The hypersphere has the equation The Volume of a Hypersphere x 2 y 2 x 2 w 2 = 2 if centered at the origin (,,,) and has a radius of in four dimensional space. We approach the project of determining its volume inductively by first considering volumes of spheres in lower dimensions. We start with the sphere of radius in one dimension, centering it on the number line so that the center is at the origin, (). I. The Sphere in one dimension, x 2 = 2 - dx + We can find the volume (actually length in this case) of the sphere, shown as the heavy portion of the number line from to by integrating the line segment whose length is dx. Therefore V 1 = dx = 2 II The Sphere in Two Dimensions The two dimensional sphere is really a circle. Proceeding inductively, we add one more integral to compute the volume, which in this case is an area. Considering the equation x 2 y 2 = 2 1
we solve it for y to determine the extent of the sphere in the y-direction for a given value of x between and +. Y x, 2 x 2 x, 2 x 2 The only thing different here is the addition of the vertical axis. The sphere now extends from 2 x 2 to 2 x 2 in the vertical direction as well as from to in the horizontal direction as was the situation in case I. To obtain the volume we integrate a square whose dimensions are dy by dx as follows. V 2 = 2 x 2 dy dx = 2 x 2 dx = 1 2 2 = 2 2 x 2 We could have evaluated the double integral in polar coordinates more easily as shown below. V 2 = 1 r dr d = 2 2 d = 2 III The Three Dimensional Sphere We give here the volume of the ordinary sphere computed with three integrals in order to show how to set up the 4-dimensional case with four integrals! The two integrals used in case II are kept. The third integral has 2
limits specified by the extent of the sphere in the z-direction, determined by solving the sphere equation x 2 y 2 z 2 = 2 for z. To determine the extent of the sphere in the z-direction for a given pair of values (x,y), we solve for z to obtain 2 x 2 y 2 and 2 x 2 y 2. Z x, y, 2 x 2 y 2 X Y x, y, The diagram above shows the first octant of the sphere and the upper extent. Unseen is the lower extent at x, y, 2 x 2 y 2. These are the limits assigned to the third integral. We proceed to integrate the cube whose dimensions are dx by dy by dz. The triple integral set up is shown below. 2 y 2 2 x 2 y 2 = dz dy dx 2 y 2 2 x 2 y 2 We begin to compute the volume by doing the dz integral and converting the dx and dy integrals to polar coordinates. 3
= 2 = 2 2 x 2 2 x 2 2 x 2 y 2 dy dx 2 r 2 r dr d = 2 3 3 d = 4 3 3 IV The Four Dimensional Hypersphere At last we consider the volume for the hypersphere whose equation is x 2 y 2 z 2 w 2 = 2 This contains the w 2 term indicating that the hypersphere extends in another direction in space. This is a direction not specified by the x-direction, the y- direction, or the z-direction. To determine, analytically, the extent of the hypersphere in the w-direction for a given triple of values (x,y,z), we solve the hypersphere s equation for w. We get that w extends from 2 x 2 y 2 z 2 to 2 x 2 y 2 z 2. These will be the limits of the fourth integral in the expression which will be used to calculate the volume of the hypersphere. We integrate a hypercube of dimensions dx by dy by dz by dw by setting up the following four-integral expression. 2 x 2 2 x 2 y 2 2 x 2 y 2 z 2 = dw dz dy dx 2 x 2 2 x 2 y 2 2 x 2 y 2 z 2 The dw integral is easy to perform after which we utilize cylindrical coordinates, (r, θ, z), by converting to polar coordinates the region over which the first two integrals apply. 4
The inner integral, = 2 2 x 2 2 x 2 2 x 2 y 2 2 x 2 y 2 z 2 dz dy dx 2 x 2 y 2 = 2 2 r 2 2 r 2 z 2 dz r dr d 2 r 2 2 r 2 2 r 2 z 2 dz, deserves special attention. With 2 r 2 respect to z, r is a constant. Let K 2 = ( 2 - r 2 ), so that K is not negative. Then the integral becomes 2 r 2 2 r 2 K 2 r 2 z 2 dz = K 2 z 2 dz = 1 K 2 K 2 = 1 2 2 r 2 as the second integral is the area under a semi-circle of radius K. Substituting this result, we continue computing the volume of the hypersphere. = 2 1 2 2 r 2 r dr d = = 1 4 4 d 2 r r 3 dr d = 1 2 2 4 For readers who are interested in much more information on the hypersphere, check the website http://www.geocities.com/jsfhome/think4d/hyprsphr/hsphere.html There you can read an extensive discussion on this topic. 5