multivariable function

Question

Sketch the region bounded by the surfaces. $z=2\sqrt{x^2+y^2}$ ,z=2

I get $x^2+y^2=z^2/4$ so do I need to insert the z and it will become $x^2+y^2=1$ ? And the graph that I will get is circle?

Answer 1

Expanding the comment into an answer. Note that $z=2$ is a plane, it has all points of the form $(x,y,2)$ on it. Now, consider $z$ as a parameter on your other surface. Plugging in, say, z=0 gets you the $0=2\sqrt{x^2+y^2}$, which has only $(0,0)$ as the solution. Any $z<0$ has no solution. As z increases, you get bigger and bigger circles. As you noted $x^2+y^2=z^2/4=(z/2)^2$, as $z$ increases you get circles of bigger and bigger radii. Once you hit $z=2$ your growing circles "hits" the plane that bounds it on top and stops.