Vector Calculus: An Introduction

As a first example look at the function $f(x,y,z)=x^2/4+y^2/9+z^2$ defined on ${\mathbb{R}}^3\text{.}$ To find the level surface, say to the level $c=1\text{,}$ we can look at cross-sections along some coordinate plains. Let us take the plane $z=0\text{.}$ Then $f(x,y,0)=x^2/4+y^2/9=1\text{,}$ which is the equation of an ellipse with axis of length $2$ and $3\text{.}$ Similarly we consider $y=0\text{,}$ which leads to the ellipse $f(x,0,z)=x^2/4+z^2=1$ with axis $2$ and $1\text{.}$ Doing the same on the third coordinate plane, $x=0\text{,}$ we get the ellipse $f(x,y,z)=y^2/9+z^2$ with axis $3$ and $1\text{.}$ The resulting surface is the ellipsoid shown in Figure 2.9.

We next consider some level sets of $f(x,y,z)=x^2+y^2-z^2\text{.}$ Observe that the function is rotationally symmetric about the $z$-axis. Hence the level surfaces $f^{-1}[c]$ can be obtained by revolving the curve $f(x,0,z)=c$ about the $z$-axis.

We determine the surfaces for the levels $c=0,\pm 1\text{.}$ If $c=0$ then $f(x,0,z)=0=x^2-z^2$ is given by $z=\pm x\text{.}$ Hence $f^{-1}[0]$ is a double cone centred about the $z$-axis as shown in Figure 2.11. For $c=1$ we have $f(x,0,z)=1=x^2-z^2\text{,}$ implying that $x=\pm \sqrt{1+z^2}\text{.}$ Revolving about the about the $z$-axis we get the corresponding surface in Figure 2.12. Proceeding similarly for $c=-1$ we get $f(x,0,z)=-1=x^2-z^2\text{,}$ or $z=\pm \sqrt{1+x^2}\text{.}$ These are the same hyperbolas as before but turned by ${90}^{\circ }\text{.}$ As a result the corresponding level surface of $f$ shown in Figure 2.13 consists of two shells.