Parameterize the curve of intersection

Question

Parameterize the curve of intersection of $x^2 + y^2 = 4$ and $z = y^2$

$\space$

This is what I have so far:

$\space\space\space\space$ $\vec r$(t) = < x(t) , y(t) , z(t) >

Let $y = t$. So, $z(t) = t^2$ and $x(t) = \sqrt{4-t^2}$

I ended up with the parametrization of

$\space\space\space$ $\vec r$(t) = < $\sqrt{4-t^2}$ , $t$ , $t^2$ >

$\space$

Answer 1

The first equation is misleading at first since you need to consider the fact that it is in $\mathbb{R}^3$ i.e $\{(x,y,z): x^2+y^2 = 4, z \in \mathbb{R}\}$ i.e you get a infinite cylinder. The other is parabolic cylinder and so their intersection looks like a "potato chip". The picture is given below, I zoomed in a bit.

You can describe points on this cylinder by $(2 \cos t, 2 \sin t, z)$. Now you are just adding $z = y^2$ i.e $z = 4 \sin^2t$ and so the parametrization is:

$$\gamma(t) = (2 \cos t, 2 \sin t, 4 \sin^2 t)$$