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You are not logged in. #1 20051222 04:16:48
hexagon area with calculusMake sure you integrate perpendicular to the side of the polygon. 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 (20051222 07:35:42) igloo myrtilles fourmis #2 20060109 09:03:11
Re: hexagon area with calculusJohn, you do not need calculus for finding the area of polygons. #3 20060109 09:32:43
Re: hexagon area with calculusJohn already worked that out all by himself, here. Isn't he clever? Why did the vector cross the road? It wanted to be normal. #4 20060109 09:49:20
Re: hexagon area with calculusI 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. 