Re: Correction on confusion with triple integral...

• To: mathgroup at smc.vnet.net
• Subject: [mg32511] Re: [mg32488] Correction on confusion with triple integral...
• From: Tomas Garza <tgarza01 at prodigy.net.mx>
• Date: Thu, 24 Jan 2002 05:21:06 -0500 (EST)
• References: <200201230600.BAA18200@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

Wow! You've really hit on something weird, which has to do - I guess - with
the fact that the lower limit in the innermost integral, i.e., 1/Cos[x], is
indeterminate at x = ArcCos[1/4]. If you evaluate the innermost integral you
get

In[1]:=
Integrate[r^2, {r, 1/Cos[x], 4}]
Out[1]=
64/3 - Sec[x]^3/3

so that the integrand in the middle integral is

In[2]:=
Sin[x]*Integrate[r^2, {r, 1/Cos[x], 4}]
Out[2]=
Sin[x]*(64/3 - Sec[x]^3/3)

The fishy thing is that the value of the definite integral of this last
function is different according to whether it is obtained using Integrate or
NIntegrate!

In[3]:=
Integrate[Sin[x]*(64/3-Sec[x]^3/3),{x,0,ArcCos[0.25]}]
Out[3]=
43/2
In[4]:=
NIntegrate[Sin[x]*(64/3-Sec[x]^3/3),{x,0,ArcCos[0.25]}]
Out[4]=
13.5

where the second one is correct. Let's hope that one of the Mathematica
gurus will give us some explanation.

Tomas Garza
Mexico City

----- Original Message -----
To: mathgroup at smc.vnet.net
Subject: [mg32511] [mg32488] Correction on confusion with triple integral...

> My apologies.  After reading Bob's message I realized I had a typo.  The
> limits on r were supposed to be [1/Cos[x],4], not [0,1/Cos[x]].  So, here
is
> the corrected problem...
>
> Please consider the Integral[r^2Sin[x]drdxdy] with limits as follows:
>
> [0,2Pi]for y, [0,ArcCos[1/4]] for x and [1/Cos[x],4] 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?
>

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