Rotation of a catenary in $\mathbb{R}^5$ If you rotate a catenary in $\mathbb{R}^3$, then you get a catenoid.
To show: If you rotate the same catenary in $\mathbb{R}^5$, then you get a 4-dimensional hypersurface.
I'm not sure, if I got this right. How do you rotate a plane curve in ${\mathbb{R}}^5$? And which axis should I take?
Can I do this by matrix multiplication?
Thank you for your help.
 A: First, you have to be more precise in your description of the catenoid. 
Definition A catenoid is the surface of revolution in $\mathbb{R}^3$ formed by revolving a catenary about its directrix. 
Note that it is extremely important to specify the fact that you are rotating about the directrix! If you translate a catenary you will get another catenary. But the surface of revolution in cylindrical coordinates given by 
$$ r(z) = \cosh z + 1 $$
is no longer what we would consider a catenoid: for starters, it is not a minimal surface! It is, however, a rotation of a catenary. You can get even stranger examples if you rotate the catenary about lines which are not even parallel to the directrix of the catenary. 
Now that we have fixed the correct definition of the catenoid, the generalisation to higher dimension is immediate. What you want to consider is the hypersurface of revolution formed by rotating a catenary about its directrix. Here rotation means the following:
Let $V$ be a vector subspace with $k$ dimensions inside $\mathbb{R}^n$. Then the subgroup of $SO(n)$ which acts on $V$ as the identity is isomorphic to $SO(n-k)$. By rotating a point $p\in \mathbb{R}^n$ about $V$ we mean the orbit of $p$ under the action of this $SO(n-k)$. 
In the case of hypersurfaces of revolution, there is a simpler description: let $\gamma$ be a curve in the upper half plane of $\mathbb{R}^2$, we can write it as $\gamma = \{f(x,y) = 0\}$.  Its corresponding hypersurface of revolution _formed by revolving $\gamma$ about the $x$ axis in $\mathbb{R}^n$ is the set 
$$ \Sigma := \{ (x_1,\ldots,x_n)\in\mathbb{R}^n | f(x_n, r_n) = 0, r_n = \sqrt{x_1^2 + \cdots + x_{n-1}^2} \} $$
In cylindrical coordinates $(r,\omega,z)\in \mathbb{R}_+ \times\mathbb{S}^{n-2} \times \mathbb{R}$, we can describe $\Sigma$ more simply as
$$ \Sigma := \{ f(z,r) = 0 \} $$
In your case, $\gamma$ is given by $y = \cosh x$, the catenary as a graph over its directrix. Hence the corresponding surface of revolution is 
$$ \Sigma := \{ r = \cosh z \} $$
or
$$ \Sigma := \{ x_1^2 + x_2^2 + \cdots + x_{n-1}^2 = \left( \cosh x_n\right)^2 \} $$
