Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
+1 for nice question – dato datuashvili Jan 11 '12 at 20:16
Are you aware of the fact that local isometries preserve geodesics? Can you see how that would be useful here? – Alex Youcis Jan 12 '12 at 17:51
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? – Nadori Jan 14 '12 at 16:42

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.