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: 31


$D=$ {$(x,y) | x^2+y^2 \neq 1$}

Work Step by Step

As we are given that $f(x,y,z)=\dfrac{(1+x^2+y^2)}{(1-x^2-y^2)}$ The function $f(x,y,z)=\dfrac{(1+x^2+y^2)}{(1-x^2-y^2)}$ represents a rational function which is continuous on its domain $D$. Thus, $ 1-x^2-y^2 \neq 0$ or, $x^2+y^2 \neq 1$ Hence, Domain: $D=$ {$(x,y) | x^2+y^2 \neq 1$}
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