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: 2

Answer

$0 \le r \le 3$ and $0 \le \theta \le 2 \pi $

Work Step by Step

We have $ z=9-x^2-y^2 \implies z=9-r^2$ Apply cylindrical coordinates. $ x=r \cos \theta ;\\ y= r \sin \theta \\ z \gt 0$ and $ r=(r \cos \theta) i+( r\sin \theta) j+(9-r^2) k $ But $9-r^2 \ge 0 \implies r^2 \leq 9$ So, $-3 \le r \le 3$ and $0 \le \theta \le 2 \pi $ We can restrict $r$ to $0 \le r \le 3$ (since the negative values repeat the same points).
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