Re: 4-dimensional Integral




A N Niel wrote:
> In article <1125085061.811875.156520@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> Randy Poe <poespam-trap@xxxxxxxxx> wrote:
>
> > Hsuan-Yeh Chang wrote:
> > > Dear All,
> > >
> > > Anyone wants to try this simple integral in four dimension:
> > >
> > > \int dx dy dz dt (x^2 + y^2 + z^2 + t^2)
> > >
> > > where the integration is restricted to the space:
> > >
> > > { (x, y, z, t) | 1 >= x^2 + y^2 + z^2 + t^2 }
> >
> > Volume of the unit 4-sphere.
> >
> > http://mathworld.wolfram.com/Hypersphere.html
> > http://www.mathreference.com/ca-int,hsp.html
> >
> > - Randy
> >
>
> without the integrand, that would be it

You're right, I answered too fast.

Another poster's suggestion sounds like the right way
to go. Convert to hyperspherical coordinates with
r^2 = x^2 + y^2 + z^2 + t^2, integrated from r=0 to 1.
The trick will be figuring out the correct transformation
of the volume element dx dy dz dt.

- Randy

.



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