Assume a curve $\gamma(t) = (r(t), 0, h(t))$, and its revolution surface: $X(t, \theta) = (r(t)\cos\theta, r(t)\sin\theta, h(t))$.

After some calculations I find my first and second fundamental form to have expressions identical to this pdf, at page 02, that is: $$ M_I = \begin{bmatrix} \dot{r}^2 + \dot{h}^2 & 0 \\ 0 & r^2 \\ \end{bmatrix} ,\quad\quad\quad M_{II} = \frac{1}{\sqrt{\dot{r}^2 + \dot{h}^2}} \begin{bmatrix} -\ddot r\dot h + \ddot h\dot r & 0 \\ 0 & r\dot h \\ \end{bmatrix} $$

Now, it is my understanding that, gaussian curvature calculates: $$ k = \frac{\det M_{II}}{\det M_I} = \frac{r\dot h\left(-\ddot r\dot h + \ddot h\dot r\right)}{r^2\left(\dot{r}^2 + \dot{h}^2\right)\sqrt{\dot{r}^2 + \dot{h}^2}} = \frac{\dot h}{r}\frac{-\ddot r\dot h + \ddot h\dot r}{\left(\dot{r}^2 + \dot{h}^2\right)^{3/2}} $$

However, everywhere seems to say that correct result is: $$ k = \frac{\dot h}{r}\frac{-\ddot r\dot h + \ddot h\dot r}{\left(\dot{r}^2 + \dot{h}^2\right)^2} $$

I fail to see why is it squared, and not $3/2$ from the determinant calculation. However, after some physical checking, I know the later to be correct. I might be making a tiny arithmetic mistake somewhere, but, I couldn't find it.

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    $\begingroup$ Err... I just found it ~1min after posting. Actually, $\det(\lambda A) = \lambda^2\det A$, and not $\lambda\det A$. Silly me. It took me almost an hour in this. =(. $\endgroup$ – Physicist137 Aug 7 '17 at 17:41
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    $\begingroup$ since you found the mistake, you can convert your comment into an answer and accept it :-) $\endgroup$ – Ivo Terek Aug 7 '17 at 19:32

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