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Area under a helix

  1. Mar 31, 2005 #1
    I have been long thinking on how one would set up an integral to find the area under a helix (the area between the helix and the line it is encircling). I couldn't get a good expression for dA, because the infinitessimal cuts were all bent, and I wasn't sure how I could accurately approximate them with flat sections.

    I was planning on finding a general form for the area under a graph as it was being rotated around an axis (in a helix-kind-of-way), but I was stumped at the most basic case >.<

    Any help appreciated.
     
  2. jcsd
  3. Mar 31, 2005 #2

    DaveC426913

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    It seems to me, that any arbitrary section of a helix (say a section that rotates through 90 degrees) will be equivalent in area to that of an arc with the same outer and inner edge lengths.

    See attached pic.

    Don't take my word on it though. I'm not schooled in this area, and this is just intuition.
     
    Last edited: Nov 28, 2006
  4. Mar 31, 2005 #3
    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 = [tex]\sqrt{v^2 + 1}[/tex]. To get the area of the portion of this surface between u=0, u=2[tex]\pi[/tex], v=0, and v=1, we would evaluate the integral [tex]\int_0^1 \int_0^{2\pi} \sqrt{v^2 + 1} du dv[/tex].
     
    Last edited: Mar 31, 2005
  5. Mar 31, 2005 #4
    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

    [tex]dL = \sqrt{dx^2+2a^2-2a\cos{\frac{2\pi dx}{b}}}[/tex]

    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.
     
  6. Mar 31, 2005 #5
    Regarding the surface area, given that the integrand is independent of u, the iterated integral is just a product of two single variable integrals.
    Regarding arc length, there's no need to resort to infinitesimal series. See http://mathworld.wolfram.com/ArcLength.html . What path did you follow to get your expression ?
     
    Last edited: Mar 31, 2005
  7. Mar 31, 2005 #6
    It was a bit overcomplicated how I got the arclength expression. I figured [tex]dL=\sqrt{(dx)^2+(dy)^2}[/tex]. 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 [tex]dy=\sqrt{2a^2-2a\cos{d\theta}}[/tex], where [tex]d\theta=\frac{2\pi dx}{b}[/tex]. Then I plugged that into my original equation.
     
  8. Apr 1, 2005 #7
    A good way to do calculus on a helix is with paremeterization:

    [tex]x = cos(t) [/tex]

    [tex]y = sin(t) [/tex]

    [tex]z = t [/tex]

    Then the element of arclength is:

    [tex] dL = \sqrt{(\frac{dx}{dt})^2 +(\frac{dy}{dt})^2 +(\frac{dz}{dt})^2} dt[/tex]

    Which is pretty easy to integrate.
     
  9. Apr 1, 2005 #8
    ... I feel stupid >.<

    Of course that's how you'd do it.

    Hmm... after working with it, it seems to simplify to [tex]dL=\sqrt{2}dt[/tex]. When you integrate that, it just comes to [tex]L=b\sqrt{2}[/tex] where b is the length of the portion of the helix. Very simple indeed. Thank you.
     
  10. Aug 2, 2011 #9
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