Answer
$D= x^2+y^2\leq 25$.
Work Step by Step
Since $f(x,y)=\sqrt{25-x^2-y^2}$ is defined for all $x,y$ satisfying that $ x^2+y^2\leq 25$. Hence the domain of $f$ is the closed disc $D= x^2+y^2\leq 25$.
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 12 - Functions of Several Veriables - 12.2 Graphs and Level Curves - 12.2 Exercises - Page 882: 13
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