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Re: Integrate this how?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg116354] Re: Integrate this how?
  • From: "Kevin J. McCann" <Kevin.McCann at umbc.edu>
  • Date: Fri, 11 Feb 2011 04:19:31 -0500 (EST)
  • References: <ij0e6p$9bc$1@smc.vnet.net>

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?
>


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