How to parametrize $\left(4-\sqrt{x^2+y^2}\right)^2 +z^2=1$

Question

How would I parametrize $$\left(4-\sqrt{x^2+y^2}\right)^2 +z^2=1$$

I am really struggling to parametrize this surface.

Here is what I observed the surface is $$(4-r)^2+z^2=1$$ so perhaps we can try and use some kind of polar coordinates to parametrize the surface that would normally involve getting a surface of the form $r=f(\theta,z)$.

But I really don't know how to proceed with this one.

Any help?

Answer 1

Based on our comment discussion, I would proceed as follows. Our surface is given by the equation $$ \left(4 - \sqrt{x^2 + y^2}\right)^2 + z^2 = 1. $$

Note that we have a thing squared plus a thing squared equals 1. This reminds me of $\cos^2 \theta + \sin^2 \theta = 1$. So, I'd like to parameterize so that I get something like $$ \underbrace{\left(4 - \sqrt{x^2 + y^2}\right)^2}_{\cos^2\theta} + \underbrace{z^2}_{\sin^2\theta} = 1. $$

Well, parameterizing $z$ is easy. Just let $z = \sin \theta$. A little more work is required for $x$ and $y$. So far for $x$ and $y$ all we have is $ 4 - \sqrt{x^2+y^2} = \cos\theta$. We can rewrite this as $$4 - \cos\theta = \sqrt{x^2 + y^2}.$$

Thinking in terms of polar coordinates as you were, if $x = r\cos\phi$ and $y = r\sin\phi$, then $\sqrt{x^2 + y^2} = r$. Note that I am ignoring issues about positive vs. negative square roots here. Anyway, in our case we clearly want $r = 4 - \cos\theta$. This therefore gives us $x = (4-\cos\theta)\cos\phi$ and $y = (4-\cos\theta)\sin\phi$.

So the final parameterization is: $$ (x,y,z) = \big((4-\cos\theta)\cos\phi, \ (4-\cos\theta)\sin\phi, \ \sin\theta\big)$$

Note that this is different from the provided answer you mentioned, but both parameterizations are correct (except for the $z$-coordinate being $\sin b$ -- I still maintain that the $z$-coordinate is supposed to be $\sin a$ in the answer you mentioned). You can verify them both by plugging them into the equation. I believe they have $4+$ where I have $4-$ because where I said $$ \left(4 - \sqrt{x^2+y^2}\right)^2 = \cos^2\theta \Rightarrow 4 - \sqrt{x^2+y^2} = \cos\theta, $$ they said $$ \left(4 - \sqrt{x^2+y^2}\right)^2 = \cos^2\theta \Rightarrow 4 - \sqrt{x^2+y^2} = -\cos\theta. $$

Answer 2

COMMENT.- My intention is to highlight the kind of your curve whose parameterization with spherical coordinates has been already given. Your equation is clearly equivalent to $$15+x^2+y^2+z^2=8\sqrt{x^2+y^2}$$ patent of which is that your curve is a quadric.

The quite interesting (at least for me) is that this quadric must be (birrationally) equivalent to an elliptic curve because of the graphic below which shows a torus with a hole, so of genus $1$, and rotational symmetry about the z-axis (see at the figure below two photos of this beauty).

In the case of a single-holed torus, (i.e. of genus $1$ and there are torus of genus $g>1$ which have $g$ holes) as ours, are well known both the cartesian and the parametric equations. $$(c-\sqrt{x^2+y^2})^2+z^2= r^2$$ where $c$ and $r$ are the center of the hole and its radius respectively. Furthermore the parametric are $$\begin{cases}x=(c+r\cos \theta)\cos \phi\\y=(c+r\cos \theta)\sin \phi \\z=r\sin \theta\\ \phi, \theta\in[0,2\pi[\end{cases}$$ You can see clearly that your equation corresponds to a particular case of this general fact.