# Don't know what trollpi is?

1. Mar 28, 2013

### mjordan2nd

If you don't know what trollpi is, you can see it here:

http://qntm.org/files/trollpi/piequals4.png [1]

It seems to me that the flaw in this problem is that while the shape may converge to a circle, the way you measure the perimeter is ds=dx+dy=1+(dy/dx) not ds=sqrt(1+(dy/dx)^2 )dx as usual. So I tried using this new definition of the arclength in the first quadrant with r=0.5 and integrating expecting to get 1. I put the following into mathematica:

Integrate[1 - x/Sqrt[0.25 - x^2], {x, 0, 0.5}]

This did not give me 1. Can someone explain why?

Last edited: Mar 29, 2013
2. Mar 29, 2013

### mjordan2nd

I think I may have figured out an answer to this. I started thinking that possibly the reason this is not 1 is because dy and dx can be positive or negative, so not all pieces of the arclength would add -- some would subtract. So I tried to do this

Integrate[Sqrt[(1 - x/Sqrt[1/4 - x62])2], {x, 0, 1/2}]

This was not correct either, however, since what I had effectively done was make ds=|dy + dx|. What I really needed was |dy|+|dx|=ds. Realizing that in quadrant I dy/dx was always negative all I needed to do was to flip the sign in my original integral. Plugging in

Integrate[1 + x/Sqrt[0.25 - x^2 ], {x, 0, 0.5}]

gave me 1, like I was expecting. It seems that my analysis was wrong because I effectively dropped a negative.

If someone could confirm that this makes sense, it would be appreciated.