[Date Index] [Thread Index] [Author Index]
Re: Integrate this how?
I just tried NIntegrate on this in version 8. I first defined a function f[t], which is your arc length Simplify'ed. I do get some warnings about slow convergence, but the result is 48.37. Just estimating the length as a series of triangles and lines gives me about something around 50. Kevin On 2/10/2011 5:21 AM, BoLe wrote: > Curve is closed (t from -Pi to Pi) and in polar form > > r[t_]:=(1+Sin[t]) (1+9/10 Cos[8 t]) (1+1/10 Cos[24 t]) (9/10+5/100 > Cos[200 t]) > > Would like to find out its arc length, so the integral of > > Sqrt[r[t]^2+D[r[t],t]^2] > > Version 7 complains but returns a somewhat reasonable number > (something around 50; drew together with a circle of the same length), > version 8 says nothing but gives just too low a number (five and > something). How can this be? New Mathematica should have no problem > with such high oscillatory integrands, shouldn't she? >