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# sphere

# 1 Sphere

A *sphere* is defined as the locus of the points in three dimensions that are equidistant from a particular point called the *center*. Note that the center of a sphere is unique.

It is generally assumed that the sphere is embedded in real-valued space ($\mathbb{R}^{3}$) unless otherwise stated.

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

A *unit sphere* is a sphere with radius 1.

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

The formula for the surface area of a sphere with radius $r$ is

$A=4\pi r^{2}.$ |

# 2 Generalization

A sphere can be generalized to $n$ dimensions. For $n>3$, a generalized sphere is called a *hypersphere* (when no value of $n$ is given, one can generally assume that “hypersphere” means $n=4$). In the same manner, the definitions of center, radius, and unit sphere can also be generalized to $n$ dimensions.

The formula for an $n$-dimensional sphere is

${x_{1}}^{2}+{x_{2}}^{2}+\dots+{x_{n}}^{2}=r^{2}$ |

where $r$ is the length of the radius. Note that when $n=2$, the formula reduces to the formula for a circle, so a circle is a 2-dimensional “sphere”. A one dimensional sphere is a pair of points (filled-in, it would be a line)!

The volume of an $n$-dimensional sphere with radius $r$ is

$V(n,r)=\frac{\pi^{{\frac{n}{2}}}r^{n}}{\Gamma(\frac{n}{2}+1)}$ |

where $\Gamma(n)$ is the gamma function. Curiously, for any fixed $r$, the volume of the $n$-d sphere approaches zero as $n$ approaches infinity. Contrast this to the volume of an $n$-d box, which always has a volume in proportion to $s^{n}$ (with $s$ the side length of the box) which increases without bound when $s\geq 1$. Note that, for any positive integer $n$ and for any radius $r$, $V(n,r)=V(n,1)r^{n}$. Also note that the volume of the $n$-d unit sphere $V(n,1)$ has a maximum precisely at $n=5$.

To illustrate how to use the formula for $V(n,r)$ and to provide some evidence of the claims made about $V(n,r)$, the values $V(4,1)$, $V(5,1)$, and $V(6,1)$ will be calculated here.

$\begin{array}[]{lll|lll|ll}V(4,1)&=\displaystyle\frac{\pi^{{\frac{4}{2}}}1^{4}% }{\Gamma(\frac{4}{2}+1)}&&V(5,1)&=\displaystyle\frac{\pi^{{\frac{5}{2}}}1^{5}}% {\Gamma(\frac{5}{2}+1)}&&V(6,1)&=\displaystyle\frac{\pi^{{\frac{6}{2}}}1^{6}}{% \Gamma(\frac{6}{2}+1)}\\ &&&&&&&\\ &=\displaystyle\frac{\pi^{2}}{\Gamma(3)}&&&=\displaystyle\frac{\pi^{2}\sqrt{% \pi}}{\Gamma(\frac{7}{2})}&&&=\displaystyle\frac{\pi^{3}}{\Gamma(4)}\\ &&&&&&&\\ &=\displaystyle\frac{\pi^{2}}{2}&&&=\displaystyle\frac{\pi^{2}\sqrt{\pi}}{% \frac{15}{8}\sqrt{\pi}}&&&=\displaystyle\frac{\pi^{3}}{6}\\ &&&&&&&\\ &\approx 4.9348&&&=\displaystyle\frac{8\pi^{2}}{15}&&&\approx 5.1677\\ &&&&&&&\\ &&&&\approx 5.2638&&&\end{array}$

# 3 Topological Treatment

In topology and other contexts, spheres are treated slightly differently. Let the $n$-*sphere* be the set

$S^{n}=\{x\in\mathbb{R}^{{n+1}}:||x||=1\}$ |

where $||\cdot||$ can be any norm, usually the Euclidean norm. Notice that $S^{n}$ is defined here as a subset of $\mathbb{R}^{{n+1}}$.

Thus, $S^{0}$ is two points on the real line:

$S^{1}$ is the unit circle:

$S^{2}$ is the unit sphere in the everyday sense of the word. It might seem like a strange naming convention to say, for instance, that the $2$-sphere is in three-dimensional space. The explanation is that $2$ refers to the sphere’s “intrinsic” dimension as a manifold, not the dimension to whatever space in which it happens to be immersed.

Sometimes this definition is generalized even more. In topology we usually fail to distinguish homeomorphic spaces, so all homeomorphic images of $S^{n}$ into any topological space are also called $S^{n}$. It is usually clear from context whether $S^{n}$ denotes the specific unit sphere in $\mathbb{R}^{{n+1}}$ or some arbitrary homeomorphic image.

## Mathematics Subject Classification

51M05*no label found*00A05

*no label found*

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## Comments

## volume?

I thought there was some problem defining volume on R^3 and this is where banach tarski paradox came from, on the other hadn, volume for a sphere is a known fact from elementary school, and this puzzles me a bit, what's going on?

f

G -----> H G

p \ /_ ----- ~ f(G)

\ / f ker f

G/ker f

## Re: volume?

drini writes:

> I thought there was some problem defining volume on

> R^3 and this is where banach tarski paradox came

> from, on the other hadn, volume for a sphere is a

> known fact from elementary school, and this puzzles

> me a bit, what's going on?

The Banach-Tarski Paradox shows that there's no way to give *every* bounded subset of R^3 a volume (at least, not if you want the volume to have the sort of properties you would expect a volume to have). But Lebesgue measure gives every a volume to every reasonably nice subset of R^3.

## Re: volume?

That last sentence was supposed to read "But Lebesgue measure gives a volume to every reasonably nice subset of R^3."

## Re: volume?

Well, there is really no problem. Remember how you defined area back in the middle school (when it is taught in your country). Draw the figure on a fine square grid, count the number of squares, multiply by the area of each square, and voila -- this is your area. Of course, what happens here is known to students of calculus as an integration, i.e., as the squares get smaller and smaller the sum of the areas of the squares covered by the figure approaches some limit, which we call area... or it does not approach any limit. That is the precisely the so-called problem in defining the area (or volume for that matter): the limit might not exist. You can try to cicumvent the problem by defining more and more elaborate limiting processes (Lebesgue integral is one of them) that allow you define area for more and more sets, but the problem is that you can't have a definition that works for any set. To make the explanation easier to follow I will suppose that we want to define a length of a set on the line in such a way that it satisfies some very natural conditions:

1. The length of a larger set is larger.

2. The length of a line segment [0,1) is 1.

3. If you translate your set by any amount in any direction, then the length will be preserved.

4. If you have countably many disjoint sets that the length of their union is the sum of their lengths (example you have [0,1/2) and [1/2,3/4), [3/4,7/8), ... then you want their union [0,1) to have the length 1/2+1/4+1/8+...=1).

Then there is no definition length that would satisfy these restriction. To see that consider an equivalence relation on [0,1) that sets numbers x and y to be equal in under this relation if x-y is a rational number. From each equivalence class, pick number, and let A be the collection of such chosen numbers. What is the length of A? Well, it can't be zero, because if we consider the translations of A by every rational number between -1 and +1 then they are pairwise disjoint, and their union contains [0,1), and so by the condition one it has to be at least 1, but by the condition 4 it has to be 0. On the other hand, it can't be positive because the union of all translations as above is contained in [-1,2) which is a union of [-1,0), [0,1) and [1,2) all of which should have length 1 by the condition 3; however, it is a countable union of sets of positive and same length, and so the length better be infinite).

I guess you might argue that the condition 4 is too strong, and you only want finite unions as opposed to countable ones. I guess you could do that. But then however you have a psychological problem that you cannot say that the length of [0,1) is the same as the length of the union of [0,1/2),[1,1+1/4),[2,2+1/8), and so forth.

Regarding Banach-Tarski paradox. This is a construction conceptually similar to the construction presented above. The set for which the limit in the definition of volume is partitioned into a number of parts for some of which the limit does not exist, then they are rearranged in a way that leaves no ``dirt'' behind like the above when the union is diffused between containing [0,1] and being contained in [-1,3] and you really can't say much about where it is in between. Why 3 is the smallest dimension where you can do such a trick in, is something I do not know much about, and so I hope someone else will shed more light on this.

Boris