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Re: Integrate this how?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg116344] Re: Integrate this how?
  • From: Simon <simonjtyler at gmail.com>
  • Date: Fri, 11 Feb 2011 04:17:43 -0500 (EST)
  • References: <ij0e6p$9bc$1@smc.vnet.net>

> 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?

I find that both 7 and 8 give basically the same results for the arc-
length, approx 48.37.
You can get rid of the error messages by bumping up MaxRecursion and
WorkingPrecision (but it will make the integral slower).
Or maybe try simplifying the integrand first.
You get a value of 5.65487 from doing this integral
NIntegrate[r[t], {t, -Pi, Pi}]
in either 7 or 8.


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