Moments and Centers of Mass: Learn It 3

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Moments and Centers of Mass: Learn It 3

Center of Mass of a Region Bounded by Two Functions

We can extend our approach to find centroids of more complex regions. Suppose our region is bounded above by the graph of a continuous function $f(x)$ and below by a second continuous function $g(x)$ , as shown in the figure.

This figure is a graph of the first quadrant. It has two curves. They are labeled f(x) and g(x). f(x) is above g(x). In between the curves is a shaded region labeled “R”. The shaded region is bounded to the left by x=a and to the right by x=b. — Figure 7. A region between two functions.

Again, we partition the interval $\left[a,b\right]$ and construct rectangles. A representative rectangle is shown in the following figure.

This figure is a graph of the first quadrant. It has two curves. They are labeled f(x) and g(x). f(x) is above g(x). In between the curves is a shaded rectangle. — Figure 8. A representative rectangle of the region between two functions.

The centroid of each rectangle is:

$({x}_{i}^{*},\frac{f({x}_{i}^{*})+g({x}_{i}^{*})}{2}).$

In the development of the formulas for the mass of the lamina and the moment with respect to the $y$ -axis, the height of each rectangle is $f(x)−g(x)$ . For the $x$ -axis moment, multiply the area by the distance of the centroid from the $x$ -axis.

Summarizing these findings, we arrive at the following theorem.

center of mass of a lamina bounded by two functions

Let $R$ denote a region bounded above by the graph of a continuous function $f(x),$ below by the graph of the continuous function $g(x),$ and on the left and right by the lines $x=a$ and $x=b,$ respectively. Let $\rho$ denote the density of the associated lamina. Then we can make the following statements:

The mass of the lamina is
$m=\rho {\displaystyle\int }_{a}^{b}\left[f(x)-g(x)\right]dx.$
The moments ${M}_{x}$ and ${M}_{y}$ of the lamina with respect to the $x$ – and $y$ -axes, respectively, are
${M}_{x}=\rho {\displaystyle\int }_{a}^{b}\frac{1}{2}({\left[f(x)\right]}^{2}-{\left[g(x)\right]}^{2})dx\text{ and }{M}_{y}=\rho {\displaystyle\int }_{a}^{b}x\left[f(x)-g(x)\right]dx.$
The coordinates of the center of mass $(\overline{x},\overline{y})$ are
$\overline{x}=\frac{{M}_{y}}{m}\text{ and }\overline{y}=\frac{{M}_{x}}{m}.$

Let $R$ be the region bounded above by the graph of the function $f(x)=1-{x}^{2}$ and below by the graph of the function $g(x)=x-1.$ Find the centroid of the region.

The region is depicted in the following figure.

This figure is a graph. It has two curves. They are labeled f(x)=1-x^2 and g(x)=x-1. In between the curves is a shaded region. The shaded region is bounded to the left by x=a and to the right by x=b. — Figure 9. Finding the centroid of a region between two curves.

The graphs of the functions intersect at $(-2,-3)$ and $(1,0),$ so we integrate from $−2$ to $1$ . Once again, for the sake of convenience, assume $\rho =1.$

First, we need to calculate the total mass:

\begin{array}{cc} m & = ρ \int_{a}^{b} [f (x) - g (x)] d x \\ = \int_{- 2}^{1} [1 - x^{2} - (x - 1)] d x = \int_{- 2}^{1} (2 - x^{2} - x) d x \\ = {[2 x - \frac{1}{3} x^{3} - \frac{1}{2} x^{2}] |}_{- 2}^{1} = [2 - \frac{1}{3} - \frac{1}{2}] - [- 4 + \frac{8}{3} - 2] = \frac{9}{2} . \end{array}

$\begin{array}{cc}\hfill m& =\rho {\displaystyle\int }_{a}^{b}\left[f(x)-g(x)\right]dx\hfill \\ & ={\displaystyle\int }_{-2}^{1}\left[1-{x}^{2}-(x-1)\right]dx={\displaystyle\int }_{-2}^{1}(2-{x}^{2}-x)dx\hfill \\ & ={\left[2x-\frac{1}{3}{x}^{3}-\frac{1}{2}{x}^{2}\right]|}_{-2}^{1}=\left[2-\frac{1}{3}-\frac{1}{2}\right]-\left[-4+\frac{8}{3}-2\right]=\frac{9}{2}.\hfill \end{array}$

Next, we compute the moments:

\begin{array}{cc} M_{x} & = ρ \int_{a}^{b} \frac{1}{2} ({[f (x)]}^{2} - {[g (x)]}^{2}) d x \\ = \frac{1}{2} \int_{- 2}^{1} ({(1 - x^{2})}^{2} - {(x - 1)}^{2}) d x = \frac{1}{2} \int_{- 2}^{1} (x^{4} - 3 x^{2} + 2 x) d x \\ = \frac{1}{2} {[\frac{x^{5}}{5} - x^{3} + x^{2}] |}_{- 2}^{1} = - \frac{27}{10} \end{array}

$\begin{array}{cc}\hfill {M}_{x}& =\rho {\displaystyle\int }_{a}^{b}\frac{1}{2}({\left[f(x)\right]}^{2}-{\left[g(x)\right]}^{2})dx\hfill \\ & =\frac{1}{2}{\displaystyle\int }_{-2}^{1}({(1-{x}^{2})}^{2}-{(x-1)}^{2})dx=\frac{1}{2}{\displaystyle\int }_{-2}^{1}({x}^{4}-3{x}^{2}+2x)dx\hfill \\ & =\frac{1}{2}{\left[\frac{{x}^{5}}{5}-{x}^{3}+{x}^{2}\right]|}_{-2}^{1}=-\frac{27}{10}\hfill \end{array}$

and

\begin{array}{cc} M_{y} & = ρ \int_{a}^{b} x [f (x) - g (x)] d x \\ = \int_{- 2}^{1} x [(1 - x^{2}) - (x - 1)] d x = \int_{- 2}^{1} x [2 - x^{2} - x] d x = \int_{- 2}^{1} (2 x - x^{4} - x^{2}) d x \\ = {[x^{2} - \frac{x^{5}}{5} - \frac{x^{3}}{3}] |}_{- 2}^{1} = - \frac{9}{4} . \end{array}

$\begin{array}{cc}\hfill {M}_{y}& =\rho {\displaystyle\int }_{a}^{b}x\left[f(x)-g(x)\right]dx\hfill \\ & ={\displaystyle\int }_{-2}^{1}x\left[(1-{x}^{2})-(x-1)\right]dx={\displaystyle\int }_{-2}^{1}x\left[2-{x}^{2}-x\right]dx={\displaystyle\int }_{-2}^{1}(2x-{x}^{4}-{x}^{2})dx\hfill \\ & ={\left[{x}^{2}-\frac{{x}^{5}}{5}-\frac{{x}^{3}}{3}\right]|}_{-2}^{1}=-\frac{9}{4}.\hfill \end{array}$

Therefore, we have

\bar{x} = \frac{M_{y}}{m} = - \frac{9}{4} \cdot \frac{2}{9} = - \frac{1}{2} and \bar{y} = \frac{M_{x}}{y} = - \frac{27}{10} \cdot \frac{2}{9} = - \frac{3}{5} .

$\overline{x}=\frac{{M}_{y}}{m}=-\frac{9}{4}·\frac{2}{9}=-\frac{1}{2}\text{ and }\overline{y}=\frac{{M}_{x}}{y}=-\frac{27}{10}·\frac{2}{9}=-\frac{3}{5}.$

The centroid of the region is $(\text{−}(1\text{/}2),\text{−}(3\text{/}5)).$