Steinmetz has gone hyper

Here's the Monte Carlo again. Randomly and uniformly generate points within a 4D hypercube of side length 2. Keep a count of the total number of points within the hypercube, as well as the number of points within the hypercube AND within the two regions. The volume within the two regions is equal to the volume of the hypercube multiplied by the ratio of the two counts. The volume turns out to be $2 \pi$

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import math
import random

N = 10**6   # Number of random points generated within hypercube

count = 0   # Number inside hypercube AND inside both regions

Volume_ref = 2.0**4.0  # Volume of hypercube of side length 2

for j in range(0,N):    # Generate points randomly and uniformly
                        # within hypercube
    x = random.uniform(-1.0,1.0)
    y = random.uniform(-1.0,1.0)
    z = random.uniform(-1.0,1.0)
    w = random.uniform(-1.0,1.0)

    if x**2.0 + y**2.0 + z**2.0 <= 1.0:

        if y**2.0 + z**2.0 + w**2.0 <= 1.0:    # If inside both regions

            count = count + 1  # increment count

print Volume_ref * float(count) / float(N)  # Volume is count ratio times hypercube volume

The $4$ -volume is defined by the equations $|x|\,,\,|w| \,\le \, \sqrt{1-(y^2+z^2)}\;,\; y^2 +z^2 \;\le 1$ and so the $4$ -volume is $\iint_{y^2+z^2 \le 1}\,4(1 - y^2 - z^2)\,dy\,dz \; = \; 4\int_0^{2\pi}\int_0^1 (1-r^2)\,r\,dr\,d\theta \; = \; 8\pi\int_0^1(r - r^3)\,dr \; = \; \boxed{2\pi}$ I would go with $8\pi$ as the hypersurface area, being the derivative of $2\pi R^4$ evaluated at $R=1$ .

For the sake of variety, let me find the "surface area" $S$ first:

$S=4\int\int_{x^2+y^2+z^2=1}\sqrt{1-y^2-z^2}\ dS$ $=4\int\int_{x^2+y^2+z^2=1}|x|\ dS$ $=4\int\int_{x^2+y^2+z^2=1}\frac{1}{2}dS$ $=4\times 4\pi\times \frac{1}{2}$ $=8\pi$ . Now we can argue as Mark does (but backwards): If the radii are $r$ , then the "surface area" is $8\pi r^3$ , so that the volume (the integral) is $2\pi r^4$ . For $r=1$ we get $\boxed{2\pi}$ .

If you have trouble understanding the first equation, consider the analogous equation $S=\int_{x^2+y^2=1}\sqrt{1-y^2}\ ds$ for the surface area of the Steinmetz solid in $\mathbb{R}^3$ given by $x^2+y^2\leq 1$ and $y^2+z^2\leq 1$ .