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Re: Volume integrals in Content MathML

From: JB Collins <joebmath@yahoo.com>
Date: Tue, 4 Nov 2008 10:40:08 -0800 (PST)
To: David Carlisle <davidc@nag.co.uk>
Cc: www-math@w3.org
Message-ID: <852530.83152.qm@web57611.mail.re1.yahoo.com>
Dear David,

That certainly looks more promising - a little more general.
But, is it really still biased towards Cartesian-like coordinates?

Assume instead that we need to integrate over a two dimensional domain on the surface of a sphere. A natural basis might be two angle coordinates - a finite domain: not quite RxR. Certainly we can map a sphere to RxR, a la Riemann. But indicating the specifics of the domain seems to restrict to a class of bases.

Also, in general, a volume element is dxdydz = sqrt(det(g))*du1du2du3 where from a general transformation from Cartesian (x,y,z) to (u1,u2,u3) we pick up a factor of the Jacobian, the square root of the determinant of the metric tensor.
Actually, looked at this way, it seems clear that specifying the domain really is like specifying classes of coordinate bases, i.e., domain(u1) x domain(u2) x domain(u3), which is not always RxRxR.

As I understand it, what is usually meant by the domain with the integral notation

$\int \rho({\bf r}) d^3{\bf r}$

is "the whole 3 dimensional domain on which &rho is defined", without reference to any particular basis or restricted class of bases.

Joe Collins
--- On Tue, 11/4/08, David Carlisle <davidc@nag.co.uk> wrote:
From: David Carlisle <davidc@nag.co.uk>
Subject: Re: Volume integrals in Content MathML
To: joebmath@yahoo.com
Cc: www-math@w3.org
Date: Tuesday, November 4, 2008, 7:08 AM

  <bvar><ci> r </ci></bvar>
   <apply> <in/>
      <ci> r </ci>
   <apply><cartesianproduct/><reals/> <reals/>
    \rho({\bf r})...

mean the right thing?


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Received on Tuesday, 4 November 2008 18:40:50 UTC

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