Parametrizing the surface $x^2 = 1-z$ and $y^2 = z$

Question

I am given the following exercise:

Find the parametrization of the surface $C: x^2 = 1 - z$ and $y^2 = z$

I got to the following answer:

\begin{cases} x &= \sin (t)\\ y &= \cos (t)\\ z &= \cos^2 (t) \end{cases}

Unfortunately, there's no way to evaluate if my answer is correct on the textbook. Could someone please verify if that's the case?

Thank you.

Answer 1

Let check directly

• $x^2=1-z \implies \sin^2 t=1-\cos^2 t$
• $y^2=z \implies \cos^2 t=\cos^2 t$

and note that it is a curve defined by the intersection of two surfaces.

Answer 2

Adding up the equations, $x^2+y^2=1$. Any parameterization that fulfills this constraint can do.

E.g. your solution, or

$$\begin{cases}x=\dfrac{1-u^2}{1+u^2},\\y=\dfrac{2u}{1+u^2}, \\z=\dfrac{4u^2}{(1+u^2)^2},\end{cases}$$

or

$$\begin{cases}x=t,\\y=\pm{\sqrt{1-t^2}}, \\z=1-t^2,\end{cases}$$

...