How do I parameterise $x^2-y^2+z^2=0$?

Question

How do I parameterise $x^2-y^2+z^2=0$ where $y\in [0,1]$ ?

Here's my thought process right now, but I'm not sure:

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

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

Let $y=u$

Then can you just parametrise it like you would a circle?

Answer 1

Yes, so $(x,y,z)=(u\cos t, u, u\sin t)$ is a parametrization, where $u\in[0,1]$ and $t\in[0,2\pi)$. Since it is a surface, we have two parameters, $u$ and $t$.

Answer 2

$y=\sin t$, $ t \in [0,π/2].$

Then $x^2+z^2 = \sin^2 t$ ;

$x=\sin t \cos s$; $z= \sin t \sin s$,

where $s \in [0,2π).$