Moments and Centers of Mass: Learn It 4

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Moments and Centers of Mass: Learn It 4

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.

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.

The region is depicted in the following figure.

This figure is a graph of the function f(x)=4-x^2. It is an upside-down parabola. The region under the parabola above the x-axis is shaded. The curve intersects the x-axis at x=-2 and x=2. — Figure 10. We can use the symmetry principle to help find the centroid of a symmetric region.

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

First, we calculate the total mass:

\begin{matrix} m & = ρ \int_{a}^{b} f (x) d x = \int_{- 2}^{2} (4 - x^{2}) d x = {[4 x - \frac{x^{3}}{3}] |}_{- 2}^{2} = \frac{32}{3} . \end{matrix}

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

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

\begin{matrix} M_{x} & = ρ \int_{a}^{b} \frac{{[f (x)]}^{2}}{2} d x = \frac{1}{2} \int_{- 2}^{2} {[4 - x^{2}]}^{2} d x = \frac{1}{2} \int_{- 2}^{2} (16 - 8 x^{2} + x^{4}) d x = \frac{1}{2} {[\frac{x^{5}}{5} - \frac{8 x^{3}}{3} + 16 x] |}_{- 2}^{2} = \frac{256}{15} . \end{matrix}

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

Then we have

¯ ¯ ¯ y = \frac{M_{x}}{y} = \frac{256}{15} \cdot \frac{3}{32} = \frac{8}{5} .

$\overline{y}=\frac{{M}_{x}}{y}=\frac{256}{15}·\frac{3}{32}=\frac{8}{5}.$

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

Watch the following video to see the worked solution to this example.

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You can view the transcript for this segmented clip of “2.6 Moments and Centers of Mass” here (opens in new window).

Theorem of Pappus

The theorem of Pappus for volume 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.

Theorem of Pappus for volume

Let $R$ be a region in the plane and let $l$ be a line in the plane that does not intersect $R$ . Then the volume of the solid of revolution formed by revolving $R$ around $l$ is equal to the area of $R$ multiplied by the distance $d$ traveled by the centroid of $R$ .

Proof

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 $\left[a,b\right],$ and for which the axis of revolution is the $y$ -axis. In this case, the area of the region is $A={\displaystyle\int }_{a}^{b}\left[f(x)-g(x)\right]dx.$ 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, $\overline{x},$ which is

¯ ¯ ¯ x = \frac{M_{y}}{m},

$\overline{x}=\frac{{M}_{y}}{m},$

where

m = ρ \int_{a}^{b} [f (x) - g (x)] d x and M_{y} = ρ \int_{a}^{b} x [f (x) - g (x)] d x .

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

Then,

d = 2 π \frac{ρ \int_{a}^{b} x [f (x) - g (x)] d x}{ρ \int_{a}^{b} [f (x) - g (x)] d x}

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

and thus

d \cdot A = 2 π \int_{a}^{b} x [f (x) - g (x)] d x .

$d·A=2\pi {\displaystyle\int }_{a}^{b}x\left[f(x)-g(x)\right]dx.$

However, using the method of cylindrical shells, we have

V = 2 π \int_{a}^{b} x [f (x) - g (x)] d x .

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

So,

V = d \cdot A

$V=d·A$

and the proof is complete.

$_\blacksquare$

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.

The region and torus are depicted in the following figure.

This figure has two graphs. The first is the x y coordinate system with a circle centered on the x-axis at x=4. The radius is 2. The second figure is the x y coordinate system. The circle from the first image has been revolved about the y-axis to form a torus. — Figure 13. Determining the volume of a torus by using the theorem of Pappus. (a) A circular region R in the plane; (b) the torus generated by revolving R about the y-axis.

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 $A·d=32{\pi }^{2}$ units³.