Counter Intuitive result for Surface area of Helical Ribbon

[jonthebaptist]

I am solving for the surface area of a helical ribbon that I represent as a ruled surface, the curve being the helix and the rulings being in the vertical $$\sigma\left(t,\varphi\right)=\left(\begin{array}{ccc} r\cos t, & r\sin t, & \omega t+\varphi\end{array}\right)$$
I solve for the terms in First Fundamental Form, $$E=r^{2}+\omega^{2}$$
$$F=\omega$$
$$G=1$$
Solving the surface area integral [tex]A=\iint\sqrt{EG-F^{2}}dtd\varphi[\tex]

I get an area equal to the product of the total angular rotation times the radius times the width of the ribbon. This is the area of a rectangle with a length equal to the circumference of the circle that is projected by the helix onto the x-y plane. What perplexes me is that I would think that if this ribbon was isometric to a rectangle it would be one with a length equal to the arc-length of the helix, not the projected circle.

 

[tiny-tim]

welcome to pf!

hi jonthebaptist! welcome to pf!

… What perplexes me is that I would think that if this ribbon was isometric to a rectangle it would be one with a length equal to the arc-length of the helix, not the projected circle.

slice the ribbon finely vertically, and let all the bits drop to the same level