Example 3.2. Find the domain of the function $z=$ $=\sqrt{8-|x-8|-|y|}$.
Solution. The domain of the given function is the set of points $P(x, y) \in E^{2}$, satisfying the condition $8-|x-8|-|y| \geqslant 0$. From this, it follows that $|y| \leqslant 8-|x-8|$.

This inequality is equivalent to the double inequality $-8+|x-8| \leqslant y \leqslant 8-|x-8|$. Solving it:

\[
\begin{aligned}
& y \leqslant 8-|x-8| \Rightarrow |x-8| \leqslant 8-y \Rightarrow \\
& \quad \Rightarrow -8+y \leqslant x-8 \leqslant 8-y \Rightarrow y \leqslant x \leqslant 16-y \text{ (a) }
\end{aligned}
\]

Similarly,

\[
\begin{aligned}
y \geqslant & -8+|x-8| \Rightarrow |x-8| \leqslant y+8 \Rightarrow \\
& \Rightarrow -y-8 \leqslant x-8 \leqslant y+8 \Rightarrow -y \leqslant x \leqslant y+16 \text{ (b) }
\end{aligned}
\]

Next, we find the intersection of the sets of points $P(x, y)$ in the plane $X O Y$ that satisfy the above double inequalities (a) and (b).

For this, in the plane $X O Y$, we draw the lines $y=x$ and $y=16-x$. Then, since according to inequality (a) $y \leqslant x$ and $y \leqslant 16-x$, this inequality will be satisfied by all points in the plane $X O Y$ lying below the lines $y=x$ and $y=16-x$. Next, in the plane $X O Y$, we draw the lines $y=-x$ and $y=x-16$. Clearly, inequality (b) is satisfied by the points in the plane $X O Y$ lying above the lines $y=x$ and $y=16-x$, since according to inequality (b) $y \geqslant -x$ and $y \geqslant x-16$. The intersection of the obtained sets of points in the plane $X O Y$ is a square with vertices $O(0,0), A(8,8), B(16,0), C(8,-8)$ (Fig. 3.1). Therefore, the domain of the function can be represented as

\[
\begin{aligned}
D(f)=\{(x, y): & y \leqslant x \leqslant 16-y, 0 \leqslant y \leqslant 8\} \cup \\
& \cup\{(x, y):-y \leqslant x \leqslant y+16, -8 \leqslant y \leqslant 0\}
\end{aligned}
\]

and the range $E(f)=\{z: z \in R_{+}\}$.

Unlike functions of one variable, which can be geometrically represented, for functions of several variables, geometric representation at the same level of clarity is only possible for functions of two variables $z=f(x, y)$. For this, a rectangular coordinate system $X O Y$ is chosen in three-dimensional Euclidean space. At each point $P(x, y) \in D(f)$, a perpendicular to the plane $X O Y$ is drawn, and the value $z=f(x, y)$ is marked on it. The geometric locus of points obtained in this way is the spatial graph of the function of two variables, which in general represents some surface, the equation of which is the equality $z=f(x, y)$.

In the case of functions of three variables $u=f(x, y, z)$, geometric interpretation at the same level of clarity is already impossible. To obtain a representation of the behavior of a function of three variables, level surfaces are used.

A level surface of the function $y=f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ is the set of points from the domain of the function $D(f)$, satisfying the condition

\[
\left\{\left(x_{1}, x_{2}, \ldots, x_{n}\right): f\left(x_{1}, x_{2}, \ldots, x_{n}\right)=C\right\} \subset D(f)
\]

where $C$ is a constant. Therefore, the level surfaces of the function $f$ are sets of points from the domain of the function where it takes specified constant values; each constant value of the function corresponds to its own level surface. For $n=2$, we have level lines.