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Spheres in 4-D (Posted on 2003-12-20) Difficulty: 5 of 5
Derive the formula for the 4-D volume of a hypersphere.

See The Solution Submitted by Brian Smith    
Rating: 3.2000 (5 votes)

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Some Thoughts Generalized answer: Derivation of volume of a n-dimensional hypershere | Comment 2 of 6 |
Holy Abstract theoretical question batman! An n-dimensional hypershere is also known as an "n-sphere". It is expressed by the equation (x_1)²+(x_2)²+(x_n)²=r² whereby generalized extension of the distance formula has r labeled as the radius of said "n-sphere". Familiarity with a "n-sphere" brings about the notion that it actually merely a sequence of (n-1) spheres of variable radius, let's call them r(t)=√(r²-t²) and let t uniformly vary between -r and +r. Thus the volume of a "n+1" sphere, denoted V(n+1)=Integral(t=-r to t=r) of V(n)r(t)dt. One should be convinced that the volume of an "n-sphere" is proportional to r^n. So lets denote this constant as k(n) for various values of n, then V(n) = k(n)*r^n. We arrive at V(n+1)= k(n)*Integral(t=-r to t=r) of r(t)^ndt. Substituing for r(t) we obtain V(n+1) = k(n)*r^n * Integral(t=-r to t=r) of (1-(t/r)²)^(n/2)dt. Letting u = (t/r) the mess reduces to V(n+1) = r*V(n)*Integral(u=-1 to u=1) of (1-u²)^(n/2)du. How to evaluate this integral precisely is escaping me right about now, besides I have lots of lumps of coal to deliver before Christmas. Good day you bad little boys and girls!
  Posted by SatanClaus on 2003-12-20 13:55:49
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