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: 18


$$4 \space \pi \sqrt 2$$

Work Step by Step

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