Find the domain of each of the following functions. Then graph the domain.

Solution (a) Since the expression under the radical must be nonnegative, the domain of $f$ consists of all points in the plane for which $$\begin{eqnarray*} 16-x^{2}-y^{2} &\geq &0 \\[4pt] x^{2}+y^{2} &\leq &16 \end{eqnarray*}$$

The domain is all the points inside and on the circle $x^{2}+y^{2}=16$. The shaded portion of Figure 3 illustrates the domain.

(b) Since the logarithmic function is defined for only positive numbers, the domain of $f$ is the set of points $(x,y)$ for which $y^{2}-4x>0$ or $y^{2}>4x$. To graph the domain, we start with the parabola $y^{2}=4x,$ and use a dashed curve to indicate that the parabola is not part of the domain. The parabola $y^{2}=4x$ divides the plane into two sets of points: those for which $y^{2}<4x$ and those for which $y^{2}>4x$. To find which points are in the domain, choose any point not on the parabola $y^{2}=4x$ and determine whether it satisfies the inequality. For example, the point $(2,0)$ is not in the domain, since $0^{2}<(4)(2)$. The set of points for which $y^{2}>4x$, the domain of $f$, is shaded in Figure 4.

Figure 3 The domain of $f(x, y)=\sqrt{16-x^2-y^2}$.

Figure 4 The domain of $f (x,y) = \ln (y^{2} -4x )$.