How to parametrize a cubic plane curve

Question

I've got a cubic surface, but I have no idea how to parametrize it.

If there is no general method, maybe there is one for a specific surface — z^2=xy(x+y-1).

I need to define this surface with the following equations: x=f1(u, v), y=f2(u, v), z=f3(u, v). The simpler f1, f2, f3 are, the better. Polynomials or rational functions will do for me.

Answer 1

As $z^2\geq 0,$ the surface has $3$ sheets, one sheet with $x\geq 0,\;y\geq 0,\;x+y-1\geq 0$, one sheet with $x\leq 0,\; y\geq 0,\; x+y-1\leq 0$ and one sheet with $x\geq 0,\; y\leq 0,\; x+y-1\leq 0.$

For the first sheet, we can set \begin{eqnarray} x_1 &=&(1+v^2) u \\ y_1 &=&(1+v^2)(1-u) \\ z_1 &=& v\,(1+v^2)\, \sqrt{u(1-u)} \end{eqnarray} with $u\in [0,1]$ and $v\in\mathbb{R}.$

The second sheet can be derived from the first one by setting \begin{eqnarray} x_2 &=& 1-x_1-y_1 \\ y_2 &=& x_1 \\ z_2 &=& z_1 \end{eqnarray} because then we have $$ x_2y_2(x_2+y_2-1) = (1-x_1-y_1)x_1(1-x_1-y_1+x_1-1) \\ = (1-x_1-y_1)x_1(-y_1) = x_1y_1(x_1+y_1-1) $$ Analogously, we can set the following for the third sheet \begin{eqnarray} x_3 &=& y_1 \\ y_3 &=& 1-x_1-y_1 \\ z_3 &=& z_1 \end{eqnarray} Rational solution: \begin{eqnarray} x_1 &=& \frac{(1+v^2) (1-u^2)^2}{(1+u^2)^2} \\ \\ y_1 &=& \frac{(1+v^2) (2u)^2}{(1+u^2)^2} \\ \\ z_1 &=& \frac{v\,(1+v^2)\, (1-u^2)(2u)}{(1+u^2)^2} \end{eqnarray} with $u\in [0,1]$ and $v\in\mathbb{R}.$

Answer 2

In this case, z is a function of x and y, so you can set f1 = u, f2 = v, and f3 = uv(u+v-1). There is no strategy for this in general, you just have to be creative.

Answer 3

For the case of an explicit relation $$z=f(x,y)$$ the parameterization is immediate:

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

If you prefer,

$$\begin{cases}x=u,\\y=v,\\z=f(u,v).\end{cases}$$

Answer 4

Monge's form with another individual parametrization:

$$ x= u+\frac12, y=v+\frac12, z= \pm \sqrt {(u+\frac12)(v+\frac12)(u+v)};$$