# mean curvature and polar tangential angle

Is it possible to express the mean curvature of a surface of revolution in terms of the first derivative of the polar tangential angle?

To be specific: Let $r=u(\theta)$ be a polar curve in the first quadrant of the xy-plane. The surface of revolution is generated by rotating this curve around the y axis. Let $\psi$ be the polar tangential angle, the angle between the tangent to the curve at a point and the ray from the origin to the point. Can the mean curvature be expressed as a first derivative of $\psi$?

Any suggestions or references appreciated.

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I was assuming the question meant something like "is there a formula for the mean curvature in terms of $\theta$, $u$, $\psi$, and first derivatives of these?". Is that correct? –  Matt Noonan Jan 23 '10 at 3:45

Yes, it is true $H=F(u,\theta,\psi,\psi')$.

You ask for suggestion: Calculate both principle curvatures, $k_1=k_1(u,\theta,\psi)$ and $k_2=k_2(u,\theta,\psi,\psi')$. Here $k_2$ is the curvature of original curve $r=u(\theta)$ and $k_1$ is principle curvature in the normal direction.

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Is this really the easiest approach? See my comment to Matt's answer. –  Deane Yang Jan 23 '10 at 15:49
The surface of revolution has reflection symmetry. So the principle directions are given. –  Anton Petrunin Jan 23 '10 at 15:59
Yes, you're right. And is it obvious that $u'$ is not needed? –  Deane Yang Jan 23 '10 at 16:15

To elaborate a little on the other answers, here's one approach: If you know the tangential angle $\psi$, then you know the angle $\pi/2 - \psi$ between the point and the outer unit normal, i.e. the Gauss map. Using $\theta$, $u(\theta)$, and $\pi/2 - \psi(\theta)$, you can write down an explicit formula for the Gauss map. Differentiate it to get the second fundamental form and take its trace (ADDED: You'll need to compute the first fundamental form to do this).

But this appears to require derivatives of both $u$ and $\psi$. It's possible that $u'$ disappears from the final formula, but I wouldn't know. (ADDED: I see from a comment that $u'$ is allowed in the formula. In that case, it all definitely works.)

ADDED: Forget what I wrote above. It's easier than that. As Anton points out, the principal curvature directions are the obvious ones. Therefore, it is easy to figure out the principal curvatures in terms of the generating curve (and the circle carved out by each point on the curve). The final answer depends on $\psi'$, $u$, and $u'$.

I now see why Anton was so brief. You really should work it all out yourself. It's a straightforward exercise.

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