What shape does this 3D graph form? What shape does the graph of $x^2 + y^2 = z^4$ equal? The graph is 3D, with x,y,z axis.
Particularly, what shape does it form when z is under the restriction $0<= z <= 1$
I know that $x^2 + y^2 = z^2$ gives a cone shape, and that makes sense and is easy to imagine in my head. (There are increasingly smaller/bigger circles stacked ontop of eachother for each z value)
However, I can't seem to figure out what this graph means. Is it taking the area of a cone and using that as the side of a square?
Thank you for the help in advance!
 A: The horizontal slices (fixed $z$) of $x^2+y^2=z^4$ i.e. $r=z^2$ are similar to those of $x^2+y^2=z^2$ i.e. $r = z$ ($z\in[0,1]$ is given). If $z=z_0$ then in $r = z_0^2$  describes a circle of radius $z_0^2$. So the only difference is in how the radius changes as you vary $z$.
The difference is that when $z<1$, the circles will be smaller than those for the cone, and when $z>1$, the circles will be larger. At exactly $z=1$ the two surfaces have the same slice ($r=1$).
Here is a plot in $r$-$z$ space of $r=z$ and $r=z^2$ (ignore the bottom half of the graph):

If we rotate the red curve around the $z$-axis, we get the 3D plot of your surface (again, ignore the bottom half):

As for the area, this is a surface of revolution; generally, if we are rotating $r=f(z)$ on $z\in[a,b]$ around the $z$ axis then the surface area is
$$ A = 2\pi \int_a^b f(z) \sqrt{1+ f'(z)^2}dz $$
An explanation for this formula can be found here (found on google). In our specific case, this is
$$A = 2\pi \int_0^1 z^2 \sqrt{1+4z^2} dz =  \dfrac{18\sqrt{5} - \operatorname{arsinh}2}{32}\pi.$$
