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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 $l= \frac{(-f(v)\cos(u),-f(v)\sin(u),0)}{\sqrt{f^2(v)(1+f'^2(v))}}$,

$m=\frac{(-f'(v)\sin(u),f'(v)\cos(u),0)}{\sqrt{f^2(v)(1+f'^2(v))}}$,

$n=\frac{f''(v)\cos(u),f''(v)\sin(u),1)}{\sqrt{f^2(v)(1+f'^2(v))}}$

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.

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  • $\begingroup$ I guess you have forgotten to take dot product with the normal vector (something like that) $\endgroup$
    – user99914
    Feb 5, 2015 at 3:19
  • $\begingroup$ Yes that is exactly what I have forgotten to do. $\endgroup$
    – tellap
    Feb 5, 2015 at 16:28

1 Answer 1

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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 $$N(u,v) = \frac{\operatorname{sign}(f(v))}{\sqrt{1 + f'(v)^2}}(\cos(u), \sin(u), -f'(v))$$

Since

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

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  • $\begingroup$ 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, 2015 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, 2015 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, 2015 at 17:07
  • $\begingroup$ No. $N$ is a vector valued function (with unit length). $\endgroup$
    – kobe
    Feb 5, 2015 at 17:10
  • $\begingroup$ Ok so I don't understand how the numerator became f(v) $\endgroup$
    – tellap
    Feb 5, 2015 at 17:15

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