Chapter 6.6, Problem 52E

(a).

To determine

To Sketch: The region bounded by two curves.

Explanation of Solution

Given:

• The two curves are $y=x^n$ and $y=x^m$ .
• The bounds are $x=0$ and $x=1$
• $m$ and $n$ are integers such that $0\le n<m$

Sketch: The region $R$ is the area enclosed inside the green curve in the sketch below.

  

(b).

To determine

To Compute: The centroid of a region bounded by two curves.

Answer to Problem 52E

The centroid is at $\left(\left[\frac{(n+1)(m+1)}{(n+2)(m+2)}\right],\left[\frac{(n+1)(m+1)}{(2n+1)(2m+1)}\right]\right)$

Explanation of Solution

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

Method: 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 $x=1$ .

Since $0\le n<m$ , $x^n$ is greater than $x^m$ for $0\le x\le 1$ .

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

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

  $M_x=\frac{1}{2}{{\displaystyle \int }}_0^1\left(x^{2n}-x^{2m}\right)dx=\frac{1}{2}{\left(\frac{x^{2n+1}}{2n+1}-\frac{x^{2m+1}}{2m+1})\right]}_0^1=\frac{1}{2}\left(\frac{1}{2n+1}-\frac{1}{2m+1}\right)=\frac{m-n}{(2n+1)(2m+1)}$

  $A=\underset{0}{\overset{1}{{\displaystyle \int }}}\left(x^n-x^m\right)dx={\frac{x^{n+1}}{n+1}-\frac{x^{m+1}}{m+1}]}_0^1=\frac{1}{n+1}-\frac{1}{m+1}=\frac{m-n}{(n+1)(m+1)}$

So,

  $\overline{x}=\frac{M_y}{A}=\frac{(n+1)(m+1)}{(n+2)(m+2)}$ and $\overline{y}=\frac{M_x}{A}=\frac{(n+1)(m+1)}{(2n+1)(2m+1)}$

Therefore, the centroid is at $\left(\left[\frac{(n+1)(m+1)}{(n+2)(m+2)}\right],\left[\frac{(n+1)(m+1)}{(2n+1)(2m+1)}\right]\right)$

(c).

To determine

To Find: The values of $n$ and $m$ such that the centroid of the region in the previous sub-part lies outside the region.

Answer to Problem 52E

Explanation of Solution

Given Information: The curves are $y=x^n$ and $y=x^m$ .

From the previous sub-part, the centroid of the region bounded by the curves is

  $(\overline{x},\overline{y})=\left(\left[\frac{(n+1)(m+1)}{(n+2)(m+2)}\right],\left[\frac{(n+1)(m+1)}{(2n+1)(2m+1)}\right]\right)$ .

Concept Used: If the centroid lies outside the region, then the $y$ coordinate of the greater function at $x=\overline{x}$ must be less than $\overline{y}$ . That is

  ${\overline{x}}^n<\overline{y}$

Computation: Put the values of $\overline{x},\overline{y}$ in the above condition to get

  ${\left[\frac{(n+1)(m+1)}{(n+2)(m+2)}\right]}^n<\frac{(n+1)(m+1)}{(2n+1)(2m+1)}$

Now, put various combinations of $n,m$ in the above inequality to identify appropriate values.

For example, $\left(n,m\right)=\left(2,3\right)$ violates the inequality whereas $\left(n,m\right)=\left(4,5\right)$ satisfies the inequality.