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MathGroup Archive 2002

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Re: confusion with triple integral...

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32487] Re: [mg32469] confusion with triple integral...
  • From: Tomas Garza <tgarza01 at prodigy.net.mx>
  • Date: Wed, 23 Jan 2002 01:00:06 -0500 (EST)
  • References: <200201220819.DAA07095@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Unless I'm misreading your integrand (I presume you mean (r^2)*Sin[x]), the
correct answer that Mathematica returns (in my computer, anyway) is 5 Pi,
and this may be checked easily. The integrand doesn't depend on y, so in
fact you have only a double integral whose value is equal to 2.5. On
multiplying this by the integral of dy from 0 to 2 Pi, which is equal to 2
Pi, you get 5 Pi. Now, this is displayed as either 5 Pi or 15.708...(the
numerical value of 5 Pi) according to whether I use ArcCos[1/4] or
ArcCos[0.25] (or, equivalently, ArcCos[1/4]//N) in the corresponding limit.

Tomas Garza
Mexico City


----- Original Message -----
From: "Bradley Stoll" <BradleyS at Harker.org>
To: mathgroup at smc.vnet.net
Subject: [mg32487] [mg32469] confusion with triple integral...


> Please consider the Integral[r^2Sin[x]drdxdy] with limits as follows:
> [0,2Pi]for y, [0,ArcCos[1/4]] for x and [0,1/Cos[x]] for r.  Mathematica
returns
> 27Pi, which is correct.  But, if instead of using ArcCos[1/4], I use
either
> ArcCos[0.25] or ArcCos[1/4]//N in the limits, Mathematica returns 43Pi.
Any idea
> why?j
>
> Bradley Stoll
> San Jose, CA
>



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