A surface of revolution $\Sigma $ is produced by rotating a smooth curve in the $(x, z)$ plane about the $z$-axis. We parametrize the surface:
$$(x, y,z)= r(\gamma)\cos(\theta),r(\gamma)\sin(\theta ),z(\gamma),$$
where $ \gamma $ is arc-length along the original curve, so that $\dot{r}(\gamma)^2 + \dot{z}(\gamma)^2=1$.
I have to calculate the normal vector which is given by $$n(X)=\frac{\nabla(f(X))}{|\nabla(f(X))|}.$$
Can normal be calculated using this? $$S_{\gamma}=\big(\dot{r}(\gamma)\cos(\theta),\dot{r}(\gamma)\sin(\theta ),\dot{z}(\gamma)\big), \qquad S_{\theta}=\big( – r(\gamma)-\sin(\theta),r(\gamma)\cos(\theta),0\big).$$ So the surface normal vector is in the direction
$Q(\gamma,\theta)=S_{\theta}\times S_{\gamma}$.
[Math] How to find the normal on a surface of revolution
differential-geometrysurfaces
Best Answer
Let me change the notation a little. You are given a parametrized arc-length curve $$\alpha(v)=(f(v),g(v))$$ The surface generated by this curve $\alpha(v)$ when rotating along the $z$-axis can be parametrized as
$$\Phi(u,v)=(f(v)\cos u,f(v)\sin u,g(v))$$
The unit normal vector to the surface at a point $q\in \Phi(U)$, where $U$ is an open subset of $\Bbb{R^2}$ is
$$N(q)=\frac{\Phi_u\times \Phi_v}{|\Phi_u\times \Phi_v|}$$ where $\Phi_u$ and $\Phi_v$ are the partial derivatives of $\Phi(u,v)$.
On the other hand, the coefficients of the first fundamental form are given as $E=\langle\Phi_u,\Phi_u\rangle$, $F=\langle\Phi_u,\Phi_v\rangle$, $G=\langle\Phi_v,\Phi_v\rangle$. This yields in $E=(f(v))^2$, $F=0$ and $G=(f'(v))^2+(g'(v))^2=1$, since $\alpha(v)$ is parametrized by arc-length. Also, it is easy to show that $$|\Phi_u\times \Phi_v|=\sqrt{EG-F^2}$$ which finishes the calculation.