How to parameterize the intersection of these two curves?

Question

So I am wondering about how to parameterize the curve of intersection of the two surfaces

$x^2+(y-1)^2=1$ (cylinder) and $z=\sqrt{x^2+y^2}$ (cone).

Should I solve both equation for one variable and set them equal to each other?

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

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

$2y=z^2$ If I then set $z=t$ we must have that $y=\frac{1}{2}t^2$. From the equation to the cone we then have that

$x=\sqrt{z^2-y^2}=\sqrt{t^2-\frac{1}{4}t^4}$

Is this right?

Answer 1

I parameterized as follows:

For the cone $(x,y,z) = ( r \cos t, r \sin t , r ) $

Plug this into the cylinder equation:

$r^2 \cos^2 t + (r \sin t - 1)^2 = 1$

So that,

$r^2 - 2 r \sin t = 0$

Hence,

$r = 2 \sin t$

Varying $t$ over $[0, 2\pi]$ gives $r$, which together with $t$ give $x, y, z$

$x = 2 \sin t \cos t = \sin 2 t $

$y = 2 \sin^2 t = 1 - \cos 2 t $

$z = 2 \sin t $

Since $z \ge 0 $, we must have $t \in [0, \pi] $