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It's an exercise from Munkres "Analysis on Manifolds" Chapter 5, "Integrating a scalar function over a manifold". Due to the suggestion, I'm repeating the question here: Express the volume of an n-sphere in terms of the volume of an n-1 dimensional ball.

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Not everyone reads the titles. You should put your questions in the body of your posts. –  Cameron Buie Nov 13 '12 at 20:39
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Similarly to Ross's answer, there's a beautiful relation between $V_{n-2}$ and $V_n$ that can be derived using standard integration, where $V_n$ is the volume of the $n$-ball. As a hint, it involves polar double integration over the radius and "location" of the "smaller" ball. This part is slightly tricky but if you get stuck I can post it.

After this, you can use the relationship between the surface area of the boundary of the $n$-ball and the volume of the $n$-ball. If you don't know this relation, it's not hard to figure out - what does it look like for the 3-ball and the 2-sphere? The 2-ball and the 1-sphere?

You're going to want to find the formulas described above for general $R$, even if you're working with $R=1$, or the second relation won't make much sense.

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Hint: If you think about a 3-ball, you can decompose it into 2-balls (disks) parallel to the $xy$ plane stacked along $z$. Given a $z$ coordinate, what is the radius of the disk? For a unit 3-ball, we have $V_3=\int_{-1}^1 dz V_2($that radius).

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Express the volume of $S^n(a)$ in terms of the volume $B^{n-1}(a)$. [\it Hint: \rm Follow the pattern of Example 2.]

\paragraph{\bf{sln}.} We can write \begin{eqnarray} S^n(a) &=& \{ (x_1, \cdots x_{n+1}) \; , \; x_1^2 + \cdots x_{n+1}^2 = a^2 \} \\ &=& \{ (x_1, \cdots x_{n-1}) \; , \; x_1^2 + \cdots x_{n-1}^2 = a^2 \cos^2 \theta \} \\ && \times \{ (x_{n},x_{n+1}) \; , \; x_n^2 + x_{n+1}^2 = a^2 \sin^2 \theta \} \end{eqnarray} (this equality is easy to show and I will omit its proof)

Then we parametrized the sphere based on the single parameter $\theta$ which we integrate between 0 and $\pi/2$. That is

\begin{equation} v(S^n(a)) = \int_0^{\pi/2} v(S^{n-2}(a \cos \theta)) \, v(S^1(a \sin \theta)) J d \theta \end{equation} The product of the two volumes is taken because we are integrating over cross product of independent spaces (for each fixed $\theta$). The Jacobian $J=a$ comes from the transformation from rectuangular coordinates to polar $(a, \theta)$ coordinates. To get this Jacobian requires a good amount of work. It is obvious for 2D when we say that $x_1^2 + x_2^2= a^2$ and then $x_1= a \cos \theta$ and $x_2 = a \sin \theta$, then the Jacobian

\begin{equation} J = \det \left ( \frac{\partial(x_1, x_2)}{\partial(a, \rho)} \right ) = a. \end{equation}

\begin{eqnarray} v(S^n(a)) &=& a \int_0^{\pi/2} v(S^{n-2}(a \cos \theta)) \; 2 \pi (a \sin \theta)) d \theta \\ &=& 2 \pi a \int_0^a v( S^{n-2}(\rho)) d \rho \\ &=& 2 a \pi v(B^{n-1}(a)). \end{eqnarray} with the substitution $\rho = a \cos \theta$, $d \rho = - a\sin \theta d \theta$, and recognizing that

\begin{equation} v(B^{n-1}(a)) = \int_0^a v( S^{n-2}(\rho)) d \rho. \end{equation} This integral is easy to send{equation} as thinking that an onion is the union of all its concentric shells.

To verify the result let us consider a few cases. \begin{itemize} \item For $n=2$ \begin{equation} v(S^2(a)) = 4 \pi a^2, \ 2 \pi B^1(a) = (2 \pi a) (2 \pi) (a) = 4 \pi a^2 \end{equation} \item For $n=3$

\begin{equation} v(S^3(a)) = 2 \pi^2 a^3, \\ 2 \pi a B^2(a) = (2 \pi a) (\pi^2)a^2) = 2 \pi^2 a^3 \end{equation}

\item and for $n=4$ \begin{equation} v(S^4(a)) = \frac{8}{3} \pi^2 a^4, \\ 2 \pi a B^3(a) = (2 \pi a) (\frac{4}{3} \pi a^3) = \frac{8}{3} \pi^2 a^3 \end{equation} \end{itemize} So, \begin{equation} \frac{v(S^n(a))}{v(B^{n-1}(a)} = 2 \pi a. \end{equation}

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