Plotting and graphing $x^2 +y^2+ e^{z^2} =10$

Question

In an exercise from an analysis book, I came across a new set.

What can I say about the subset of $\mathbb R^3$ described by the equation $x^2 +y^2+ e^{z^2} =10$?. Let A be this set.

The function $f:x^2 +y^2+ e^{z^2}$ from $\mathbb R^3$ to $\mathbb R$ is continuous and A is closed because it is the preimage of a closed set (the set {10}. It is bounded because $0\le x^2 + y^2 \le 9$ and $0\le z ^2 \le ln(10)$. So it is a compact set. The set is symmetric with respect to the variables $x, y, z $, in the sense that $x\in A$ if and only if $-x \in A$, and the same holds for y and z. If we fix a certain value for z, we have a circumference $x^2 +y^2=10-e^{z^2}$, its center being (0,0,z). So it seems to me quite similar to an ellipsoid. However, should the exponential "kind of" make a difference?

Do you have any suggestions on how to draw it? Or do you know where to find a graph of this surface?

Answer 1

You can plot this with Mathematica and probably most any other language with graphics:

ContourPlot3D[x^2 + y^2 + Exp[z^2] == 10, {x, -3.5, 3.5}, {y, -3.5, 3.5}, {z, -3.5,3.5}]

Answer 2

A surface of revolution made flatter along $z$ axis like dutch cheese stock. Otherwise cylindrical.

$$ z= \sqrt{ \log(10-r^2)}$$