Let $f(x)$ be a smooth function. Consider a surface of revolution, \begin{equation} M(u, v) = (f(v) \cos(u), f(v) \sin(u), v). \end{equation} (a) Calculate coefficients of the first and second fundamental forms for the surface;

(b) Calculate principal curvatures κ1, κ2, the Gaussian curvature K and the mean curvature H;

(c) Find the length of the portion of the normal line contained between a point of the surface and the axis of revolution (in the present case, the z-axis).

I have obtained the first fundamental form and tried working out the second fundamental form and obtained the following: \begin{equation} \frac{1}{\sqrt{f^2(v)(1+f'^2(v))}} (l du^2 + 2mdudv+n dv^2) \end{equation} where $l= \frac{(-f(v)\cos(u),-f(v)\sin(u),0)}{\sqrt{f^2(v)(1+f'^2(v))}}$,



I am not too sure whether my values for $l, m$ and $n$ is correct because from what I know, numerator for all 3 of them should be scalar and not be in vector form. Any help would be appreciated.

  • $\begingroup$ I guess you have forgotten to take dot product with the normal vector (something like that) $\endgroup$ – user99914 Feb 5 '15 at 3:19
  • $\begingroup$ Yes that is exactly what I have forgotten to do. $\endgroup$ – tellap Feb 5 '15 at 16:28

Since $M_u = (-f(v)\sin(u), f(v)\cos(u), 0)$ and $M_v = (f'(v)\cos(u), f'(v)\sin(u), 1)$, $M_u \wedge M_v = (f(v)\cos(u), f(v)\sin(u), -f(v)f'(v))$. Thus $\|M_u \wedge M_v\| = |f(v)|\sqrt{1 + f'(v)^2}$ and the unit normal is $$N(u,v) = \frac{\operatorname{sign}(f(v))}{\sqrt{1 + f'(v)^2}}(\cos(u), \sin(u), -f'(v))$$


\begin{align}M_{uu} &= (-f(v)\cos(u), -f(v)\sin(u), 0)\\ M_{uv}&= (-f'(v)\sin(u), f'(v)\cos(u), 0)\\ M_{vv}&= (f''(v)\cos(u), f''(v)\sin(u), 0) \end{align}

we have

\begin{align}l &= M_{uu} \cdot N = -\frac{|f(v)|}{\sqrt{1 + f'(v)^2}}\\ m &= M_{uv} \cdot N = 0\\ n &= M_{vv} \cdot N = \frac{\operatorname{sign}(f(v))f''(v)}{\sqrt{1 + f'(v)^2}} \end{align}

So your second fundamental form is $$II(u,v) = -\frac{|f(v)|}{\sqrt{1 + f'(v)^2}}\, du^2 + \frac{\operatorname{sign}(f(v))f''(v)}{\sqrt{1 + f'(v)^2}}\, dv^2$$

  • $\begingroup$ Thank you very much for this, I completely forgot to take the dot product. Could you just explain to me what sign(f(v)) is because I am not familiar with this notation. $\endgroup$ – tellap Feb 5 '15 at 16:28
  • $\begingroup$ Sure. The sign of $f(v)$, which I denoted $\operatorname{sign}(f(v))$, is equal to $+1$ if $f(v)$ is positive and $-1$ if $f(v)$ is negative. $\endgroup$ – kobe Feb 5 '15 at 16:29
  • $\begingroup$ Ok that makes sense thanks. Although shouldn't $N$ be $\frac{1}{f(v) \sqrt{1+f'(v)^2}}$ ? $\endgroup$ – tellap Feb 5 '15 at 17:07
  • $\begingroup$ No. $N$ is a vector valued function (with unit length). $\endgroup$ – kobe Feb 5 '15 at 17:10
  • $\begingroup$ Ok so I don't understand how the numerator became f(v) $\endgroup$ – tellap Feb 5 '15 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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