[Date Index]
[Thread Index]
[Author Index]
Re: ArcCos[x] with x > 1
*To*: mathgroup at smc.vnet.net
*Subject*: [mg49450] Re: ArcCos[x] with x > 1
*From*: jaegerm at ibmt.fhg.de
*Date*: Tue, 20 Jul 2004 07:53:55 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
koopman at sfu.ca (Ray Koopman) wrote in cd35vb$qd5$1 at smc.vnet.net:
>> [...] Given is a cube of side length a (or, let's say, 1). What is
>> the mean distance of the cube's centre point to its surface.
>> Obviously, the result is between a / 2 (the distance to any of the
>> six centres of the six faces) and a / Sqrt[2] (the distance to any
>> of the eight corners). Sure, that problem can easily be reduced.
>> Think that sounds like a simple integration on kindergarden level?
>> So did I. But I soon got the impression I was wrong.
> Maybe because the distance from the centre to any corner is
> Sqrt[(a/2)^2 + (a/2)^2 + (a/2)^2] = (a/2)*Sqrt[3]? :)
Thanks, Ray. Of course you're right. When I wrote that, I had the
reduced form (plane square instead of a cube) in mind. This was meant
to be an estimate of the upper limit so I know the mean m lies in the
interval ]1/2;Sqrt[3]/2[ but that doesn't tell me where exactly. By
the way, m actually even lies in ]1/2;1/Sqrt[2][ :)
Magnus
Prev by Date:
**7 equations and 7 unknowns (non-linear equations)**
Next by Date:
**Re: Return different values of a list**
Previous by thread:
**Re: ArcCos[x] with x > 1**
Next by thread:
**question about Integrate**
| |