# Automated generation of a parametric plot?

Currently we are introducing integrals of manifolds in $3d$-Space in a course. We are given some set $E \subset \mathbb{R}^3$ and should evaluate the surface and scetch how it looks.

For example we should evaluate the 2 dimensional surface of the set given by

$E=\{ (x,y,z) \in \mathbb{R}^3: 0 < z = x^2 + y^2 < 2^2 \}$

of course the standard approach to solve this easy task is to first parametrize the set with for example

$\phi(u,t)=\left(\sqrt{u^2} \cos (t),\sqrt{u^2} \sin (t),u\right)$

where $t \in (0,2\pi), u \in (0,2^2)$. The plot looks like: Generated by "ParametricPlot3D[Phi[u, t], {t, 0, 2 Pi}, {u, 0, 2^2}]" in mathematica.

Using the formula for manifolds one can evaluate the surface by

A = JacobianMatrix[Phi[u, t], {u, t}]
FullSimplify[ Integrate[Sqrt[Det[Transpose[A].A]], {u, 0, 4}, {t, 0, 2*Pi}]]


Which yields $\frac{1}{6} \left(17 \sqrt{17}-1\right) \pi$

a more elegant approach would be to notice that the graph is generated by rotating the function $f(x)=\sqrt{x}$ around the x-Axis in the interval $(0,2^2)$ and evaluate

2*Pi*Integrate[f[x]*Sqrt[1 + (f'[x])^2], {z, 0, 4}]


which yields the same result.

As I like to generalize solutions to such problems I try to find a possibility to input such a set in a natural form and get both of the above results, namely the plot and the surface area without doing any work by myself. However, even generating the plot seems to be very hard without parametrizing the set by hand. An approach to relax the equality a bit and use the regionplot method fails blatantly:

    RegionPlot3D[ Abs[z - x^2 + y^2] <= 0.3 && x^2 + y^2 < 4, {x, -3, 3}, {y, -3,
3}, {z, 0, 4}]


results in the output: And I have no idea how I would let mathematica approximate the surface of the object.

Is there some generic approach to plot and determine the surface of the set, without having to parametrize it?

ContourPlot3D[z == x^2 + y^2, {x, -2, 2}, {y, -2, 2}, {z, 0, 4}]