Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.2 - Limits and Continuity - 14.2 Exercise - Page 911: 33


$D=$ {$(x,y) | x \geq 0 ; x^2+y^2 \leq 1$}

Work Step by Step

As we are given that $G(x,y)=\sqrt x+\sqrt {1-x^2-y^2}$ The function $G(x,y)=\sqrt x+\sqrt {1-x^2-y^2}$ represents a squared root function which is defined for positive numbers greater than zero. Thus, $ x \geq 0 ; 1-x^2-y^2 \geq 0$ or, $x^2+y^2 \leq 1$
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