Hyper-intriguing Ornament I - Inspired by Otto Bretscher

Another very good problem, Comrade! I should have started with this one as a gentle introduction to mine.

Although the formula $\frac{R^3\pi^2}{2}$ for the hypervolume of a hyperball of radius $R$ is well known , let's derive it from first principles, for practice. This will also help us later on: $V_{total}=2\int\int\int_{B_5} \sqrt{25-x^2-y^2-z^2} \ dV=8\pi \int_0^5 \sqrt{25-\rho^2}\ \rho^2 d\rho=\frac{625\pi^2}{2},$ where $B_r$ is the ball $x^2+y^2+z^2\leq r^2$ in $\mathbb{R}^3$ .

This hyperball has eight "caps" outside the hypercube, where the absolute value of one of the variables is $\geq 4$ . Luckily, the intersection of any two of these caps is empty, unlike here , since $4^2+4^2>5^2$ . By symmetry, it suffices to find the hypervolume of one of those caps; let's look at $w\geq 4$ . Now, on the hypersphere, $w\geq 4$ means that $\rho\leq 3$ , so that $V_{cap}=\int\int\int_{B_3} (\sqrt{25-x^2-y^2-z^2}-4) \ dV=4\pi \int_0^3 (\sqrt{25-\rho^2}-4)\rho^2 d\rho=\frac{625\pi}{2}\arcsin\left(\frac{3}{5}\right)-186\pi$ The required proportion is $\frac{V_{total}-8V_{cap}}{V_{total}}\approx 0.877\approx \boxed{88}\text{\%}$