Surface as an intersection of a sphere and a cone

Question

I have to evaluate the density of the solid bounded by the surface $$(x^{2}+y^{2}+z^{2})^{2}=x^{2}+y^{2}$$ How can I see this surface? Is it a intersection of a sphere and a cone?

Answer 1

Let $r=\sqrt{x^2+y^@+z^2}$ and $a=\sqrt{x^2+y^@}$ then the curve is the curve $r^2=a$, which is the 2D curve $x^2+y^2-x=0$ on the subset $\{x\ge 0\}$ , rotated around the axis $z$.

Answer 2

It is the surface of revolution where a circle of diameter $1$ is tangent to the revolution axis at the origin. It is a torus where the radius of revolution is equal to the radius of the circular cross section.