# 10.4: D- Volume of a Sphere in d Dimensions


I will call the volume of a d-dimensional sphere, as a function of radius, Vd(r). You know, of course, that

$V_{2}(r)=\pi r^{2}$

(two-dimensional volume is commonly called “area”) and that

$V_{3}(r)=\frac{4}{3} \pi r^{3}.$

But what is the formula for arbitrary d? There are a number of ways to find it. I will use induction on dimensionality d. That is, I will use the formula for d = 2 to find the formula for d = 3, the formula for d = 3 to find the formula for d = 4, and in general use the formula for d to find the formula for d + 1. This is not the most rigorous formal method to derive the formula, but it is very appealing and has much to recommend it.

To illustrate the process, I will begin with a well-known and easily visualized stage, namely deriving V3(r) from V2(r). Think of a 3-dimensional sphere (of radius r) as a stack of pancakes of various radii, but each with infinitesimal thickness dz. The pancake on the very bottom of the stack (z = −r) has zero radius. The one above it is slightly broader. They get broader and broader until we get to the middle of the stack (z = 0), where the pancake has radius r. The pancakes stacked still higher become smaller and smaller, until they vanish again at the top of the stack (z = +r). Because the equation for the sphere is

$x^{2}+y^{2}+z^{2}=r^{2},$

the radius of the pancake at height z0 is

$\sqrt{r^{2}-z_{0}^{2}}.$

This whole process shows that

$V_{3}(r)=\int_{-r}^{+r} d z V_{2}\left(\sqrt{r^{2}-z^{2}}\right).$

It is easy to check this integral against the known result for V3(r):

$V_{3}(r)=\int_{-r}^{+r} d z \pi\left(r^{2}-z^{2}\right)$

$=\pi\left[r^{2} z-\frac{1}{3} z^{3}\right]_{-r}^{+r}$

$=\pi\left[2 r^{3}-\frac{2}{3} r^{3}\right]$

$=\frac{4}{3} \pi r^{3}.$

So we haven’t gone wrong yet.

Now, how to derive V4(r) from V3(r)? This requires a more vivid imagination. Last time we started with a two-dimensional disk of radius r0 in (x, y) space and thickened it a bit into the third dimension (z) to form a pancake of three-dimensional volume dz V2(r0). Stacking an infinite number of such pancakes in the z direction, from z = −r to z = +r, gave us a three-dimensional sphere. Now we begin with a three-dimensional sphere of radius r0 in (w, x, y) space and thicken it a bit into the fourth dimension (z) to form a thin four-dimensional pancake of four-dimensional volume dz V3(r0). Stacking an infinite number of such pancakes in the z direction, from z = −r to z = +r, gives a four-dimensional sphere. Because the equation for the four-sphere is

$w^{2}+x^{2}+y^{2}+z^{2}=r^{2},$

the radius of the three-dimensional sphere at height z0 is

$\sqrt{r^{2}-z_{0}^{2}},$

and the volume of the four-sphere is

$V_{4}(r)=\int_{-r}^{+r} d z V_{3}\left(\sqrt{r^{2}-z^{2}}\right).$

In general, the volume of a (d + 1)-sphere is

$V_{d+1}(r)=\int_{-r}^{+r} d z V_{d}\left(\sqrt{r^{2}-z^{2}}\right).$

If we guess that the formula for Vd(r) takes the form

$V_{d}(r)=C_{d} r^{d}$

(which is certainly true for two and three dimensions, and which is reasonable from dimensional analysis), then

$V_{d+1}(r)=\int_{-r}^{+r} d z C_{d}\left(r^{2}-z^{2}\right)^{d / 2}$

$=\int_{-1}^{+1} r d u C_{d}\left(r^{2}-r^{2} u^{2}\right)^{d / 2}$

$=r^{d+1} C_{d} \int_{-1}^{+1} d u\left(1-u^{2}\right)^{d / 2}.$

This proves our guess and gives us a recursive formula for Cd:

$C_{d+1}=C_{d} \int_{-1}^{+1} d u\left(1-u^{2}\right)^{d / 2}.$

The problem below shows how to build this recursive chain up from C2 = π to

$C_{d}=\frac{\pi^{d / 2}}{\Gamma\left(\frac{d}{2}+1\right)}=\frac{\pi^{d / 2}}{(d / 2) !}.$

Thus the volume of a d-dimensional sphere of radius r is

$V_{d}(r)=\frac{\pi^{d / 2}}{(d / 2) !} r^{d}.$

D.1 (I) Problem: Volume of a d-dimensional sphere

Before attempting this problem, you should read the material concerning beta functions in an applied mathematics textbook, such as George Arfken’s Mathematical Methods for Physicists or Mary Boas’s Mathematical Methods in the Physical Sciences. (Or in the Digital Library of Mathematical Functions.)

a. Show that

$\int_{-1}^{+1}\left(1-u^{2}\right)^{d / 2} d u=B\left(\frac{1}{2}, \frac{d}{2}+1\right).$

b. Use

$B(p, q)=\frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)}$

and C2 = π to conclude that

$C_{d}=\frac{\pi^{d / 2}}{\Gamma\left(\frac{d}{2}+1\right)}.$

D.2 (I) Problem: Volume of a d-dimensional ellipse

Show that the volume of the d-dimensional ellipse described by the equation

$\left(\frac{x_{1}}{a_{1}}\right)^{2}+\left(\frac{x_{2}}{a_{2}}\right)^{2}+\left(\frac{x_{3}}{a_{3}}\right)^{2}+\cdots+\left(\frac{x_{d}}{a_{d}}\right)^{2}=1$

is

$V_{d}(r)=\frac{\pi^{d / 2}}{(d / 2) !} a_{1} a_{2} a_{3} \cdots a_{d}.$

This page titled 10.4: D- Volume of a Sphere in d Dimensions is shared under a CC BY-SA license and was authored, remixed, and/or curated by Daniel F. Styer.