Chapter 6.1, Problem 36E

To determine

To graph the astroid enclosed by the curve $x=a{\cos }^3\theta {\text{ and y=asin}}^3\theta$ and set up an integral for the area that it encloses.

Answer to Problem 36E

The area enclosed by the curve $x=a{\cos }^3\theta {\text{ and y=asin}}^3\theta$is $7.536a^{2}sq.units.$

Explanation of Solution

Given: $x=a{\cos }^3\theta {\text{ and y=asin}}^3\theta$

Calculation:

  

  $\begin{array}{l}Area=2\pi {\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}y\sqrt{{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dx}{d\theta }\right)}^2}}d\theta \\ \text{}\text{ where, }\frac{dx}{d\theta }=a\mathrm{.3.}{\cos }^2\theta \text{(-sin}\theta \text{) and }\frac{dy}{d\theta }=a\mathrm{.3.}{\sin }^2\theta \text{(cos}\theta \text{)}\\ {\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dx}{d\theta }\right)}^2{\text{= 9a}}^2{\sin }^2\theta {\text{ cos}}^4\theta {\text{ + 9a}}^2{\sin }^4\theta {\text{ cos}}^2\theta {\text{ = 9a}}^2{\sin }^2\theta {\text{ cos}}^2\theta \\ \text{Area = 2}\pi {\displaystyle \underset{0}{\overset{\pi }{\int }}\text{a}{\sin }^3\theta }.\sqrt{{\text{9a}}^2{\sin }^2\theta {\text{ cos}}^4\theta }d\theta \\ \text{ =2}\pi {\displaystyle \underset{0}{\overset{\pi /2}{\int }}\text{a}{\sin }^3\theta }.3a\sin \theta \cos \theta d\theta \text{ -2}\pi {\displaystyle \underset{\pi /2}{\overset{\pi }{\int }}\text{a}{\sin }^3\theta }.3a\sin \theta \cos \theta d\theta \\ {\text{ =6a}}^2\pi {\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\sin }^4\theta }\text{.cos}\theta \text{.d}\theta {\text{-6a}}^2\pi {\displaystyle \underset{\pi /2}{\overset{\pi }{\int }}{\sin }^4\theta }\text{.cos}\theta \text{.d}\theta \\ {\text{ =6a}}^2\pi {\left[\frac{{\sin }^5\theta }{5}\right]}_0^{\pi /2}-{\text{6a}}^2\pi {\left[\frac{{\sin }^5\theta }{5}\right]}_{\pi /2}^{\pi }={\text{6a}}^2\pi (1/5)-{\text{6a}}^2\pi (-1/5)\\ {\text{Area =6a}}^2\pi (2/5)=12\pi {\text{a}}^2/5=7.536{\text{a}}^{2}sq.units\end{array}$