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How do I parametrise the surface $x^2 + y^2 + z^2 -2x =0$?

This is the beginning of a question where we have to work out a tangent plane at a point. I understand the rest of this method but somehow cannot work out how to parametrise tricky surfaces.

The other surfaces we have to parametrise are $x^2 + y^2 - z^2 = 2y + 2z$ where $−1 ≤ z ≤ 0$ and $(4 − x^2 + y^2)^2 + z^2 = 1$.

Obviously I don't expect someone to parametrise all these surfaces for me but any help would be appreciated!

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  • $\begingroup$ Visit : math.stackexchange.com/questions/305894/… Hope this works. $\endgroup$
    – Sujit Bhattacharyya
    Feb 23, 2018 at 4:44
  • $\begingroup$ If you have quadratic surfaces, just complete the square. This (usually) tells you which surface it is, at which point you can introduce the standard parametrisations. Here $x^2 - 2x + y^2 + z^2 = 0 \iff (x-1)^2 + y^2 + z^2 = 1,$ and shifted spherical coordinates should suffice. (More generally, you can view a second degree polynomial in $\mathbb{R}^n$ as $x^T P x + c$ for a symmetric $n\times n$ matrix $P$, and you can just diagonalise this $P$ to obtain the second degree polynomial in the diagonal basis, giving you natural parametrisations.) $\endgroup$
    – stochasticboy321
    Feb 23, 2018 at 4:47

2 Answers 2

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In the first one: you can complete the square: if you add $1$ to both sides, you get $$ (x-1)^2+y^2+z^2=1. $$ So you have a sphere or radius $1$, centered at $(1,0,0)$. You can parametrize it with spherical coordinates: $$ x-1=\cos\theta\,\sin\varphi,\ \ \ y=\sin\theta\,\sin\varphi,\ \ z=\cos\varphi. $$ Thus $$ x=1+\cos\theta\,\sin\varphi,\ \ \ y=\sin\theta\,\sin\varphi,\ \ z=\cos\varphi,\ \ 0\leq\theta\leq2\pi,\ \ 0\leq\varphi\leq\pi. $$

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$$x^2 + y^2 + z^2 -2x =0$$

$$x^2 -2x+ 1+ y^2 + z^2 =1$$

$$ (x-1)^2+y^2 + z^2 =1$$

Use spherical parametrization; $$x−1=cosθsinφ, y=sinθsinφ, z=cosφ.$$

$$x=1+cosθsinφ, y=sinθsinφ, z=cosφ.$$

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