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

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.