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I have an equation x² + y² + z³ - z² = 0

This is an alpha-loop rotated around the z-axis.

Can someone please help me to convert this to three parametric equations for x, y, and z (varying for u and v)?

I ask because I am an artist trying to make models of equations (not a new idea but great fun).

I am almost there, but there is an interesting problem with my surface. The tear drop is upside down... I assume it should be rounded at the top and come to a point (a singularity???) where it meets a similar point on the top of the mountain.

The reason for my assumption is the curve (before rotation) has the shape I am struggling to describe. See: Image of curve from Wolfram

I have used parametric equations as suggested by Ted and Compacto [thanks to both].

Any idea why my curve is inside out or upside down? It should be the same as the curve but rotated around an axis.

Image of surface from equation

Here is the successful result: Success!

Paul St George
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    What have you tried? Have you tried finding parametric equations first for the curve $z^3-z^2+x^2=0$ in $xz$-space? – Ted Shifrin Feb 14 '22 at 20:18
  • True, if you know already that it's a surface of revolution, the sensible thing would be to first parametrize the curve that generates it. – Compacto Feb 14 '22 at 20:25
  • That is certainly the right curve in xz space. I have just plotted it. Could you help me to find the parametric equations for the curve? – Paul St George Feb 14 '22 at 20:46
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    For a rational expression, $$\begin{cases}x=\sqrt{v^2-v^3}\dfrac{1-u^2}{1+u^2},\y=\sqrt{v^2-v^3}\dfrac{2u}{1+u^2},\z=v.\end{cases}$$ –  Feb 15 '22 at 14:56
  • @YvesDaoust How did you obtain the rational parametrization? Using $t= \tan(u/2) $? – Compacto Feb 15 '22 at 18:21
  • @Compacto: yes, the Weierstrass substitution. –  Feb 15 '22 at 18:22

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If one knows beforehand it's a surface of revolution, then it will be generated by a curve

$$x(t) = f(t), z(t) = g(t)$$

Which is the intersection of the surface with the plane $y = 0$. Rotating this curve around the $z$ axis gives the parametric equations for the surface:

$$(x(u,v), y(u,v), z(u,v)) = (f(v)\cos (u), f(v)\sin (u), g(v))$$

So you just need to obtain the functions $f, g$ which parametrize the curve. As I said before (and was hinted by Ted in the comments), the generating curve is obtained by making $y = 0$ in the implicit equation, giving

$$x^2 = z^2-z^3 = z^2(1-z)$$

And so, $x = |z|\sqrt{1-z}$ ($x$ is assumed to be positive so it lies "on the right" of the axis and forms the part of the curve that will be rotated) so we can put $x(t) =f(t) = |t|\sqrt{1-t}, z(t) = g(t) = t$, and so the parametric equation for the surface is

$$(x(u,v), y(u,v), z(u,v)) = (|v|\sqrt{1-v}\cos (u), |v|\sqrt{1-v}\sin (u), v)$$

Compacto
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  • The equations given by Yves Daoust and Compacto give me a surface, but neither are the surface of revolution of the curve given by Ted Shifrin and shown as an image in [https://i.stack.imgur.com/aArNI.png].

    I am wondering whether the equation I gave is incorrect. Perhaps we can approach this in a different way. What, please, is the equation that gives z³ −z² + x² = 0 rotated around the axis that bisects this curve?

    Is it as I gave (x² + y² + z³ - z² = 0)? Or is it something else?

     – Paul St George Feb 15 '22 at 18:16
  • I think your equation is right. From this Wolfram Alpha output: https://www.wolframalpha.com/input?i=z%5E3+-z%5E2+%2B+x%5E2+%2B+y%5E2%3D+0 – Compacto Feb 15 '22 at 18:26
  • All is now working. In case someone wants to follow this and make their own 3D surface, the equations are as above. Care also needs to be taken with the range of u and the range of v. Start from a negative value.

    The result is shown as an edit to the original question.

     – Paul St George Feb 16 '22 at 11:08
  • Thanks for sharing your output! – Compacto Feb 16 '22 at 11:16