surface that is created by the intersection of paraboloid and plane

Question

Find the surface that is created by the intersection of the paraboloid $x^2+y^2-z=0$ and the plane $z=2$. $$x^2+y^2-z=0 \Rightarrow x^2+y^2=z$$ $$z=2$$

EDIT: I had to find the area of the surface that is created.

Can I find it using a theorem? Which formula could I use?

Answer 1

Almost right: it's in $3$-dimensional space, not $2$-dimensional. So the center is $(0,0,2)$ and the circle is in the plane $z=2$.