Volume of a parabole function with 3 axis? (4-dimensional volume)
Hi, I'm trying to figure out the formula for calculating the volume of part of a 3-axis parabole, with different lengths on each axis, but starting on (0,0,0).
My maths isn't that strong (especially terminology), so I have no idea what to google for, and please forgive me if I'm sounding like a noob, because I am ;).
Anyway, what I have been able to figure out are the calculations for the volume of the parabole when I use the same maximum value (m) for each axis:
1 axis:
(m*m)/3 * m
2 axes:
(m*m)/3 * m * (2*m)
3 axes:
(m*m)/3 * m * (2*m) * (1.5*m)
or simply m^5...
Any ideas on how to make different maximum values for x, y and z possible?
By the way: the formula for the parabole is simply x*x, (x*x+y*y), (x*x+y*y+z*z).
An example of what I'd like to do:
The volume of the graph (4-dimensional) in range (0,0,0) to (12,7,31) where w = (x*x+y*y+z*z). Brute-forcing this answer seems to give a value somewhere above 1,000,000.
Thanks!
[Edited by - Tree Penguin on August 14, 2009 6:33:19 AM]
Right, I had a collegue of mine crack this down, and he did it :] . The code is below for anyone who might need it for any reason.
Where (x1, y1, z1) to (x2, y2, z2) is a box in the 3d coordinate space of the graph.
EDIT: forgot some parenthesis.
[Edited by - Tree Penguin on August 14, 2009 10:43:53 AM]
inline float fabc(float a, float b, float c){ return (a*a*a * b * c + a * b*b*b * c + a * b * c*c*c)/3.0f;}float fXXYYZZ(float x1, float x2, float y1, float y2, float z1, float z2){ return ((fabc(x2,y2,z2) - fabc(x1,y2,z2)) - (fabc(x2,y1,z2) - fabc(x1,y1,z2))) - ((fabc(x2,y2,z1) - fabc(x1,y2,z1)) - (fabc(x2,y1,z1) - fabc(x1,y1,z1)));}
Where (x1, y1, z1) to (x2, y2, z2) is a box in the 3d coordinate space of the graph.
EDIT: forgot some parenthesis.
[Edited by - Tree Penguin on August 14, 2009 10:43:53 AM]
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