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Consider the integral $$I = \int_{||\mathbf{r}|| < d} f(x_n)\ \mathrm{d}x_1 \cdots \mathrm{d}x_n,$$ where $\mathbf{r} = (x_1, \dots, x_n)^\top$. Since $f$ only depends on $x_n$, the integral can be expressed as $$I = \int_{||\mathbf{r}|| < d} \left(\mathrm{d}x_1 \cdots \mathrm{d}x_{n-1} \right)f(x_n)\ \mathrm{d}x_n.$$ I do not know how to evaluate this integral. I think the right answer for $d = 1$ is $$I = \int_{-1}^1 (1 - x_n^2)^{(n - 1)/2} V_{n - 1} f(x_n)\ \mathrm{d}x_n,$$ where $V_n$ is the volume of the $n$ ball, as per this answer, but I am not sure where the $(1 - x_n^2)$ part comes from.

user110971
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  • See the book "Interpolation and approximation on the unit sphere" by Atkinson and Han, you'll find that formula in the very beginning. You can also consult one of the books on spherical harmonics by Müller. – Giuseppe Negro Jun 22 '17 at 21:54
  • The integral over $x_1^2+\ldots+x_{n-1}^2 + x_n^2 \leq d^2$ is equal to the integral over $x_n^2 \leq d^2$ and for a given value of $x_n$ we must integrate over $x_1^2 + \ldots + x_{n-1}^2 \leq d^2 - x_n^2$ so we have

    $$\int_{x_1^2+\ldots+x_{n-1}^2 + x_n^2 \leq d^2} f(x_n),{\rm d}x_1\cdots{\rm d}x_n = \int_{-d}^d\left(\int_{x_1^2+\ldots+x_{n-1}^2\leq d^2 - x_n^2}{\rm d}x_1\cdots {\rm d}x_{n-1}\right)f(x_n){\rm d}x_n$$

    The term in the brackets is the volume of a $n-1$ dimensional sphere of radius $r = d^2-x_n^2$.

     – Winther Jun 23 '17 at 00:25
  • @Winther actually, the radius is $\sqrt{d^2 - x_n^2}$. – user110971 Jun 23 '17 at 21:14

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