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: 34


Total area of the flower does not depend on $n$: $area = \frac{\pi }{2}$

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 $. Using Eq. (2) of Theorem 1 of Section 12.4, the total area is $area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{2\pi } {\sin ^2}\left( {n\theta } \right){\rm{d}}\theta $ Since ${\sin ^2}\left( {n\theta } \right) = \frac{1}{2}\left( {1 - \cos \left( {2n\theta } \right)} \right)$, the integral becomes $area = \frac{1}{4}\cdot\mathop \smallint \limits_0^{2\pi } \left( {1 - \cos \left( {2n\theta } \right)} \right){\rm{d}}\theta $ $area = \frac{1}{4}\left( {\theta - \frac{1}{{2n}}\sin \left( {2n\theta } \right)} \right)|_0^{2\pi }$ $area = \frac{\pi }{2}$ Hence, the total area of the flower does not depend on $n$.
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