Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.2 Exercises - Page 924: 35

Answer

{$(x,y,z) | x^2+y^2+z^2\leq 1$}

Work Step by Step

We are given that $f(x,y,z)=arcsin(x^2+y^2+z^2)$ The function $f(x,y,z)=arcsin(x^2+y^2+z^2)$ represents a trigonometric function and we know that $\sin x$ lies in between -1 and +1. Thus, $ -1 \leq x^2+y^2+z^2\leq 1$ Or: $x^2+y^2+z^2\leq 1$ Hence, {$(x,y,z) | x^2+y^2+z^2\leq 1$}
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