Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - Chapter Review Exercises - Page 638: 33


The area of one petal: $area = \frac{\pi }{{16}}$

Work Step by Step

In Exercise 9 of Section 12.4, we see that the "rose" traces out a complete curve for the interval $0 \le \theta \le 2\pi $. From the figure attached we see that it traces out two petals for the interval $0 \le \theta \le \frac{\pi }{2}$. So, by symmetry it traces out one petal for the interval $0 \le \theta \le \frac{\pi }{4}$. Using Eq. (2) of Theorem 1 of Section 12.4, the area of one petal is $area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /4} {\sin ^2}\left( {4\theta } \right){\rm{d}}\theta $ Since ${\sin ^2}\left( {4\theta } \right) = \frac{1}{2}\left( {1 - \cos \left( {8\theta } \right)} \right)$, the integral becomes $area = \frac{1}{4}\cdot\mathop \smallint \limits_0^{\pi /4} \left( {1 - \cos \left( {8\theta } \right)} \right){\rm{d}}\theta $ $area = \frac{1}{4}\left( {\theta - \frac{1}{8}\sin \left( {8\theta } \right)} \right)|_0^{\pi /4}$ $area = \frac{1}{4}\left( {\frac{\pi }{4}} \right) = \frac{\pi }{{16}}$
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