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 990: 45


$$\dfrac{\pi}{6}(17\sqrt {17}-5\sqrt {5}) $$

Work Step by Step

Apply cylindrical coordinates. $ x=r \cos \theta ;\\ y= r \sin \theta ;\\ z \gt 0$ We know that $ r(r, \theta)=xi+yj+zk $ or, $ r^2=x^2+y^2+z^2$ Now, $ r_r= \cos \theta \space i+\sin \theta \space j+2 \space k ; \\ r_{\theta}=-r\sin \theta \space i+ r\cos \theta \space j $ Also, $|r_r \times r_{\theta}|=r\sqrt {4r^2+1}$ $$ Area =\int_0^{2 \pi} \int_1^{2} (r\sqrt {4r^2+1}) \space dr \space d \theta \\=\int_0^{2 \pi} (\dfrac{17\sqrt {17}}{12}- \dfrac{5\sqrt {5}}{12}-) \space d \theta \\= \dfrac{\pi}{6}(17\sqrt {17}-5\sqrt {5}) $$
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