Maybe someone can verify my answers. The problem is as follows:

Consider a line $l$ in $\mathbb{R}^3$ and rotate it around the $Oz$ axis. Denote by $A$ the set of all such obtained surfaces.

Question 1: Write their parametrizations.

My answer: Let $S \in A$. Then there exists a line $$\phi:\mathbb{R}\rightarrow\mathbb{R}^3:t\mapsto \phi(t)=p_0+tv$$ and a matrix $R(\theta)$ corresponding to the rotation around the $Oz$ axis by an angle $0\leq \theta <2\pi$ such that $S$ is parametrized as $$\psi: [0,\theta]\times \mathbb{R}\rightarrow S:(\alpha,t)\mapsto R(\alpha)\phi(t).$$

Queston 2: Describe as many geodesics as you can for each surface.

My partial answer: I started with taking an $S$ as above and defining $\gamma(t)=\psi(\alpha,t)/||v||$, where the vector $v$ is a "direction" of the line $l$. Since $\gamma''(t)=(0,0,0)$, $\gamma$ is a geodesic on $S$.

How can I find many others?

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    $\begingroup$ Are you aware of the fact that local isometries preserve geodesics? Can you see how that would be useful here? $\endgroup$ Jan 12, 2012 at 17:51
  • $\begingroup$ Thank you. Sorry for my late reply. No, I am not aware of that. But If I use that fact, then I can work with a surface $M_1$ obtained by rotating a line of which the direction is equal to $v=(v_1,v_2,0)$. This surface is a part of $\mathbb{R}^2$ and since we know that the geodesics of a plane are exactly the lines on it, we have found ALL geodesics on $M_1$. And if it is true (I am convinced it is) that alle elements of $A$ are locally isometric then every geodesic on a surface of $A$ is just the image of a line in plane through a local isometry. How does this sound? $\endgroup$
    – Nadori
    Jan 14, 2012 at 16:42

2 Answers 2


For Question 1 :

I am assuming you already acquainted yourself with how Clairaut's law is arrived at and how to express by quadratures surface of revolution (x,y,z) parametrized on u and v in 3-space:

f(u) cos(th(u) + v ), f(u) sin(th(u) + v), g(u) where v adds rotation to this space curve around z-axis.


Answer for comment of Jan 12 2012

Sorry too late reply, but still..

Yes, that is correct. Apart from maths of it, you can see it to believe in it..Take a sheet of paper on which a straight line is drawn. Make any cone rolling up the paper arbitrarily and your line will be seen on it as a geodesic. Next, take a thin plastic ball and draw its equator which is a great circle. Cut it into two hemispheres, deforming it anyway (convexity preserved) you please. You will see the geodesics on the conoids so formed.These are isometries in action.


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