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#1 2005-12-22 04:16:48
- John E. Franklin
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hexagon area with calculus
Make sure you integrate perpendicular to the side of the polygon.
I mean the x-axis cuts a side of the polygon in half.
The 12 in the following equation is the perimeter.
The reason for the :sqrt 3 in the denominator is because all of the
thousands of hexagons inside one another are similar or proportional.
So you can look at the biggest one, and when x is :sqrt 3, then
you want 12dx for that skinny donut area.
The thing that interests me is what happens at all the vertices?
It is surprising it works.
Last edited by John E. Franklin (2005-12-22 07:35:42)
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#2 2006-01-09 09:03:11
- irspow
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Re: hexagon area with calculus
John, you do not need calculus for finding the area of polygons.
Area = ns² / (4tan[180/n]) for anypolygon with equal sides numbering greater than 3
n = number of sides
s = length of side
This is used with an assumption of degrees being used and not radians for the trigonomic function. I am sure that you can figure out how to convert this to use for radians.
#3 2006-01-09 09:32:43
- mathsyperson
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Re: hexagon area with calculus
John already worked that out all by himself, here. Isn't he clever? 
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It wanted to be normal.
#4 2006-01-09 09:49:20
- irspow
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Re: hexagon area with calculus
I know that John is clever. He is always trying to work things out on his own. A great way of really understanding the logic behind many concepts. I will provide the proof over on the other thread because he seemed to want that there.