Chapter 7, Problem 28RE

To determine

To find:The perimeter of the bow-tie-shaped region.

Answer to Problem 28RE

The perimeter of the bow-tie-shaped region enclosed between the graphs is $5.245$ .

Explanation of Solution

Given information:

  $\begin{array}{l}y=x^3-x\\ y=x-x^3\end{array}$

Calculation:

Let's graph the given area and find the point of intersection.

  

To find the point of intersection from the given equations,

  $\begin{array}{c}{x}^3-x=x-x^3\\ 2x^3-2x=0\\ 2x\left(x^2-1\right)=0\\ 2x\left(x-1\right)\left(x+1\right)=0\\ x=\pm 1,0\end{array}$

The formula for length of the continuous curve is given by,

  $L={\displaystyle {\int }_a^b\sqrt{1+{\left(y\text{'}\right)}^2}dx}$

Now finding the derivative,

  $\begin{array}{l}y=x^3-x\\ y\text{'}=3x^2-1\end{array}$

We can limit the integration to a single region segment because both the equations are even ( $y$ -axis symmetry) and $x^3-x=x-x^3\in \left[0,1\right]$ ( $x$ -axis symmetry).

  $4{\displaystyle {\int }_0^1\sqrt{1+{\left(3x^2-1\right)}^2}dx\approx 5.245}$

Therefore, the perimeter of the bow-tie-shaped region enclosed between the graphs is $5.245$ .