Volumes In Arbitrary Dimension

The motivation for this module is to find the volume (often referred to as hypervolume) of an object in dimension . This has physical meaning for , but what happens for ?

The cube in dimension

Consider the unit cube (i.e. the cube of side length 1) in dimensions, sometimes called the -hypercube or just the -cube. Formally, this is defined to be the set of -tuples (i.e. lists of length ) such that for all . For , these are familiar figures: the point, line segment, square, and cube, respectively.

Now, consider some of the various measurements for each of these cubes.

Volume of the cube

For , the cube is just a point, and volume is defined to just be the number of points. So a single point has volume 1.

For , the cube is a line segment. The volume in one dimension is just length, so the one dimension cube has volume 1.

For , the cube is a square of side length 1. In two dimensions, volume is area, so the cube in two dimensions has volume .

For , the cube is a (traditional) cube of side length 1, which has (traditional) volume .

For higher values of , this pattern continues. The intuition is that each additional dimension adds an extra factor of 1, thus the volume of each unit -cube is 1.

Surface area of cubes

Consider the surface area of the cube in dimension . As with volume, this has physical meaning for and .

For , the surface area of a square is really its perimeter, which is 4.

For , the surface area is the total area of the faces which bound the cube. There are 6 faces each with area 1, so the surface area is 6.

In general, the dimension cube will have boundary faces, and each face is a cube of dimension , so the surface area (really the hypervolume of the boundary) is .

Other features

The diagonal of the -cube can be defined to be the distance from to . Using the distance formula, one finds that the diagonal of the -cube is .

The number of corners is fairly easy to count. For , the number of corners is 1, 2, 4, and 8 respectively. Since the -cube can be thought of as two copies of the -cube, one can show by induction that there are corners in the -cube.


A simplex is a generalization of a triangle or a pyramid. In dimension , the simplex is defined to be the set of -tuples such that and . This can be thought of as the corner of the dimension cube where the sum of the coordinates is less than 1. Here are the simplices of dimension :

Volume of spheres in arbitrary dimension

Now, consider a sphere of radius in dimensions. This is the set of points such that . Let be the volume of the sphere of radius in dimensions (as above, volume means length, area, volume, hypervolume for , respectively). With some careful integration and induction, one finds that

Now, note that as (and stays fixed), the volume goes to 0 (since factorial grows faster than exponentials).


  • Consider a four-dimensional box (or "rectangular prism") with side-lengths , , , and . What is the 4-dimensional volume of this box?
  • What is the "diameter" -- i.e., the farthest distance between two points -- in this 4-d box? Hint: think in terms of diagonals.
  • High-dimensional objects are everywhere and all about. Let's consider a very simple model of the space of digital images. Assume a planar digital image (such as that captured by a digital camera), where each pixel is given values that encode color and intensity of light. Let's assume that this is done via an RGB (red/green/blue) model. Though there are many RGB model specifications, let us use one well-suited for mathematics: to each pixel on associates three numbers , each taking a value in .
Since the red/green/blue values are independent, each pixel has associated to it a 3-d cube of possible color values. Consider a (fairly standard) 10-megapixel camera. If I were to consider the "space of all images" that my camera can capture, what does the space look like? How many dimensions does it have? Note: there's no calculus in this problem...just counting!
  • Consider an -dimensional "hypercube" of all side-lengths equal to . Its -dimensional volume is, clearly, . Now consider what happens when you shrink the hypercube's side-lengths by percent (concentrically, so that the shrunken cube has the same center as the original) and remove it from the original cube. By subtracting the -dimensional volume of this slightly smaller hypercube, conclude how much volume remains in the -percent outer "shell".
  • In the previous question, what happens to the volume of the -percent shell as ?
  • We have seen that the -dimensional volume of a unit radius ball in dimension converges to zero as . But what about a really large ball? For a ball of radius meters in dimension , what is the limit as of its volume? (in unit of meters-to-the-)
  • For the brave: so, as , the volume of the -ball all concentrates near the surface shell. OK, you've got that. Now answer this: what proportion of the volume is concentrated along the "equatorial plane"? Let's make that specific. Recall, we computed the volume as , where
We can compute the volume of the equatorial slice of thickness (for some small but fixed ) as
So, here is the (hard!) problem. Compute the limit as of the ratio of to :
If you can do this (a very big if...) you will get a surprise...