Chapter 6.6, Problem 51E

(a).

To determine

The coordinates of centroid of a region bounded by two curves.

Explanation of Solution

Given Information:

• The two curves are $y=f\left(x\right)$ and $y=g\left(x\right)$ .
• The interval on the $x$ -axis is $a\le x\le b$ .
• The function $f$ is always greater than the function $g$ for all values of $x$ in the interval.

Concept Used:

The coordinates of center of mass of a discrete mass system of $n$ particles are

  $\left(\overline{x},\overline{y}\right)=\left(\frac{M_y}{m},\frac{M_x}{m}\right)$

where

  $M_y={\displaystyle \sum }_{i=1}^n{x}_i{m}_i$ , $M_x={\displaystyle \sum }_{i=1}^n{y}_i{m}_i$ and $m={\displaystyle \sum }_{i=1}^n{m}_i$

The quantities $M_x$ and $M_y$ are called the moments of the system about the respective axes.

Divide the area between the two curves into several vertical rectangles of width $\text{Δ}x$ .

The height of each such rectangle at any value of $x$ is $f\left(x\right)-g\left(x\right)$ .

The centroid of such a rectangle is located at half the height.

So, its center of mass is at $\left(x,g\left(x\right)+\frac{f\left(x\right)-g\left(x\right)}{2}\right)=\left(x,\frac{f\left(x\right)+g\left(x\right)}{2}\right)$ .

Now, the moments $M_y,M_x$ are the sum of the moments of all the individual rectangles about the respective axes.

So,

  $M_y={\displaystyle \sum }_a^{b}x\cdot \rho \left(f\left(x\right)-g\left(x\right)\right)\Delta x=\rho \underset{a}{\overset{b}{{\displaystyle \int }}}x\left[f\left(x\right)-g\left(x\right)\right]dx$

  $M_x={{\displaystyle \sum }}_a^b\frac{\left[f\left(x\right)+g\left(x\right)\right]}{2}\cdot \rho \left[f\left(x\right)-g\left(x\right)\right]\Delta x=\rho {{\displaystyle \int }}_a^b\frac{1}{2}\left[f{(x)}^2-g{(x)}^2\right]dx$

  $A\left(orm\right)={\displaystyle \sum }_a^b\rho \left[f\left(x\right)-g\left(x\right)\right]\times \Delta x=\rho \underset{a}{\overset{b}{{\displaystyle \int }}}\left[f\left(x\right)-g\left(x\right)\right]dx$

Therefore

  $\left(\overline{x},\overline{y}\right)=\left(\frac{{{\displaystyle \int }}_a^{b}x\left[f\left(x\right)-g\left(x\right)\right]dx}{{{\displaystyle \int }}_a^b\left[f\left(x\right)-g\left(x\right)\right]dx},\frac{{{\displaystyle \int }}_a^b\frac{1}{2}\left(\left[f{\left(x\right)}^2-g{\left(x\right)}^2\right]\right)dx}{{{\displaystyle \int }}_a^b\left[f\left(x\right)-g\left(x\right)\right]dx}\right)$

(b).

To determine

To Calculate: The location of centroid the region bounded between two given curves.

Answer to Problem 51E

The centroid of the region is at $\left(\frac{1}{2},\frac{2}{5}\right)$

Explanation of Solution

Given Information: The two curves are $y=x$ and $y=x^2$ .

Method: From the previous sub-part, the location of coordinates of center of mass between two curves $f\left(x\right),g\left(x\right)$ where $f\left(x\right)\le g\left(x\right)$ for $a\le x\le b$ is

  $\left(\overline{x},\overline{y}\right)=\left(\frac{{{\displaystyle \int }}_a^{b}x\left[f\left(x\right)-g\left(x\right)\right]dx}{{{\displaystyle \int }}_a^b\left[f\left(x\right)-g\left(x\right)\right]dx},\frac{{{\displaystyle \int }}_a^b\frac{1}{2}\left(\left[f{\left(x\right)}^2-g{\left(x\right)}^2\right]\right)dx}{{{\displaystyle \int }}_a^b\left[f\left(x\right)-g\left(x\right)\right]dx}\right)$

Calculation: The two given curves intersect at $x=0$ and $x=1$ .

Also, $x\ge x^2$ when $0\le x\le 1$ .

So, put $f\left(x\right)=x$ , $g\left(x\right)=x^2$ , $a=0$ and $b=1$ in the above formula to get

  $M_y=\underset{0}{\overset{1}{{\displaystyle \int }}}x\cdot \left(x-x^2\right)dx={\frac{x^3}{3}-\frac{x^4}{4}]}_0^1=\frac{1}{12}$

  $M_x=\frac{1}{2}{{\displaystyle \int }}_0^1\left(x^2-x^4\right)dx=\frac{1}{2}{\left(\frac{x^3}{3}-\frac{x^5}{5})\right]}_0^1=\frac{1}{15}$

  $A=\underset{0}{\overset{1}{{\displaystyle \int }}}\left(x-x^2\right)dx={\frac{x^2}{2}-\frac{x^3}{3}]}_0^1=\frac{1}{6}$

Therefore, the centroid is at $\left(\frac{\frac{1}{12}}{\frac{1}{6}},\frac{\frac{1}{15}}{\frac{1}{6}}\right)=\left(\frac{1}{2},\frac{2}{5}\right)$