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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!

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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.

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  • $\begingroup$ $\alpha_2(t) = (t,a)$ 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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