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