1
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ Not sure what the sentence "Where can a parametrisation $z(x, y)$ be introduced?" means: by restricting $z > 0$ say, then you can choose the $+$ sign in your formula and get a parametrisation? $\endgroup$ – NickD Oct 21 '19 at 21:55
  • $\begingroup$ I suppose it is meant in the way that you can not just restrict to one sign for the root. $\endgroup$ – 3nondatur Oct 21 '19 at 21:57
2
$\begingroup$

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 $z\colon D\longrightarrow\mathbb R$.

$\endgroup$
2
$\begingroup$

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 $\frac18$:

enter image description here

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 $(x,y,z(x,y))$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.