Question about equations of 3D solids

[Gianmarco]

I was working on a problem on this domain:
$$E=[x,y,z)\:s.t. \: \sqrt{x^2+y^2}\leq z\leq \sqrt{3x^2+3y^2},\: x^2+y^2+z^2\leq 2]$$
and at some point I wanted to find the intersection of the internal cone($\sqrt{3x^2+3y^2}=z$) with the sphere of radius $\sqrt{2}$ to find the height z of the circle. I did so by equating the two in cylindrical coordinates as follows:
$$3x^2+3y^2-z^2=x^2+y^2+z^2-2\\x=rcos\theta,\: y=rsin\theta,\: z=z\\2r^2-2z^2+2=0 \rightarrow z^2=r^2+1$$
This result confuses me. The intersection of a cone along the z-axis and a sphere centered in (0,0,0) should be a circle around the z-axis(and so with a fixed z). The radius($\sqrt{x^2+y^2}$ is of course also fixed. Then why do I get a function of z and r? Isn't the intersection of a cone and a sphere a circle?

 

[BvU]

I did so by equating the two in cylindrical coordinates as follows

That's not the way it works for an intersection. Both sides should be zero, so you have two equations:$$
3x^2+3y^2-z^2=0 \\x^2+y^2+z^2-2 = 0 $$ and from this you have to find your way out...

 

[Gianmarco]

That's not the way it works for an intersection. Both sides should be zero, so you have two equations:$$
3x^2+3y^2-z^2=0 \\x^2+y^2+z^2-2 = 0 $$ and from this you have to find your way out...

You are a genius. :D thank you!

 

[BvU]

You're welcome !