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I'm trying to wrap my head around the two-dimension recursion formula for which there are few threads here, but which weren't clarifying.

The Wikipedia article speaks of:

The volume of the ball can therefore be written as an iterated integral of the volumes of the (n − 2)-balls over the possible radii and azimuths.

Why are the $(n-2)$ balls iterated over radii and azimuths?


I tried to experiment with

$$\frac{2\pi r^2}{n}V_{n-2}(R)$$

which for $n=4$:

$$\frac{2\pi r^2}{4}\pi r^2=\frac{\pi^2r^4}{2}$$

which is the volume of a 4-sphere.

But I don't understand the integrations and why the go from $0$ to $2 \pi$ and $0$ to $r$. Nor why the integral is claimed to be the same for all $V_n$s and $V_{n-2}$s (surely the elements that are integrated are of different shape?).

I.e. why is the volume integral

$$V_4 (R) = \int_0^{2\pi} \int_0^R V_{n-2}$$

Also what is the $r$ in

$$\int_0^{2\pi} \int_0^R V_{n-2}(\sqrt{R^2-r^2})\,\color{red}r\, dr d\theta$$

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  • $\begingroup$ It would be much more clear if you wrote out the integration that you had trouble with. $\endgroup$ – Bobson Dugnutt Feb 17 '16 at 18:01
  • $\begingroup$ may be going from n=0 to n=2 will help you. To get to the $\pi r^2$ you use same integration. $\endgroup$ – Moti Feb 17 '16 at 18:14
  • $\begingroup$ @Moti May be, but the reasoning is definitely not the same for $V_0$ and $V_2$ elements that are used in the integration. $\endgroup$ – mavavilj Feb 17 '16 at 19:25
  • $\begingroup$ @mavavilj Take a look here $\endgroup$ – alexjo Feb 17 '16 at 21:00

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