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# 7.6: Calculating Centers of Mass and Moments of Inertia

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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 $$\PageIndex{1}$$ shows a point $$P$$ as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass.

To find the coordinates of the center of mass $$P(\bar{x},\bar{y})$$ of a lamina, we need to find the moment $$M_x$$ of the lamina about the $$x$$-axis and the moment $$M_y$$ about the $$y$$-axis. We also need to find the mass $$m$$ of the lamina. Then

$\bar{x} = \dfrac{M_y}{m}$

and

$\bar{y} = \dfrac{M_x}{m}.$

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 $$\bar{x} = \dfrac{M_y}{m}$$ and $$\bar{y} = \dfrac{M_x}{m}$$ give the centroid of the lamina.

Suppose that the lamina occupies a region $$R$$ in the $$xy$$-plane and let $$\rho (x,y)$$ be its density (in units of mass per unit area) at any point $$(x,y)$$. Hence,

$\rho(x,y) = \lim_{\Delta A \rightarrow 0} \dfrac{\Delta m}{\Delta A}$

where $$\Delta m$$ and $$\Delta A$$ are the mass and area of a small rectangle containing the point $$(x,y)$$ and the limit is taken as the dimensions of the rectangle go to $$0$$ (see the following figure).

Just as before, we divide the region $$R$$ into tiny rectangles $$R_{ij}$$ with area $$\Delta A$$ and choose $$(x_{ij}^*, y_{ij}^*)$$ as sample points. Then the mass $$m_{ij}$$ of each $$R_{ij}$$ is equal to $$\rho (x_{ij}^*, y_{ij}^*) \Delta A$$ (Figure $$\PageIndex{2}$$). Let $$k$$ and $$l$$ be the number of subintervals in $$x$$ and $$y$$ respectively. Also, note that the shape might not always be rectangular but the limit works anyway, as seen in previous sections.

Hence, the mass of the lamina is

$m =\lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l \rho(x_{ij}^*,y_{ij}^*) \Delta A = \iint_R \rho(x,y) dA.$

Let’s see an example now of finding the total mass of a triangular lamina.

Example $$\PageIndex{1}$$: Finding the Total Mass of a Lamina

Consider a triangular lamina $$R$$ with vertices $$(0,0), \, (0,3), \, (3,0)$$ and with density $$\rho(x,y) = xy \, kg/m^2$$. Find the total mass.

Solution

A sketch of the region $$R$$ is always helpful, as shown in the following figure.

Using the expression developed for mass, we see that

$m = \iint_R \, dm = \iint_R \rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} xy \, dy \, dx = \int_{x=0}^{x=3} \left[ \left. x \dfrac{y^2}{2} \right|_{y=0}^{y=3} \right] \, dx = \int_{x=0}^{x=3} \dfrac{1}{2} x (3 - x)^2 dx = \left.\left[ \dfrac{9x^2}{4} - x^3 + \dfrac{x^4}{8} \right]\right|_{x=0}^{x=3} = \dfrac{27}{8}.$

The computation is straightforward, giving the answer $$m = \dfrac{27}{8} \, kg$$.

Exercise $$\PageIndex{1}$$

Consider the same region $$R$$ as in the previous example, and use the density function $$\rho (x,y) = \sqrt{xy}$$. Find the total mass.

$$\dfrac{9\pi}{8} \, kg$$

Now that we have established the expression for mass, we have the tools we need for calculating moments and centers of mass. The moment $$M_z$$ about the $$x$$-axis for $$R$$ is the limit of the sums of moments of the regions $$R_{ij}$$ about the $$x$$-axis. Hence

$M_x = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^*)m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^*) \rho(x_{ij}^*,y_{ij}^*) \Delta A = \iint_R y\rho (x,y) dA$

Similarly, the moment $$M_y$$ about the $$y$$-axis for $$R$$ is the limit of the sums of moments of the regions $$R_{ij}$$ about the $$y$$-axis. Hence

$M_x = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (x_{ij}^*)m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^*) \rho(x_{ij}^*,y_{ij}^*) \Delta A = \iint_R x\rho (x,y) dA$

Example $$\PageIndex{2}$$: Finding Moments

Consider the same triangular lamina $$R$$ with vertices $$(0,0), \, (0,3), \, (3,0)$$ and with density $$\rho (x,y) = xy$$. Find the moments $$M_x$$ and $$M_y$$.

Solution

Use double integrals for each moment and compute their values:

$M_x = \iint_R y\rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} x y^2 \, dy \, dx = \dfrac{81}{20},$

$M_y = \iint_R x\rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} x^2 y \, dy \, dx = \dfrac{81}{20},$

The computation is quite straightforward.

Exercise $$\PageIndex{2}$$

Consider the same lamina $$R$$ as above and use the density function $$\rho (x,y) = \sqrt{xy}$$. Find the moments $$M_x$$ and $$M_y$$.

$$M_x = \dfrac{81\pi}{64}$$ and $$M_y = \dfrac{81\pi}{64}$$

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 $$\bar{x}$$ and the y-coordinate by $$\bar{y}$$. Specifically,

$\bar{x} = \dfrac{M_y}{m} = \dfrac{\iint_R x\rho (x,y) dA}{\iint_R \rho (x,y)dA}$

and

$\bar{y} = \dfrac{M_x}{m} = \dfrac{\iint_R y\rho (x,y) dA}{\iint_R \rho (x,y)dA}$

Example $$\PageIndex{3}$$: center of mass

Again consider the same triangular region $$R$$ with vertices $$(0,0), \, (0,3), \, (3,0)$$ and with density function $$\rho (x,y) = xy$$. Find the center of mass.

Solution

Using the formulas we developed, we have

$\bar{x} = \dfrac{M_y}{m} = \dfrac{\iint_R x\rho (x,y) dA}{\iint_R \rho (x,y)dA} = \dfrac{81/20}{27/8} = \dfrac{6}{5},$

$\bar{y} = \dfrac{M_x}{m} = \dfrac{\iint_R y\rho (x,y) dA}{\iint_R \rho (x,y)dA} = \dfrac{81/20}{27/8} = \dfrac{6}{5}.$

Therefore, the center of mass is the point $$\left(\dfrac{6}{5},\dfrac{6}{5}\right).$$

Analysis

If we choose the density $$\rho(x,y)$$ 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,

$x_c = \dfrac{M_y}{m} = \dfrac{\iint_R x \, dA}{\iint_R dA} = \dfrac{9/2}{9/2} = 1,$

$y_c = \dfrac{M_x}{m} = \dfrac{\iint_R y \, dA}{\iint_R dA} = \dfrac{9/2}{9/2} = 1.$

Notice that the center of mass $$\left(\dfrac{6}{5},\dfrac{6}{5}\right)$$ is not exactly the same as the centroid $$(1,1)$$ of the triangular region. This is due to the variable density of $$R$$ If the density is constant, then we just use $$\rho(x,y) = c$$ (constant). This value cancels out from the formulas, so for a constant density, the center of mass coincides with the centroid of the lamina.

Exercise $$\PageIndex{3}$$

Again use the same region $$R$$ as above and use the density function $$\rho (x,y) = \sqrt{xy}$$. Find the center of mass.

$$\bar{x} = \dfrac{M_y}{m} = \dfrac{81\pi/64}{9\pi/8} = \dfrac{9}{8}$$ and $$\bar{y} = \dfrac{M_x}{m} = \dfrac{81\pi}{9\pi/8} = \dfrac{0}{8}$$.
Once again, based on the comments at the end of Example $$\PageIndex{3}$$, we have expressions for the centroid of a region on the plane:
$x_c = \dfrac{M_y}{m} = \dfrac{\iint_R x \, dA}{\iint_R dA} \, \text{and} \, y_c = \dfrac{M_x}{m} = \dfrac{\iint_R y \, dA}{\iint_R dA}.$