Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Section 16.5 - Surfaces and Area - Exercises 16.5 - Page 989: 17


$\dfrac{\pi \sqrt 5}{2}$

Work Step by Step

We have $ r_r= \cos \theta \space i+\sin \theta \space j-\dfrac{ \sin \theta}{2} \space k ;\\r_{\theta}=-r \sin \theta \space i+r \cos \theta j-\dfrac{r \cos \theta}{2} k $ Also, $|r_r \times r_{\theta}|=\dfrac{\sqrt 5r}{2}$ and $0 \le r \le 1$; $0 \le \theta \le 2\pi $ Now, $ Area=\int_0^{2 \pi} \int_0^{1} \dfrac{\sqrt 5r}{2} \space dr \space d \theta \\= \int_0^{2 \pi} [ \dfrac{\sqrt 5r^2}{4}]_0^{1} \space d \theta \\=\dfrac{\pi \sqrt 5}{2}$
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