Before diving into integral calculus, let's develop some intuition. Enclose the sphere inside a cube whose coordinates run from -1 to 1. In 2 dimensions the circle takes up almost all of the square's area, π out of 4 to be exact. Even in 3 dimensions the sphere takes up most of the cube. This is not true in 100 dimensions. The sphere touches the cube along the axes, but watch what happens as we follow the diagonal line from the origin to the far corner of the cube. When all coordinates are 0.1 we reach the boundary of the sphere. That's only a tenth of the distance to the corner of the cube. The remaining 90% is all empty space. We will not be surprised to see the volume of the sphere drop to 0 as n approaches infinity.

In one dimension the "sphere" is a line segment of length 2. In two dimensions the sphere is a circle, having area π. Proceed by induction on n.

Use polar coordinates to describe the first two dimensions of an n dimensional sphere. For every point r,θ, there is a sphere in n-2 dimensions with radius sqrt(1-r2). Let v be the volume of the unit sphere in n-2 dimensions. Scale by the radius, and the volume of the sphere at a distance r from the origin is v×(1-r2)½(n-2). Integrate with respect to r, and then θ. Don't forget the extra factor of r, for polar coordinates.

∫∫ vr(1-r2)½(n-2)

We are integrating r times a power of 1-r2. The result is -v(1-r2)½n/n. Evaluate at 0 and 1 and get v/n. Then integrate with respect to θ and bring in a factor of 2π. Therefore vn = 2π/n×vn-2.

Set n = 3 and confirm the volume of the 3-sphere = 4π/3. The volume of the 4-sphere is π2/2, and the volume of the 5-sphere is 8π2/15. For larger values of n, n exceeds 2π, hence the volume of the n-sphere approaches zero.

The volume of an ellipsoid is the volume of the corresponding unit sphere, vn, multiplied by the lengths of the semi-axes. If the sphere has radius r it's volume is vnrn.

As the radius increases, the sphere encloses an ever-larger volume. The surface of the sphere is always perpendicular to its outward motion. Therefore the surface area is the derivative of volume. Differentiate rn to get an extra factor of n. If an is the surface area then an = nvn. In other words, the surface area of the unit hypersphere is volume times dimension. The circumference of the unit circle is 2 times π, or 2π, as expected. The surface area of the unit sphere is 3 times 4π/3, or 4π. Here is the volume and surface area for the first 10 dimensions. As you can see, the greatest volume occurs at dimension 5, but the greatest surface area occurs at dimension 7. Like volume, surface area approaches 0 as n approaches infinity.

Dimension | Volume | Area |
---|---|---|

1 | 2.0000 | 2.0000 |

2 | 3.1416 | 6.2832 |

3 | 4.1888 | 12.5664 |

4 | 4.9348 | 19.7392 |

5 | 5.2638 | 26.3189 |

6 | 5.1677 | 31.0063 |

7 | 4.7248 | 33.0734 |

8 | 4.0587 | 32.4697 |

9 | 3.2985 | 29.6866 |

10 | 2.5502 | 25.5016 |