Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 14 - Section 14.1 - Functions of Several Variables - 14.1 Exercise - Page 946: 6

Answer

(a) 0 (b) All points above cone $z=\sqrt{x^2+y^2}$. $D=\{(x,y,z)\in\mathbb{R}^3|z>\sqrt{x^2+y^2}\}$

Work Step by Step

(a) $f(4, -3, 6)=\ln(6-\sqrt{4^2+(-3)^2})=0$ (b) The value inside the natural logarithm must be greater than 0, which means $z>\sqrt{x^2+y^2}$. The expression inside the square root is always non-negative because we have the sum of two squares. Thus, we have that the domain of f is all points above the cone $z=\sqrt{x^2+y^2}$.
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