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: 35


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

Work Step by Step

As 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 which cannot be negative and exists only for positive numbers. Also, $\sin x$ function lies in between -1 and +1. Thus, $ -1 \leq x^2+y^2+z^2\leq 1$ or, $x^2+y^2+z^2\neq -1 $ This means that $x^2+y^2+z^2\leq 1$ Hence, Domain: $D=$ {$(x,y,z) | x^2+y^2+z^2\leq 1$}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.