이차곡면과 곡면 그리기

Since the level curves are circles centred at the origin (in the $xy$-plane), when $z$ is a constant, the equation will have the form $x^2+y^2=c$ for some constant. That is, our equation looks like $$x^2+y^2=g(z),$$ where $g(z)$ is a function depending only on $z$.

Because our cross-sections are so nicely symmetric, we know the intersection of the figure with the left side of the $yz$-plane as well: $z=3(-y-1)=-3(y+1)$ (when $z\ge0$) and $z=-3(-y-1)=3(y+1)$ (when $z<0$). Below is the intersection of our surface with the $yz$ plane.

Setting $x=0$, our equation becomes $y^2=g(z)$. Looking at the right side of the $yz$ plane, this should lead to: $\left.\begin{cases} z=3(y-1) &\text{if } z\geq 0,\ y\ge 1\\ z=-3(y-1) &\text{if } z< 0,\ y\ge 1 \end{cases}\right\}$. That is:

$$\begin{aligned} |z|&=3(y-1)\\ \frac{|z|}{3}+1&=y\\ \left(\frac{|z|}{3}+1\right)^2&=y^2 &(*) \end{aligned}$$

A quick check: when we squared both sides of the equation in $(*)$, we added another solution, $\frac{|z|}{3}+1=-y$. Let's make sure we haven't diverged from our diagram.

$$\begin{aligned} &&& \left(\frac{|z|}{3}+1\right)^2=y^2\\ &\Leftrightarrow&& \underbrace{\frac{|z|}{3}+1}_{\text{positive}}=\pm y\\ &\Leftrightarrow&& \begin{cases} \frac{|z|}{3}+1 =y & y>0\\[0.05in] \frac{|z|}{3}+1 =-y & y<0 \end{cases}\\[0.1in] & \Leftrightarrow&& \begin{cases} \frac{|z|}{3}+1 =y & y\ge1\\[0.05in] \frac{|z|}{3}+1 =-y & y\le-1 \end{cases}\\[0.1in] &\Leftrightarrow&& \begin{cases} |z| =3(y-1) & y\ge1\\[0.05in] |z|=-3(y+1) & y\le-1 \end{cases}\\[0.1in] &\Leftrightarrow&& \begin{cases} z =\pm \underbrace{3(y-1)}_{\text{positive}} & y\ge1\\ z=\pm \underbrace{3(y+1)}_{\text{negative}} & y\le-1 \end{cases}\\[0.1in] &\Leftrightarrow&& \begin{cases} z =3(y-1) & y\ge1,\ z\ge0\\ z =-3(y-1) & y\ge1,\ z\le0\\ z=-3(y+1) & y\le-1,\ z \ge 0\\ z=3(y+1) & y\le-1,\ z\le 0 \end{cases} \end{aligned}$$

This matches our diagram eactly. So, all together, the equation of the surface is $$x^2+y^2=\left( \frac{|z|}{3}+1\right)^2$$