Hyper-intriguing Ornament II

Another brute force:

$\text{NIntegrate}\left[\text{Boole}\left[w^2+x^2+y^2+z^2\leq 25\right],\{x,-3,3\},\{y,-3,3\},\{z,-3,3\},\{w,-3,3\}\right]$ is $1280.94538849$ .

$\frac{\pi ^{d/2} r^d}{\Gamma \left(\frac{d}{2}+1\right)}$ is $\frac{625 \pi ^2}{2}$ is $3084.251375340424$ . The result of the division is $0.415318089418$ .

I will post a solution based on Monte Carlo integration, because it is wonderfully simple conceptually, albeit computationally intensive. Here is the process:

1) Populate many points (like ten million) randomly and uniformly within the hyperball. I actually do this by populating them within a circumscribing hypercube of side length 10, and taking the subset which falls within the hyperball. Keep count of the number of points in the hyperball.

2) Keep another count of the points that are both within the hyperball AND within the hypercube.

3) Since the points are populated uniformly as a function of hypervolume, the ratio of the counts gives the ratio of the volumes. The beauty of this approach is that it transforms integration into a counting exercise. The percentage comes out to about 42.

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import math
import random

N = 10**7

numsphere = 0
numcube = 0

for j in range(0,N):

    x = -5.0 + 10.0 * random.uniform(0.0,1.0)
    y = -5.0 + 10.0 * random.uniform(0.0,1.0)
    z = -5.0 + 10.0 * random.uniform(0.0,1.0)
    w = -5.0 + 10.0 * random.uniform(0.0,1.0)

    if x**2.0 + y**2.0 + z**2.0 + w**2.0 <= 25.0:

        numsphere = numsphere + 1

        if math.fabs(x) < 3.0 and math.fabs(y) < 3.0 and math.fabs(z) < 3.0 and math.fabs(w) < 3.0:

            numcube = numcube + 1


print 100.0 * float(numcube) / float(numsphere)

Since my time is very limited, I resorted to "brute force," the way a busy engineer might. The portion of the hypervolume in the first hexadecant (?) is $V=\int_{0}^{3}\int_0^3\int_{0}^{\min(3,\sqrt{25-x^2-y^2})}\min\left(3,\sqrt{25-x^2-y^2-z^2}\right)\ dz\ dy\ dx\approx 80.059$ The proportion we seek is $\frac{16V}{V_{total}}\approx 0.4153\approx \boxed{42}\text{\%}$ where $V_{total}=\frac{625\pi^2}{2}$ as we found in an earlier problem . I look forward to seeing more conceptual solutions, but the problem does appear to be a bit "messy."