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The motivation for this module is to find the volume (often referred to as hypervolume) of an object in dimension ($n$). This has physical meaning for ($n\leq 3$), but what happens for ($n \geq 4$)?

The cube in dimension ($n$)

Consider the unit cube (i.e. the cube of side length 1) in ($n$) dimensions, sometimes called the ($n$)-hypercube or just the ($n$)-cube. Formally, this is defined to be the set of ($n$)-tuples (i.e. lists of length ($n$)) ($(x_1,x_2,\ldots,x_n)$) such that ($0 \leq x_i \leq 1$) for all ($1 \leq i \leq n$). For ($n=0,1,2,3$), 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 ($n=0$), 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 ($n=1$), the cube is a line segment. The volume in one dimension is just length, so the one dimension cube has volume 1.

For ($n=2$), the cube is a square of side length 1. In two dimensions, volume is area, so the cube in two dimensions has volume ($ w \times h = 1 \times 1 = 1$).

For ($n=3$), the cube is a (traditional) cube of side length 1, which has (traditional) volume ($ l \times w \times h = 1 \times 1 \times 1 = 1$).

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

Surface area of cubes

Consider the surface area of the cube in dimension ($n$). As with volume, this has physical meaning for ($n=2$) and ($n=3$).

For ($n=2$), the surface area of a square is really its perimeter, which is 4.

For ($n=3$), 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 ($n$) dimension cube will have ($2n$) boundary faces, and each face is a cube of dimension ($n-1$), so the surface area (really the hypervolume of the boundary) is ($2n$).

Other features

The diagonal of the ($n$)-cube can be defined to be the distance from ($(0,0,\ldots,0)$) to ($(1,1,\ldots,1)$). Using the distance formula, one finds that the diagonal of the ($n$)-cube is ($\sqrt{n}$).

The number of corners is fairly easy to count. For ($n=0,1,2,3$), the number of corners is 1, 2, 4, and 8 respectively. Since the ($n$)-cube can be thought of as two copies of the ($(n-1)$)-cube, one can show by induction that there are ($2^n$) corners in the ($n$)-cube.


A simplex is a generalization of a triangle or a pyramid. In dimension ($n$), the simplex is defined to be the set of ($n$)-tuples ($(x_1,x_2,\ldots,x_n)$) such that ($0 \leq x_i \leq 1$) and ($ \sum x_i \leq 1$). This can be thought of as the corner of the ($n$) dimension cube where the sum of the coordinates is less than 1. Here are the simplices of dimension ($n = 0,1,2,3$):

Volume of spheres in arbitrary dimension

Now, consider a sphere of radius ($r$) in ($n$) dimensions. This is the set of points ($(x_1,x_2,\ldots,x_n)$) such that ($x_1^2+x_2^2+\ldots+x_n^2 \leq r^2$). Let ($V_n(r)$) be the volume of the sphere of radius ($r$) in ($n$) dimensions (as above, volume means length, area, volume, hypervolume for ($n=1,2,3,\ldots$), respectively). With some careful integration and induction, one finds that

(:latex:) \begin{equation*} V_n(r) = \begin{cases} \frac{\pi^k}{k!} r^n & \hbox{if $n=2k$} \\ \frac{2^n \pi^k k!}{n!}r^n & \hbox{if $n=2k+1$} \end{cases}. \end{equation*} (:latexend:)

Now, note that as ($n \rightarrow \infty$) (and ($r$) stays fixed), the volume goes to 0 (since factorial grows faster than exponentials).


  • Consider a four-dimensional box (or "rectangular prism") with side-lengths ($1$), ($1/2$), ($1/3$), and ($1/4$). 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 ($(R,G,B)$), each taking a value in ($ [0,1] $).
Since the red/green/blue values are independent, each pixel has associated to it a 3-d cube of possible color values. Consider a (fairly standard) 10-megapixel camera. If I were to consider the "space of all images" that my camera can capture, what does the space look like? How many dimensions does it have? Note: there's no calculus in this problem...just counting!
  • Consider an ($n$)-dimensional "hypercube" ($C$) of all side-lengths equal to ($1$). Its ($n$)-dimensional volume is, clearly, ($1$). Now consider what happens when you shrink the hypercube's side-lengths by ($1$) percent (concentrically, so that the shrunken cube has the same center as the original) and remove it from the original cube. By subtracting the ($n$)-dimensional volume of this slightly smaller hypercube, conclude how much volume remains in the ($1$)-percent outer "shell".
  • In the previous question, what happens to the volume of the ($1$)-percent shell as ($n\to\infty$)?
  • We have seen that the ($n$)-dimensional volume of a unit radius ball in dimension ($n$) converges to zero as ($n\to\infty$). But what about a really large ball? For a ball of radius ($R=10^{10}$) meters in dimension ($n$), what is the limit as ($n\to\infty$) of its volume? (in unit of meters-to-the-($n^{th}$))
  • For the brave: so, as ($n\to\infty$), the volume of the ($n$)-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 ($V_n$) as ($I_n\cdot V_{n-1}$), where
($ \displaystyle I_n = \int_{\theta=-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^n\theta\, d\theta. $)
We can compute the volume of the equatorial slice of thickness ($2\epsilon$) (for some small but fixed ($\epsilon>0$)) as
($ \displaystyle V_{n,2\epsilon} = V_{n-1}\int_{-\epsilon}^\epsilon\cos^n\theta\, d\theta . $)
So, here is the (hard!) problem. Compute the limit as ($n\to\infty$) of the ratio of ($V_{n,2\epsilon}$) to ($V_n$):
($ \displaystyle \lim_{n\to\infty} \frac{V_{n,2\epsilon}}{V_n} = \lim_{n\to\infty} \frac{1}{I_n}\int_{-\epsilon}^\epsilon\cos^n\theta\, d\theta . $)
If you can do this (a very big if...) you will get a surprise...
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Page last modified on October 15, 2014, at 11:10 AM