# Area under a helix

by Moo Of Doom
Tags: helix
 P: 506 Hmm, let's figure out the area element for your helical surface. I covered it with coordinates (u,v) such that v is the distance from the spiral axis and u was the distance from the "bottom" of the helix measured on the surface of the helical surface along a helix of radius v. Flattening this out to the familiar R^2, we get the set R x (0,1) for a helical surface with min radius of 0 and max radius of 1. Using this set, a parametrization of your helical surface would then be x = v*cos(u), y=v*sin(u), z=u. Then we have the area element dA = $$\sqrt{v^2 + 1}$$. To get the area of the portion of this surface between u=0, u=2$$\pi$$, v=0, and v=1, we would evaluate the integral $$\int_0^1 \int_0^{2\pi} \sqrt{v^2 + 1} du dv$$.
 P: 367 Area under a helix Ouch, an integral of an integral, I feared it was going to be something like this... Thanks. On a related note, how would you do arclength? I got it down to $$dL = \sqrt{dx^2+2a^2-2a\cos{\frac{2\pi dx}{b}}}$$ where dL is a portion of the arclength, dx is a portion of the helix (along the u-axis in hypermorphism's system), a is the radius of the helix, and b is the rate of curvature (measured in distance along u per revolution). But I couldn't get the dx out of the cosine, so I was stuck there... P.S. I was sure I was posting this in the Calculus board... strange how it ended up here... sorry about that.
 P: 367 It was a bit overcomplicated how I got the arclength expression. I figured $$dL=\sqrt{(dx)^2+(dy)^2}$$. But I had to get dy in terms of dx, so I used an isosceles triangle and the rate of curvature in order to calculate dy. I found $$dy=\sqrt{2a^2-2a\cos{d\theta}}$$, where $$d\theta=\frac{2\pi dx}{b}$$. Then I plugged that into my original equation.
 P: 1,295 A good way to do calculus on a helix is with paremeterization: $$x = cos(t)$$ $$y = sin(t)$$ $$z = t$$ Then the element of arclength is: $$dL = \sqrt{(\frac{dx}{dt})^2 +(\frac{dy}{dt})^2 +(\frac{dz}{dt})^2} dt$$ Which is pretty easy to integrate.
 P: 367 ... I feel stupid >.< Of course that's how you'd do it. Hmm... after working with it, it seems to simplify to $$dL=\sqrt{2}dt$$. When you integrate that, it just comes to $$L=b\sqrt{2}$$ where b is the length of the portion of the helix. Very simple indeed. Thank you.