Chapter 8.4, Problem 22E

To determine

To find the length of the asteroid by finding the length of half of the first quadrant portion, and multiply it by 8.

Answer to Problem 22E

The length of asteroid is 2.360

Explanation of Solution

Given information:

The given equation is $y={\left(1-x^{\frac{2}{3}}\right)}^{\frac{3}{2}},\frac{\sqrt{2}}{4}\le x\le 1$

Formula used:

If a smooth curve begins at $\left(a,c\right)$ and ends at $\left(b,d\right)$ , $a<b,c<d$ , then the length of the curve

  $\begin{array}{l}L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}\text{ if }y\text{ is smooth function of }x\text{ on }\left[a,b\right]\\ L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}dy}\text{ if }x\text{ is smooth function of }y\text{ on }\left[c,d\right]\end{array}$

Calculation :

Evaluate $\frac{dy}{dx}$

  $\begin{array}{l}\frac{dy}{dx}=\frac{3}{2}{\left(1-x^{\frac{2}{3}}\right)}^{\frac{1}{2}}\left(-\frac{2}{3}{x}^{-\frac{1}{3}}\right)\\ \frac{dy}{dx}=-x^{-\frac{1}{3}}{\left(1-x^{\frac{2}{3}}\right)}^{\frac{1}{2}}\end{array}$

Which is continuous on $\left[\frac{\sqrt{2}}{4},1\right]$ therefore,

  $\begin{array}{l}L={\displaystyle \underset{-\frac{\pi }{3}}{\overset{0}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}\\ L={\displaystyle \underset{-\frac{\pi }{3}}{\overset{0}{\int }}\sqrt{1+{\left(-x^{-\frac{1}{3}}{\left(1-x^{\frac{2}{3}}\right)}^{\frac{1}{2}}\right)}^2}dx}\\ L={\displaystyle \underset{-\frac{\pi }{3}}{\overset{0}{\int }}\sqrt{1+\frac{{\left(1-x^{\frac{2}{3}}\right)}^2}{x^{\frac{2}{3}}}}dx}\\ L={\displaystyle \underset{-\frac{\pi }{3}}{\overset{0}{\int }}\sqrt{\frac{x^{\frac{5}{3}}-x+\sqrt[3]{x}}{x}}dx}\\ L={\frac{\sqrt[3]{{\left(2x^{\frac{2}{3}}-1\right)}^2+3}\left(2x^{\frac{2}{3}}-1\right)+9\ln \left(\left|2x^{\frac{2}{3}}+\sqrt{{\left(2x^{\frac{2}{3}}-1\right)}^2+3}-1\right|\right)}{16}]}_{-\frac{\pi }{3}}^0\\ L=0.295\left[\text{substitute the upper and lower bound}\right]\end{array}$

Now the length of asteroid is equal to length of half of first quadrant multiplied by 8.

  $\begin{array}{l}\text{length of astroid }=8\times 0.295\\ \text{length of astroid }=2.360\end{array}$

Therefore,

The length of asteroid is 2.360