## Thomas' Calculus 13th Edition

a) for all (x,y,z) such that $z \gt x^2+y^2+1$; b) For all $(x,y,z )$ such that $z \ne \sqrt {x^2+y^2}$
a) Take all the positive values for $\ln n$. Thus, for all (x,y,z) such that $z - x^2-y^2-1 \gt 0$ b) There can be no zero in the denominator.Therefore,for all $(x,y,z )$ such that $z \ne \sqrt {x^2+y^2}$