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Re: Bug of Integrate

  • To: mathgroup at
  • Subject: [mg83028] Re: Bug of Integrate
  • From: danl at
  • Date: Thu, 8 Nov 2007 06:06:05 -0500 (EST)
  • References: <fg4dfv$6c3$><fgs908$4m9$>

On Nov 7, 5:53 am, Miguel <misv... at> wrote:
> [...]
> By other hand, I think everybody knows the solution of the following
> problem:
> "Derive the formula for the circunference of a circle of radius "a" by
> computing the length of the arc     x=a cost; y= a sint  for 0<=t<=Pi
> "
> L=Integrate[Sqrt[1+(y'[t]/x'[t])^2]*x'[t],{t,0,2Pi}]

Not quite. It needs an absolute value for the last factor.

> I have tried to resolve with version 5.2 and version 6.0.1. The
> results have been differents.
> When the result is unknown, version 6.0.1 is reliable?

It seems straightforward to verify via numerical computation.

In[1]:= x = a*Cos[t]; y = a*Sin[t];

In[2]:= ll = Integrate[Sqrt[1+(D[y,t]/D[x,t])^2]*Abs[D[x,t]], {t,

Out[2]= 2 a Pi

In[3]:= a = 2;

In[5]:= NIntegrate[Sqrt[1+(D[y,t]/D[x,t])^2]*Abs[D[x,t]], {t,0,2*Pi}]

Out[5]= 12.5664

Without the Abs[] one (correctly) obtains 0 for both exact and numeric

By the way, when dealing with a parametrized curve, it is often
simpler to avoid implicitly treating y as a function of x; instead
just work directly with the parametrization.

In[7]:= a=.

In[8]:= ll = Integrate[Sqrt[D[x,t]^2+D[y,t]^2], {t,0,2*Pi},

Out[8]= 2 a Pi

Daniel Lichtblau
Wolfram Research

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