Finding Area Under a Helix: A Stumper Problem

  • Thread starter Moo Of Doom
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In summary: It seems that the contact area between two helical surfaces is quite complicated to calculate. In the thread you linked, someone suggests using parametrization to simplify the problem.
  • #1
Moo Of Doom
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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.
 
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  • #2
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.
 
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  • #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].
 
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  • #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.
 
  • #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 ?
 
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  • #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.
 
  • #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.
 
  • #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.
 
  • #9
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1. What is the purpose of finding the area under a helix?

The purpose of finding the area under a helix is to calculate the total surface area of a three-dimensional helix. This can be useful in various fields such as engineering, physics, and mathematics.

2. How is the area under a helix calculated?

The area under a helix is calculated by using the formula A = πr√(r^2 + h^2), where r is the radius of the helix and h is the height of the helix. This formula takes into account the curved surface area of the helix as well as the area of the circular base.

3. What are some real-world applications of finding the area under a helix?

One real-world application of finding the area under a helix is in designing spiral staircases. The area under a helix can also be used to calculate the surface area of a coiled wire or spring, which is important in manufacturing. In physics, the area under a helix can be used to calculate the energy of a spinning object.

4. Are there any limitations to using the formula for finding the area under a helix?

Yes, the formula for finding the area under a helix assumes that the helix is a perfect mathematical shape, which may not be the case in real-world scenarios. Additionally, the formula does not take into account any irregularities or imperfections in the helix, which may affect the accuracy of the calculation.

5. Can the area under a helix be negative?

No, the area under a helix cannot be negative. This is because the area is a measurement of space and cannot have a negative value. If the calculation results in a negative number, it is likely due to an error in the input values or the formula being used.

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