Integration Methods Beyond 3 Dimensions
Date: 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 Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.