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 >

**References**:**confusion with triple integral...***From:*Bradley Stoll <BradleyS@Harker.org>