Variable change in integral with spherical coordinates in $N$-dimensinal space I am having difficulty with this problem.
Given $f(x)$ is smooth enough, $x \in \mathbb{R}^N$. The spherical coordinate of $x$ is $x = \left(r, \sigma\right)$, $r = \left\|x\right\|$, $\sigma \in S^{N-1}\left(0,r\right)$. Assuming $f(x) = u(Cr)$, for some constant $C$.
So how can I transform $\int\limits_{B(0,1)}f(x)dx$ to the term of some integral including $u(r)$ respected to $dr$?
I think we can write
$$\int\limits_{B(0,1)}f(x)dx = \int\limits_{0}^1 \int\limits_{S^{N-1}}u(Cr)drd\sigma = \int\limits_{0}^1 u(Cr)dr \int\limits_{S^{N-1}}d\sigma = \int\limits_{0}^1 u(Cr) \cdot 4\pi r^2 dr = 4 \dfrac{\pi}{C^3} \int\limits_{0}^C t^2 u(t) dt.$$
I feel that something does not add up here, but I don't know what it is.
Thank you very much for your help.
 A: I know, that the answer is done and accepted, but I would like suggest one fine way, different from suggested by me in comments (I started it and have yet hidden), in hope it will be useful for somebody.
Firstly suppose we would like to calculate volume
$$V_n=\int\limits_{x_1^2+x_2^2+\cdots+x_n^2 \leqslant R^2}dx_1\cdots dx_n$$
Taking new coordinats $x_1=R\xi_1,\cdots, x_n=R\xi_n$ it's easy to obtain $$V_n=\beta_nR^n$$
where $\beta_n$ is volume of $n$-dimensional sphere with radius $1$.
Then we will have
$$\beta_n=\int\limits_{x_1^2+x_2^2+\cdots+x_n^2 \leqslant 1}dx_1\cdots dx_n=\int\limits_{-1}^{1}dx_n\int\limits_{x_1^2+x_2^2+\cdots+x_n^2 \leqslant 1-x_n^2}dx_1\cdots dx_{n-1}$$
On right hand side inner integral is $(n-1)$-dimensional sphere with radius $\sqrt{1-x_n^2}$ and, therefore, is equal to $\beta_{n-1}\big(\sqrt{1-x_n^2}\big)^{n-1}$. Now using famous Beta function we  have
$$\beta_n = \beta_{n-1}\cdot \sqrt\pi \cdot\frac{\Gamma \left(\frac{n+1}{2} \right)}{\Gamma \left(\frac{n+2}{2} \right)} =\cdots = \frac{\pi^{\frac n2}}{\Gamma \left(\frac{n}{2}+1 \right)}$$
where we used $\beta_1=2$.
A: The volume element ($d\vec x$ or $dV$) in $n$-dimensional spherical coordinates is $r^{n-1}\, dr\, d\sigma$, where $d\sigma$ is a volume form on $S^{n-1}$. (In the familiar cases, for two dimensions $dx\, dy = r\, dr\, d\theta$ and for three dimensions $dx\, dy\, dz = (\rho^2\, d\rho)(\sin \phi\, d\phi\, d\theta)$.)
So it should be:
$$ \int_{B^n(0,1)} f(x)\, dx = \int_{S^{n-1}} \int_0^1 u(Cr)\, r^{n-1}\, dr\, d\sigma = \frac{1}{C^n} \int_{S^{n-1}} d\sigma \cdot \int_0^C t^{n-1} u(t)\, dt $$
The unit hypersphere volume is
$$ \int_{S^{n-1}} d\sigma = \frac{2 \pi^{n/2}}{\Gamma\left(\frac{n}{2}\right)} $$
or in equivalent forms for even and odd $n$,
$$ \int_{S^{2k-1}} d\sigma = \frac{2 \pi^k}{(k-1)!} $$
$$ \int_{S^{2k}} d\sigma = \frac{2 (k!) (4\pi)^k}{(2k)!} $$
