Spheres in 4-D (Posted on 2003-12-20)

	Submitted by Brian Smith
Rating: 3.2000 (5 votes)
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Formula: V = pi^2 * R^4 / 2 Let the hypersphere be centered at the origin. A hyperplane perpendictular to an axis creates a cross section which is a 3-D sphere. If the hypersphere has a radius R and the axis is cut x units away from the origin, then the 3-D volume of the cross section is 4*pi*(sqrt(R^2-x^2))^3. The 4-D volume of the hypersphere can be found by integrating the cross section from -R to R. V = integral{-R to R},( 4/3*pi*( sqrt(R^2-x^2) )^3 ) dx V = 4/3 * pi * integral{-R to R},(R^2-x^2)^(3/2) dx To evaluate this integral, a trig substitution can be used. R*sin a = x (R*cos a) da = dx R*sin a = R -> a = pi/2 R*sin a = -R -> a = -pi/2 V = 4/3 * pi * integral{-pi*2 to pi*2},((R^2-R^2*(sin a)^2)^(3/2))*(R*cos a) da V = 4/3 * pi * R^4 * integral{-pi*2 to pi*2},(cos a)^4 da After using identities (cos a)^2 = (1+cos(2a))/2 and (sin a)^2 = (1-cos(2a))/2 V = 1/6 * pi * R^4 * integral{-pi*2 to pi*2},(cos(4a) + 4*cos(2a) + 3) da Evaluating the ingetral yeilds: integral{-pi*2 to pi*2},(cos(4a) + 4*cos(2a) + 3) da = (1/4)*sin(4a) + 2*sin(2a) + 3a From that, V = 1/6 * pi * R^4 * ( ((1/4)*sin(2*pi) + 2*sin(pi) + 3*pi/2) - ((1/4)*sin(-2*pi) + 2*sin(-2*pt) - 3*pi/2) ) V = 1/6 * pi * R^4 * ( (0 + 0 + 3*pi/2)-(0 + 0 - 3*pi/2) ) = 1/2 * pi^2 * R^4

	Subject	Author	Date
	Puzzle Thoughts	K Sengupta	2023-03-05 21:11:52
	re: Answer is Calculus	Charlie	2003-12-28 11:19:08
	Answer is Calculus	CC	2003-12-28 05:07:09
	re: Generalized answer: Derivation of volume of a n-dimensional hypershere	SatanClaus	2003-12-21 14:26:38
	solution	Charlie	2003-12-20 14:10:20
	Generalized answer: Derivation of volume of a n-dimensional hypershere	SatanClaus	2003-12-20 13:55:49
	Hint	Penny	2003-12-20 12:00:02