El uso de homogéneamente distribuido multivariante de las variables
Taeke proporciona un enlace a un artículo que el texto a continuación se hace más intuitiva explicando específicamente 2-norma y 1-norma de los casos.
2-norma $\Vert x \Vert_2 \leq r$
ejemplo de dirección
Usted puede utilizar este resultado http://mathworld.wolfram.com/HyperspherePointPicking.html
Un multivariante Gaussiano distribuido variable $X$ (con la identidad de la matriz de covarianza) sólo depende de la distancia, o la suma de los cuadrados.
$$f(X_1,X_2,...,X_n) = \prod_{1\leq i \leq n} \frac{1}{\sqrt{2\pi}}e^{\frac{1}{2}x_i^2} = \frac{1}{\sqrt{2\pi}}e^{\frac{1}{2}\sum_{1 \leq i \leq n} x_i^2} $$
Thus $\frac{X}{\Vert X \Vert_2}$ is uniformly distributed on the surface of the n-dimensional-hypersphere.
sample distance
To complete you only need to sample the distance, to change the homogeneous distribution on the sphere to a homogeneous distribution in a ball. (which is more or less similar as your linked example for disk point picking)
If you would simply sample $r$ as a uniform distribution then you would have a relatively higher density near the center (the volume scales as $r^n$ so a fraction $r$ of the points would end up in a volume $r^n$, which is more dense near the center and would not mean a uniform distribution)
If instead you use the $n$-th root of a variable sampled from a uniform distribution, then you get an even distribution.
1-norm $\Vert x \Vert_1 \leq r$
direction
In this case you sample $X$ from the Laplace distribution instead of the Gaussian distribution and divide by the 1-norm. The $\frac{X}{\vert X \vert_1}$ is uniformly distributed on the n-dimensional 1-norm sphere.
I have no formal proof, just intuition
(since the pdf is independent from position, you will expect for any infinitesimal area/volume with the same 1-norm to have the same probability $f(x) dV$ and when you collapse this to the unit surface the same $f(x) dA$)
but testing with simulations looks good.
library(rmutil)
x <- abs(rlaplace(20000))
y <- abs(rlaplace(20000))
z <- abs(rlaplace(20000))
rn <- abs(x)+abs(y)+abs(z)
xi <- (x/rn)
yi <- (y/rn)
zi <- (z/rn)
plot(sqrt(0.5)*(xi-yi),
sqrt((0.5-0.5*(xi+yi))^2+zi^2),
pc=21,bg=rgb(0,0,0,0.02), col=rgb(0,0,0,0),cex=1)
distance
The distance goes similar as with the 2-norm case (the volume still scales as $r^n$).
p-norm $\Vert x \Vert_p \leq r$
In this case, if you wish to follow the same principle, you would need to sample from distributions with $f(x) \propto e^{\vert x \vert^p}$ (I hypothesize). These are generalized normal distributions and probably relate to the distribution $G()$ mencionado por Taeke.