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I'm working out some examples for surfaces in differential geometry. I was working out simple rotational surface, but I think I've done something wrong. Let $\gamma\left(t\right)$ a curve parametrized with length of arc and given by $$\gamma\left(t\right)=\left(a\left(t\right),\,0,\,b\left(t\right)\right)$$ Let us consider the rotational surface given by $$\varphi\left(\theta,t\right)=R_{z}\left(\theta\right)\gamma\left(t\right)=\left(a\left(t\right)\cos\theta,\,a\left(t\right)\sin\theta,\,b\left(t\right)\right).$$. I then have tangent and normal vectors $$ \frac{\partial\varphi}{\partial\theta}= \left(-a\left(t\right)\sin\theta,\,a\left(t\right)\cos\theta,\,0\right),$$ $$\frac{\partial\varphi}{\partial t}= \left(\dot{a}\left(t\right)\cos\theta,\,\dot{a}\left(t\right)\sin\theta,\,\dot{b}\left(t\right)\right),$$ $$N= \left(\dot{b}\left(t\right)\cos\theta,\,\dot{b}\left(t\right)\sin\theta,\,-\dot{a}\left(t\right)\right).$$ So calculating the first and second fundamental form I should have $$E=\dot{a}\left(t\right)^{2}+\dot{b}\left(t\right)^{2}, \,\,F=0, \,\,G=a\left(t\right)^{2},$$ $$e=a\left(t\right)\left(\dot{a}\left(t\right)\ddot{b}\left(t\right)-\dot{b}\left(t\right)\ddot{a}\left(t\right)\right), f=0, g=a\left(t\right)^{2}\dot{b}\left(t\right).$$ And curvature $$K=\frac{a\left(t\right)\dot{b}\left(t\right)\left(\dot{a}\left(t\right)\ddot{b}\left(t\right)-\dot{b}\left(t\right)\ddot{a}\left(t\right)\right)}{\dot{a}\left(t\right)^{2}+\dot{b}\left(t\right)^{2}}.$$ Now If I try with the torus as a special case then $$\gamma\left(t\right)=\left(R+r\cos\left(t\right),\,0,\,r\sin\left(t\right)\right).$$ And first and second fundamental forms are $$E=r^{2}, \,\,F=0, \,\,G=\left(R+r\cos\left(t\right)\right)^{2},$$ $$e=r^{2}\left(R+r\cos\left(t\right)\right), f=0, g=\left(R+r\cos\left(t\right)\right)^{2}r\cos\left(t\right).$$ And therefore curvature is $$K=r\cos\left(t\right)\left(R+r\cos\left(t\right)\right).$$ What am I doing wrong?

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  • $\begingroup$ Didn't check all, but (1) the third component of $\frac{\partial \varphi}{\partial \theta}$ should be zero, (2) it seems that you did not normalize $N$. $\endgroup$ – user99914 Jul 27 '16 at 14:50
  • $\begingroup$ indeed the third component was 0, I copied wrong $\endgroup$ – Dac0 Jul 27 '16 at 15:02
  • $\begingroup$ For the normalization of N, I forgot to write that the curve was parametrized for length of arc so the the formula of N should be ok... $\endgroup$ – Dac0 Jul 27 '16 at 16:28
  • $\begingroup$ Did you really checked that the length of $N$ is one? Which curve was parametrized by arc length? $\endgroup$ – user99914 Jul 27 '16 at 17:27
  • $\begingroup$ Yes I did, the curve $\gamma$ is the one parametrized by lenght of arc. Indeed we have $\dot{a}\left(t\right)^{2}+\dot{b}\left(t\right)^{2}=1$ $\endgroup$ – Dac0 Jul 27 '16 at 23:26
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The key issue was that $\dot{a}\left(t\right)^{2}+\dot{b}\left(t\right)^{2}=1$ because the curve was parametrized for lenght of arc. Then we had $$ \frac{\partial\varphi}{\partial t}= \left(\dot{a}\left(t\right)\cos\theta,\,\dot{a}\left(t\right)\sin\theta,\,\dot{b}\left(t\right)\right),$$ $$\frac{\partial\varphi}{\partial\theta}= \left(-a\left(t\right)\sin\theta,\,a\left(t\right)\cos\theta,\,0\right),$$ $$ N= \left(\dot{b}\left(t\right)\cos\theta,\,\dot{b}\left(t\right)\sin\theta,\,-\dot{a}\left(t\right)\right).$$ First fundamental form was $$ E=1, \,\,F=0, \,\,G=a\left(t\right)^{2},$$ and second fundamental form had an error $$e=\dot{a}\left(t\right)\ddot{b}\left(t\right)-\dot{b}\left(t\right)\ddot{a}\left(t\right), f=0, g=a\left(t\right)\dot{b}\left(t\right).$$ Now the curvature is $$K=\frac{\dot{b}\left(t\right)\left(\dot{a}\left(t\right)\ddot{b}\left(t\right)-\dot{b}\left(t\right)\ddot{a}\left(t\right)\right)}{a\left(t\right)}.$$ Case of the Torus: $$ \gamma\left(t\right)=\left(R+r\cos\left(t\right),\,0,\,r\sin\left(t\right)\right).$$ And if we want $\gamma\left(t\right)$ to be parametrized for lenght of arc we need$r^{2}=1$ So that $$E=1, \,\,F=0, \,\,G=\left(R+r\cos\left(t\right)\right)^{2},$$ $$e=1, f=0, g=\left(R+r\cos\left(t\right)\right)r\cos\left(t\right).$$ And the curvature is correctly $$K=\frac{r\cos\left(t\right)}{R+r\cos\left(t\right)}.$$

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