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Integration Methods Beyond 3 DimensionsDate: 11/12/96 at 12:50:27 From: Rick Mendoza Subject: Integration Methods Beyond 3 Dimensions Are there any resources on the WWW which are relative to calculating areas of, for instance, a hypercube? I have taken an advanced Calculus course at SDSU (San Diego State University), but this curiosity was apparently beyond the scope of any professor's personal interest. Sincerely, Rick Mendoza
Date: 11/13/96 at 12:42:15
From: Doctor Tom
Subject: Re: Integration Methods Beyond 3 Dimensions
Hi Rick,
I don't know about resources on the Web, but you might consider
looking in some different advanced calculus texts.
The easiest way to think about it is to notice the pattern as the
dimensions increase. See what the problem looks like in 1, 2, and 3
dimensions, and it's usually clear how to go to 4 (or more).
For example, to think about a 4-D hypercube, think about "cubes" in 1,
2, and 3 dimensions first. In 1-D, it's a line segment, and the
"volume" is the length. In 2-D, it's a square, and the "volume" is
the area -- length times width. In 3-D, it's a cube, and the volume
is the volume -- length times width times height. So in 4-D, the
volume will be length times width times height times extent (I don't
know what you want to call the 4th dimension, so I just made up the
word "extent").
Or another way to think about it is by looking at a mathematical
definition of "cubes" in 1 through 4 dimensions:
1-D: a < x < b
2-D: a < x < b and
c < y < d
3-D: a < x < b and
c < y < d and
e < z < f
4-D: a < x < b and
c < y < d and
e < z < f and
g < w < h (I use w for the 4th coordinate)
The "volumes" of the examples above are:
1-D: (b-a)
2-D: (b-a)(d-c)
3-D: (b-a)(d-c)(f-e)
4-D: (b-a)(d-c)(f-e)(h-g)
You can do the same thing for other shapes. For example, what
is the definition of 'sphere' in the first four dimensions? Here are
some reasonable definitions:
1-D: x^2 < r^2
2-D: x^2 + y^2 < r^2
3-D: x^2 + y^2 + z^2 < r^2
4-D: x^2 + y^2 + z^2 + w^2 < r^2
To find the volumes of these spheres, just integrate over the
appropriate ranges:
1-D: int(-r, r, dx)
2-D: int(-r, r, dx)int(-sqrt(1-x^2), sqrt(1-x^2), dy)
3-D: int(-r, r, dx)int(-sqrt(1-x^2), sqrt(1-x^2), dy)
int(-sqrt(1-x^2-y^2), sqrt(1-x^2-y^2), dz)
4-D: int(-r, r, dx)int(-sqrt(1-x^2), sqrt(1-x^2), dy)
int(-sqrt(1-x^2-y^2), sqrt(1-x^2-y^2), dz)
int(-sqrt(1-x^2-y^2-z^2), sqrt(1-x^2-y^2-z^2), dw)
The stuff above is meant to indicate repeated integrals, and
the integrand in all cases is the constant 1.
I hope this helps some.
-Doctor Tom, The Math Forum
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