Does anyone know what kind of surface is defined by $z=\left(x^2+y^2+z^2\right)^2$?

Question

Does anyone know what kind of surface is defined by this equation? $$z=\left(x^2+y^2+z^2\right)^2$$

Thank you in advance.

Answer 1

Not sure if it has a name.

Given fixed $z,$ then:

$$x^2+y^2=\sqrt{z}-z^2$$ So you need $0\leq z\leq 1$ for $x^2+y^2\geq 0.$ At $z=0$ and $z=1$ the set of solutions is the single point $(x,y)=(0,0).$ For any other $z$ in that range, the cross section curve is a circle.

Topologically, it is a sphere.

Its interior is convex. This is because the second derivative of $\sqrt z-z^2$ is negative.

For small $\epsilon>0$ the cross section at $z=1-\epsilon$ has radius approximately $\sqrt{3\epsilon/2},$ and the cross section at $z=\epsilon$ is of radius $\approx \sqrt[4]{z},$ because when $z=\epsilon$, $\sqrt z-z^2\approx \sqrt z.$

This means that the surface is flat at both ends, like a sphere of radius $\frac{3}4$ and center $(0,0,1/4)$ near $z=1$ and an even flatter side near $z=0.$

When $z=1/2$ the radius of the circle cross-section is at its largest, which is $\frac12\sqrt{2\sqrt 2-1}\approx 0.676.$