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Is the volume of a convex, solid $n-$dimensional body (containing the origin) equal to the $cE[R^n]$ where $c$ is the volume of a unit $n-$ball and $R$ is the distance between the origin and a point on the body located at a random angle (distributed according to the rotation-invariant distribution)?

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    I don't quite understand what you mean by "$R$ is the distance between the origin and a point on the body located at a random angle". I guess by $E[R^n]$ you are taking the expectation value, but over what probability space? There are many points of the body at a random angle. Do you mean $R$ to be the "radius" at the angle? In other words, do you mean that your body is equal to the set in polar coordinates $\{ (r,\omega); r < R(\omega) \}$ where $\omega\in \mathbb{S}^{n-1}$ ?2011-08-24
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    @Willie: I think "point on the body" is intended to mean "point on the surface of the body", so that would correspond to your definition in polar coordinates.2011-08-24
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    @Joriki: it would indeed, and it would also correlate with the formula below written by Sasha.2011-08-24

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You are correct.

To work out correct result, change coordinates to spherical coordinates:

$$ V = \int_S \mathrm{d}V = \int_{\mathbb{S}^{n-1}} \mathrm{d} \Omega(\alpha) \int_0^{R(\alpha)} r^{n-1} \mathrm{d} r = \int_{\mathbb{S}^{n-1}} \mathrm{d} \Omega(\alpha) \frac{1}{n} R(\alpha)^{n} = \frac{S_{n-1}}{n } \int_{\mathbb{S}^{n-1}} \frac{\mathrm{d} \Omega(\alpha)}{S_{n-1}} R(\alpha)^{n} $$

The last form is $ V = \left( \frac{S_{n-1}}{n} \right) \cdot \mathbb{E}( R^n )$.

But $c = \frac{S_{n-1}}{n } $ is exactly the volume of the unit ball in $\mathbb{R}^n$.

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    @Didie:r With $\alpha$ meaning collective coordinate on the unit sphere $S^{n-1}$.2011-08-25