This question came up while trying to calculate the density of states for an ideal gas in statistical mechanics. After some physics, I've found that I need to evaluate the following integral:
$$I=\int \delta\left (\sum_{i=1}^{N}\frac{|\vec{p_{i}}|^{2}}{2m}-E\right ) d^{3}p_{1}...d^{3}p_{N}$$ Where $m$ and $E$ are constants, and the integral is taken over $3N$ dimensional space.
I realise that $\sum_{i=1}^{N}\frac{|\vec{p_{i}}|^{2}}{2m}=E$, describes a $3N$ dimensional hypersphere, so $I$ would be an integration over its surface area. At this point I'm havinga it of touble visualizing the geometry of the integral.
My notes now define $P=\sum_{i=1}^{N}p_{i}^{2}$, and write $d^{3}p_i...d^{3}_N=dPP^{3N-1}d\Omega_{3N}$, where $\Omega_{3N}$, is the unit solid angle in $3N$ dimensions. From here, as the argument of the delta function has no angular dependence, the integral can be split up, to give:
$$I=\int d\Omega_{3N}\int \frac{P^{3N-1}\delta(P-\sqrt{2mE})}{P/m} dP$$
The first integral is clearly the surface area of the $3N$ dimensional hypersphere, which I've been able to calculate. The second integral is trivial, by setting $P=\sqrt{2mE}$, and evaluating the integrand.
Where I'm having trouble is in the change of variables to $P$ and $\Omega$. In my notes it is done quite arbitrarily. I think I understand it physically ($P$ is the length of the $3N$ dimensional momentum vector, which sweeps over the hypersphere), but mathematically I can't justify it. I've always had trouble with change of variables in this kind of situationa and any clarification would be much appreciated.
If you use the Fourier integral representation of the delta function,
$$\delta\left (\sum_{i=1}^{N}\frac{|\vec{p_{i}}|^{2}}{2m}-E\right ) = \frac{1}{2\pi} \int_{-\infty}^{\infty} d\omega \ e^{i \omega \left (\sum_{i=1}^{N}\frac{|\vec{p_{i}}|^{2}}{2m}-E\right )}$$
and change the order of the integrations, you will be left with Gaussian integrals over the $\vec{p}_i$-s and the final integral over $\omega$ can be solved using the $\Gamma$ function (it will just give you the surface area of the unit $3N-1$ sphere).