Cavalieri's principle for surface measures In the book by Stein and Weiss the integral $$\omega_{n-1}\int_0^{\pi/2} \sin^{n-1}\theta \,\mathrm{d}\theta$$ is investigated, where $\omega_k$ is the surface measure of the $k$-dimensional unit-sphere in $k+1$. They do it in the following way: First, they observe that $\omega_{n-1} \sin^{n-1} \theta$ is the surface measure of an $n-1$-dimensional sphere with radius $\sin \theta$, which one gets by intersecting the $n$-dimensional sphere with the hyperplane where $x_{n+1}=\cos \theta$. Integrating $\theta$ from $0$ to $\pi/2$ then yields hyperplanes whose intersections cover half the $n$-sphere, hence the integral is $\omega_n/2$.
The geometric idea is clear to me and moreover I was able to justify this argumentation by a rigorous calculation. The idea in the calculation is that of Cavalieri: Divide into slices and integrate the measures of the slices. It is well-known that this works with the Lebesgue measure and in fact it is a special instance of Fubini's theorem using characteristic functions of measurable sets.
My question is: What is the reason that it works in this case also with surface measures and are there reasonable conditions for surface measures to posses a Cavalieri principle?
 A: Reason 1: Well, we just believe it. It is geometrically so much appealing :)
Reason 2: It is again Fubini's theorem, but after you (implicitly) pullback the upper half of the $n$-sphere to a domain in $\mathbb{R}^n$. You can imagine a map that sends $\mathbb{R}^n = \mathbb{R}^{n-1} \times \mathbb{R}$ onto the $n$-sphere, embedded in $\mathbb{R}^{n+1}$, such that for fixed $\theta \in \mathbb{R}$, the horizontal $\mathbb{R}^{n-1} \times \{\theta\}$ gets mapped to the $(n-1)$-sphere at the right height. $\theta = 0$ gives the largest such $(n-1)$-sphere, the equator, and $\theta = 1$ gets crushed to the radius zero, the north pole.
Now, your original integral is transformed to an integral over a (bounded) region $U$ of $\mathbb{R}^n$, to which the usual Fubini's theorem applies:
$$ \int _U *** d\mathcal{L}^n = \int_{[0,1]} \left( \int_{U \cap \mathbb{R}^n} *** d\mathcal{L}^{n-1}  \right) d\theta \ .$$
Key Observation: Since those $(n-1)$-spheres, the parallels, are perpendicular to the meridians at all points, the Jacobian breaks up into Jacobian of the restriction to $\mathbb{R}^{n-1}$ times the Jacobian along $\{fixed \} \times \theta$, i.e. the meridians. So, we can pull out the latter part out of the inner integral and attach it to the $d \theta$ of the integration.
Now, once, $\theta$ is fixed, the inner integral can be transformed back to an integration on the image, which is the appropriate $(n-1)$-sphere. Then, you are integrating over the sphere at height $\theta$ and then integrating the resulting values against $d \theta$.
