Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.2 Limits and Continuity - 14.2 Exercises - Page 951: 33

Answer

$D=$ {$(x,y) | 0 \leq x \leq \sqrt {1-y^2}$}

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 cannot be less than $0$. Thus, $ x \geq 0 ; 1-x^2-y^2 \geq 0$ or, $x^2+y^2 \leq 1$ or, $x^2 \leq 1-y^2$ or, $x \leq \sqrt {1-y^2}$ This means that $x \geq 0,x\leq \sqrt {1-y^2}$ Hence, Domain: $D=$ {$(x,y) | 0 \leq x \leq \sqrt {1-y^2}$}
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