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Re: Bug of Integrate
*To*: mathgroup at smc.vnet.net
*Subject*: [mg83028] Re: Bug of Integrate
*From*: danl at wolfram.com
*Date*: Thu, 8 Nov 2007 06:06:05 -0500 (EST)
*References*: <fg4dfv$6c3$1@smc.vnet.net><fgs908$4m9$1@smc.vnet.net>
On Nov 7, 5:53 am, Miguel <misv... at gmail.com> 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,
0,2*Pi},
Assumptions->a>0]
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
integrals.
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},
Assumptions->a>0]
Out[8]= 2 a Pi
Daniel Lichtblau
Wolfram Research
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