Parametrising the surface $x^2+z^2 = y^3(1-y)^3$

Question

I am working on the following exercise:

Consider the surface given by the equation $x^2+z^2 = y^3(1-y)^3$. Where can a parametrisation $z(x,y)$ be introduced? Calculate $\partial z / \partial_x$ and $\partial z / \partial_y$.

REMARK: I found out that the surface looks like this here:

https://imaginary.org/gallery/herwig-hauser-classic

I reorded the equation to: $$z^2 = y^3(1-y)^3-x^2$$ , which implies

$$z = \pm\sqrt{y^3(1-y)^3-x^2}$$

So I think a parametrisation of the surface by $z(x,y)$ is not possible, every point $(x,y)$ gets mapped to two different $z$ unless $y^3(1-y)^3-x^2 = 0.$ Is there something I am missing in here?

Answer 1

No, you are missing nothing. For each $(x,y,z)$ in your surface, $(x,y,-z)$ belongs to the surface too. And both points have the same first and second coordinates. So, unless $z=0$, there are two points of the surface with the same first and second coordinates and therefore your surface is not of the form$$\left\{\bigl(x,y,z(x,y)\bigr)\,\middle|\,(x,y)\in D\right\}.$$for some subset $D$ of $\mathbb R^2$ and some function $z\colon D\longrightarrow\mathbb R$.

Answer 2

One can parameterize $$ x^2+z^2=y^3(1-y)^3 $$ on $[0,1]^2$ as $$ \left((t(1-t))^{3/2}\cos(2\pi s),\,t,\,(t(1-t))^{3/2}\sin(2\pi s)\right) $$ As given, the radius in the $y=\frac12$ plane is $\frac18$:

However, as with a lot of orientable, closed surfaces, almost all lines that intersect the surface (entering the interior) will intersect the surface again (exiting the interior). Thus, the lines for most $x$ and $y$ will intersect the surface at more than one point. This prevents a complete parameterization of the form $(x,y,z(x,y))$.