Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.2 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 673: 14

Answer

The area that the curve encloses $$ r=3-2 \cos 4 \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$ is equal to $$ A =\int_{0}^{2\pi} \frac{1}{2} r^{2} d \theta =11 \pi $$
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Work Step by Step

The area that the curve encloses $$ r=3-2 \cos 4 \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$ is equal to $$ \begin{aligned} A &=\int_{0}^{2 \pi} \frac{1}{2} r^{2} d \theta \\ &=\int_{0}^{2 \pi} \frac{1}{2}(3-2 \cos 4 \theta)^{2} d \theta \\ &=\frac{1}{2} \int_{0}^{2 \pi}\left(9-12 \cos 4 \theta+4 \cos ^{2} 4 \theta\right) d \theta \\ &=\frac{1}{2} \int_{0}^{2 \pi}\left[9-12 \cos 4 \theta+4 \cdot \frac{1}{2}(1+\cos 8 \theta)\right] d \theta \\ &=\frac{1}{2} \int_{0}^{2 \pi}(11-12 \cos 4 \theta+2 \cos 8 \theta) d \theta \\ & =\frac{1}{2}\left[11 \theta-3 \sin 4 \theta+\frac{1}{4} \sin 8 \theta\right]_{0}^{2 \pi} \\ &=\frac{1}{2}(22 \pi) \\ &=11 \pi \end{aligned} $$
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