Hair Curler Surface Calculation with Calculus

[CeeAnne]

This isn't homework. It's maths curiosity.
I have a hair curler set. The curlers are spool shape, 3 inches long, 1/2 inch diameter ends, 3/8 inch at the middle.
The surface is a surface of revolution. The curve generating the surface is a circular arc, concave on the curler. Assume a smooth surface and ignore the end area.
I have a picture of the curler at the bottom of my main webpage.
http://www.geocities.com/fran3406/Cee_Ann.html

Here's how I did it, but I want to know how to do it using Calculus. I don't know Calculus except for some of the really basics but I should be able to understand a solution for this problem.

Okay, If I roll the curler out one circumference and imagine the net that would roll back up to make the surface of the curler, I get a rectangular area which is 3 by 0.5pi minus a parabola having base 3 and height 0.5pi - 0.375pi. Imagine all the circumferences from end to end laid out next to each other to form the total area which should be a rectangle with two parabolic notches along each of the 3 inch ends. You could have one parabola of the calculated height or two parabolas half that height ... total area in each case is the same. The height of the single parabola is 0.392699081, so its area is 2/3 of 3 times 0.392699081 which equals 0.785398161 which is subtracted from 3 times 0.5pi to get a surface area of 3.926990819 square inches.
If this is correct, it simplifies to (3/4pi)/3.

 

[Andrew Mason]

Here's how I did it, but I want to know how to do it using Calculus. I don't know Calculus except for some of the really basics but I should be able to understand a solution for this problem.

You have taken a very good approach. You have to break up the complicated surface into surfaces whose areas we can compute and then add them up. Newton did it by dividing it up into infinitesimally tiny surfaces whose areas were known and then adding them up, which is what (integral) calculus is all about.

How do you know the area of a parabola? It is 2/3 base x height, but I am not aware of any geometric proof of this. The only way I know of determining it is through calculus. For example: the area of a parabola y=x² is the area of the square xy = x³ less the area under the (upside down parabola) curve (which by calculus is x³/3). So you are actually using calculus to solve this problem but didn't know it!

Now a few comments on your method. As you point out, the length of the shortest strands of the net (middle) are .375 pi and the longest (edge) are .5pi and this difference is spread out over two sides. But how do you know that the curve of the 3 inch edge of the net is a parabola? Also, does your net take into account the fact that the length of the curler along its curved surface is actually longer than the curler width?

Now using calculus: one would divide the curler up into an arbitrarily large number of arbitrarily thin slices of thickness dl and add up the surface areas. Each slice would have a surface area of cylinder of radius r and thickness dl: $A = 2\pi rdl$. Now if we knew exactly how r (1/2 thickness of curler at any given point along the curler length) varied with l (the distance of that point from one end of the curler) we could solve the problem. To find this, we have to know the radius of the circle that makes the arc of the roller's surface. Can you work that out? Work it out and post your answer. We can then go on from there to solve the problem.

AM

 

[CeeAnne]

I discovered the parabola rule, empirically, years ago while playing with simpler area under the curve problems. It seemed there ought be something for parabolas because there was an area formula for the ellipse. So, I experimented to find something similar. It wasn't complicated and it seemed to work so I adopted it. It applies only to complete parabolas and their halves but with a few tricks, parts of the parabola can be managed. I've tested it by comparison to problems solved by integration. However, this year I found it had been published in a recent Schaum's Mathematical Handbook of Formulas and Tables.

I don't know the curve of the net is a parabola. I didn't find its equation so I couldn't check values at various points. I did use known parabolas in reverse to generate what seem to be shallower parabolas tending toward the circle. It is an assumption made because it looked like a parabola and because the circle and parabola are conics.

Okay, I figured the radius for the arc on the curler's surface to be 18 and 1/32 inches, that is 18.03125.

 

[CeeAnne]

Mr. Mason, since we know the radius, and it would be the same radius for a sphere, and the curler would roll perfectly on the sphere, and if we rolled it exactly one revolution; isn't there a simple method to determine the area it rolled on? I mean, we can also determine the total surface of the sphere and the length and width of the curler's path and the angles involved from the center of the sphere. Why is it necessary to determine the edge surface of an infinitude of little discs? This begins to seem more suited to Solid or Analytic Geometry than Calculus. I need to check into how I might do this. At the least, it's a fun neato concept.

 

[Andrew Mason]

Okay, I figured the radius for the arc on the curler's surface to be 18 and 1/32 inches, that is 18.03125.

Ok. What we want to do is approximate a parabola to the curler curve. If you place the middle of the curler axis at the origin (0,0) of a graph, the curve of the curler can be approximated by a parabola in the form $$x^2=4p(y-d)$$ where p is the focus of the parabola. We have to do it so that when x=0 y = 3/16 = .1875 (the radius at the middle) and when x=1.5 (at the end) y = 4/16 = .25.

Have a look at this website to see what we are trying to do: http://mathdemos.gcsu.edu/bulbdemo/

So: d has to be equal to .1875

$$x^2 = 4p(y-.1875)$$ and

$$p = x^2/4(y-.1875)$$

Putting in the values x = 1.5 and y = .25:

$$p = 1.5^2/4(.25-.1875)= 9$$

So you see, you can approximate the circular arc of radius 18 by a parabola with focus 9.

This means the equation of our curve is:

$$x^2 = 36(y-.1875)$$ or
$$y = x^2/36+.1875$$

So by dividing the curler into little cylindrical ring sections of width dx and surface area [itex]A=2\pi ydx[/tex] and substituting our formula for y we get the following for area A of the right half of the curler:

$$A = \sum_{x=0}^{x=1.5} 2\pi(x^2/36 +.1875)dx$$

This is written in calculus form as:

$$A = \int_{0}^{1.5} 2\pi(x^2/36 +.1875)dx$$ which means the same thing.

Now the solution to this kind of equation was figured out by Newton. He showed that:

$$A = \int_{0}^{L} 2\pi(x^2/36 +.1875)dx = 2\pi(\frac{1}{3}L^3/36 +.1875L)$$

So substitute L = 1.5 and that will give you exactly 1/2 of the area (just the part to the right of the middle). So double that to get the area of the whole curler.

Check to see how it compares to what you measure if you wrap a piece of plastic wrap around it so it just covers the roller.

AM

 

[CeeAnne]

Oh, Mr. Mason, that is truly fantastic. I am so impressed. And the answers agree. This is amazing. I really do appreciate what you have done for me. Thank you ever so much. Hugs. CeeAnne

 

[CeeAnne]

How do you know the area of a parabola? It is 2/3 base x height, but I am not aware of any geometric proof of this. The only way I know of determining it is through calculus. For example: the area of a parabola y=x² is the area of the square xy = x³ less the area under the (upside down parabola) curve (which by calculus is x³/3). So you are actually using calculus to solve this problem but didn't know it!
AM

This is really neat. In Calculus - Early Transcendental Functions, 2nd edition, by Larson - Hostetler - Edwards (Chapter 4, page 322, Problem 107), the question opens with this statement:

Archimedes showed that the area of a parabolic arch is equal to 2/3 the product of the base and the height ...

Isn't that neat?

 

[Andrew Mason]

Archimedes showed that the area of a parabolic arch is equal to 2/3 the product of the base and the height ...

Isn't that neat?

Not only is it 'neat', Archimedes used what appears to be a form of calculus to prove it, 2000 years before Newton and Liebniz! See:
http://www.math.ubc.ca/~cass/archimedes/parabola.html Thanks for pointing this out.

AM