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The relation given below, given on Wikipedia https://en.wikipedia.org/wiki/N-sphere#Recurrences in recurrence section, establishes the relation between the volume of (n+1) ball and surface area of n-ball. $$S_{n}R^{n}=\frac{dV_{n+1}R^{n+1}}{dR}={(n+1)V_{n+1}R^{n}}$$ Where,$S_n$ and $V_n$ are surface area and volumes of unit balls in n dimesions. For $n=2$, $S_{n} = 2\pi$, $V_{n+1} = \frac{4 \pi}{3}$ i.e surface area of a circle and volume of a sphere respectively. Therefore the relation becomes, $$2\pi R^{2} = 4\pi R^2 = 4\pi R^2$$ I think the relation will hold if we say,

$$S_{n+1}R^{n}=\frac{dV_{n+1}R^{n+1}}{dR}={(n+1)V_{n+1}R^{n}}$$ can anyone check and confirm my understanding?

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    $\begingroup$ No way Wikipedia can be wrong! It's impossible... $\endgroup$
    – Klangen
    Oct 25, 2019 at 13:17

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What you think of as $S_n$ is what Wikipedia (and many other sources) would call $S_{n-1}.$

Here is a clue from the same Wikipedia article:

the $n$-sphere at the boundary of the $(n + 1)$-ball of radius $R$

So if $n=2,$ and if we agree that this means the $(n + 1)$-ball of radius $R$ is the interior of a three-dimensional sphere, then the $2$-sphere is a three-dimensional sphere. The unit circle is called the $1$-sphere.

If you look closer to the beginning of the article it says explicitly,

a $2$-sphere is an ordinary $3$-dimensional sphere in $3$-dimensional Euclidean space

The idea is that the dimension a sphere depends on what a small piece of the sphere looks like. If you cut out a tiny piece of a circle it's a slightly bent line segment, so it's one-dimensional. Another way to look at it is, if you're constrained to stay on the sphere, how many degrees of freedom of motion do you have? On the circle it's only one. (You can go forward or backward, that's considered one degree of freedom because it only takes one positive or negative number to say where you went.) On a sphere you can go north/south and you can also go east/west, so there are two degrees of freedom.

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  • $\begingroup$ Thanks for the answer, I think I am half way there. So would you say that surface area of an n-ball with volume $V_n$ is $S_{n-1}$ ?? $\endgroup$
    – SagarM
    Oct 25, 2019 at 14:05
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    $\begingroup$ If it's a unit $n$-ball, yes. $\endgroup$
    – David K
    Oct 26, 2019 at 1:57
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You are off by an index. $S_2$ is a 2-dimensional surface living in $\mathbb{R}^3$, so its area is $4\pi$.

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