac{\iint_R x , dA}{\iint_R dA} \quad \text{and} \quad y_c = \dfrac{\iint_R y , dA}{\iint_R dA}[/latex]
Note: [latex]\iint_R dA[/latex] gives the area of region [latex]R[/latex].
Center of Mass vs. Centroid
Centroid: Geometric center of a region (constant density)
Center of Mass: Balance point accounting for variable density
When density is uniform, these two points coincide!
Find the mass, moments, and the center of mass of the lamina of density [latex]{\rho}{({x},{y})} = {x} + {y}[/latex] occupying the region [latex]R[/latex] under the curve [latex]y=x^{2}[/latex] in the interval [latex]{0} \leq {x} \leq {2}[/latex] (see the following figure).
Figure 5. Locating the center of mass of a lamina [latex]R[/latex] with density [latex]\rho(x,y)=x+y[/latex].
First we compute the mass [latex]m[/latex]. We need to describe the region between the graph of [latex]y=x^{2}[/latex] and the vertical lines [latex]x=0[/latex] and [latex]x=2[/latex]:
Hence the center of mass is [latex]{({\overline{x}},{\overline{y}})} = {\left ({\frac{44}{27}},{\frac{100}{63}} \right )}.[/latex]
Find the centroid of the region under the curve [latex]{y} = {{e}^{x}}[/latex] over the interval [latex]{1} \leq {x} \leq {3}[/latex] (see the following figure).
Figure 6. Finding a centroid of a region below the curve [latex]y=e^{x}[/latex]
To compute the centroid, we assume that the density function is constant and hence it cancels out:
[latex]\hspace{4cm}\begin{align} x_c&=\frac{M_y}m=\dfrac{\underset{R}{\displaystyle\iint}x \ dA}{\underset{R}{\displaystyle\iint}dA}\text{ and }y_c=\frac{M_x}m=\dfrac{\underset{R}{\displaystyle\iint}y \ dA}{\underset{R}{\displaystyle\iint}dA}, \\ x_c&=\frac{M_y}m=\dfrac{\underset{R}{\displaystyle\iint}x \ dA}{\underset{R}{\displaystyle\iint}dA}=\dfrac{\displaystyle\int^{x=3}_{x=1}\displaystyle\int^{y=e^x}_{y=0}x \ dy \ dx}{\displaystyle\int^{x=3}_{x=1}\displaystyle\int^{y=e^x}_{y=0}dy \ dx}=\dfrac{\displaystyle\int^{x=3}_{x=1}xe^x{dx}}{\displaystyle\int^{x=3}_{x=1}e^x{dx}}=\frac{2e^3}{e^3-e}=\frac{2e^2}{e^2-1}, \\ y_c&=\frac{M_x}m=\dfrac{\underset{R}{\displaystyle\iint}y \ dA}{\underset{R}{\displaystyle\iint}dA}=\dfrac{\displaystyle\int_{x=1}^{x=3}\displaystyle\int_{y=0}^{y=e^x}y \ dy \ dx}{\displaystyle\int_{x=1}^{x=3}\displaystyle\int_{y=0}^{y=e^x}dy \ dx}=\dfrac{\displaystyle\int_{x=1}^{x=3}\frac{e^{2x}}2dx}{\displaystyle\int_{x=1}^{x=3}e^xdx}=\frac{\frac14e^2(e^4-1)}{e(e^2-1)}=\frac14e(e^2+1). \end{align}[/latex]
