In my case, we consider an helicoid $\varphi(u,v)= (u \; sin\, v, u \; cos\, v,v).$

It's first fundamental form is $\mathit{I} = \begin{pmatrix} E & F\\ F & G \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & u^2 + 1 \end{pmatrix}.$

Now, we have to compute the area, the lenght of the sides and the angles of the triangle, defined by: $$0< u< sinh\, v,\; 0< v< a.$$

For the area I used the formula $A = \int_{0}^{a}\int_{0}^{sinh\, v}\sqrt{EG-F^2}dudv$.

For the sides I have done the following parametrizations: $$\alpha _1(t) = (0,t);\; 0< t< a.$$ $$\alpha _2(t) = (t,a);\; 0< t< sinh\, a.$$ $$\alpha _3(t) = (sinh\, t,t);\; 0< t< a.$$ For the lenght of the curves I have that if $\alpha (t)=(u(t),v(t))$ then: $$L = \int_{a}^{b} \sqrt{E(u')^2+2Fu'v'+G(v')^2}dt.$$ And finally, if $\alpha$ and $\beta$ are two curves that intersect in the surface they form an angle $\theta$ such that: $$cos\, \theta=\frac{\mathit{I(\alpha',\beta')}}{\left | \alpha' \right |\left | \beta' \right |}$$ The problem I have is that the $G$ has an $u^2$ and the lenght of the curves and their angles depends of $u$. So there is something I'm not understanding well. I'll appreciate any help. Thanks!


You've set this up well, except for one issue. All you have to do is realize where the vertices of the triangle are (in $(u,v)$ coordinates). One is at $(0,0)$, one is at $(0,a)$, and the last is at $(\sinh a,a)$. So are your curves the correct boundary curves of the triangle? (Hint: One is wrong.) But once you know where the vertices are, there's no problem computing the first fundamental form at this points to find angles.

  • $\begingroup$ $\alpha_2(t) = (t,a)$ right? $\endgroup$ – Sergi De la Torre Apr 25 '18 at 20:37
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    $\begingroup$ Yes, that's right. :) $\endgroup$ – Ted Shifrin Apr 25 '18 at 21:01
  • $\begingroup$ @TedShifrin So it means we only need to calculate the first fundamental form at the intersection point and that makes $u$ gone? In particular of the problem, if $\alpha (t_0)=\beta(s_0)=\varphi(u_0, v_0)$ then we have $G=u_0^2 +1$. Am I correct sir ? $\endgroup$ – RopuToran Jun 15 at 15:11
  • $\begingroup$ @RopuToran I don't know what your "and that makes $u$ gone" means. But, yes, you need $E,F,G$ at the intersection point(s). $\endgroup$ – Ted Shifrin Jun 15 at 15:33
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    $\begingroup$ @TedShifrin Sorry for my confusing expression. What I mean was when we substitute $u$ with value $u_0$ on $G$, the cosine function becomes a number, not a function anymore. Thank you! $\endgroup$ – RopuToran Jun 16 at 4:56

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