Answer
(a) 0
(b) All points above cone $z=\sqrt{x^2+y^2}$. $D=\{(x,y,z)\in\mathbb{R}^3|z>\sqrt{x^2+y^2}\}$
Work Step by Step
(a) $f(4, -3, 6)=\ln(6-\sqrt{4^2+(-3)^2})=0$
(b) The value inside the natural logarithm must be greater than 0, which means $z>\sqrt{x^2+y^2}$. The expression inside the square root is always non-negative because we have the sum of two squares. Thus, we have that the domain of f is all points above the cone $z=\sqrt{x^2+y^2}$.
