Example 3.3. Find the level surfaces of the function $y=$ $=\sqrt{36-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}}$ and the value of the function at the point $P(1,1,3)$.
Solution. According to the definition of level surfaces, we have: $\sqrt{36-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}}=C$, where $C \geqslant 0$. From this, it follows that $36-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}=C^{2}$, that is, $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=$ $=36-C^{2}$ (obviously, $0 \leqslant C \leqslant 6$). The obtained equation represents the equation of a sphere in the rectangular Cartesian coordinate system $O x_{1} x_{2} x_{3}$ with its center at the origin and radius $\sqrt{36-C^{2}}$. When $\overparen{C}=6$, the level surface degenerates into the point 0.

Now let's find the value of the function at the point $P(1,1,3)$. For this, we substitute $x_{1}=1, x_{2}=1, x_{3}=3$ into the analytical expression defining the function. We get $y=\sqrt{36-1^{2}-1^{2}-3^{2}}=$ $=\sqrt{25}=5$