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: 43


$$\pi \sqrt {c^2+1}$$

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$ $$ Surface \space Area =\iint_{R} \dfrac{|\nabla f|}{|\nabla f \cdot p|} dA \\=\int_{0}^{2 \pi} \int_{0}^{1}\sqrt {c^2+1} r \space dr \space d\theta\\ =\int_{0}^{2} (\dfrac{1}{2}) \sqrt {c^2+1} \space d\theta \space \\=\pi \sqrt {c^2+1}$$
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