I have an exam in a few days and I don't know how to solve this exercise from an old exam. Can you please help me?
Let $\alpha : \mathbb R -> \mathbb R^2 $ be the periodic parametrisation of a simply closed regular curve with positive curvature. Assume $\alpha$ is parametrised by arc length and let $n$ and $b$ be its normal and binormal vector respectively. Consider the tabular surface parametrised by $ f(s,t) = \alpha (s) + rn(s)cos(t) + rb(s)sin(t)$ with $ s,t \in \mathbb R , r>0$
- Prove that such a tubular surface is a regular surface for r small enough
- Compute the first and the second fundamental form
- Compute its Gaussian curvature as a function of the curvature and torsion of $\alpha$
- Show that the area of S is $2 \pi rl $
- Compute the Euler characteristic of the surface using the Gauss-Bonnet theorem
I found that a surface is regular iff its Jacobi matrix has rang 2 or iff the gradient is $\neq 0$ So I used Frenet formula, i.e. $T(s)= \dot c(s),\; \dot T(s)=k(s)N(s),\; \dot N(s) = -T(s)k(s) + \tau(s)B(s),\; \dot B(s)=-\tau (s)N(s) \; $ where $T, N, B, k\; $and$ \;\tau $ are respectively the tangential vector, the normal vector, the binormal vector, the curvature and the torsion. Than I computed the gradient of f and I obtained: $f_s = T(s)(1-k(s)rcos(t))+r cos(t)\tau (s)b(s) - rsin(t)\tau(s)n(s)$ and $f_t = -rn(s)sin(t) + rb(s)cos(t)$. We can compare the two result and obtain $f_s = T(s)(1-k(s)rcos(t)) + \tau (s)f_t$. But I can't go on.
Sorry, I'm not able to create a matrix here so I did it in Latex and I did a screenshot.
For the second point I don't know how to simplify the computation, since my formula are very long.
For 3. I need 2. for compute the Gaussian curvature that is given by $\frac{det(II)}{det(I)}$
For 4. I use the formula $A(f)= \int_s \int_t \sqrt det(I)$ where $t\in [0,2\pi]$ and since the curve is parametrised by arc length $s\in [0,l]$ where $l$ is the length of the curve. But I'm not sure.
Than for 5. I have no Idea... there is a reasoning ora a formula to use?
Thanks