Express the volume of an n-sphere in terms of the volume of an n-1 dimensional ball 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.
 A: 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).
A: 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.
A: 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}
\end{document}
