Geometric Identities involving $π^2$

Question

Are there any known geometric identities that have $π^2$ in the formula?

Answer 1

The sphere of radius $r$ in $\mathbb R^4$ has volume (4-dimensional volume) $$ \frac{\pi^2 \; r^4}{2} $$ Here, we refer to the set $$ x_1^2 + x_2^2 + x_3^2 + x_4^2 \leq r^2. $$

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The sphere of radius $r$ in $\mathbb R^5$ has volume (5-dimensional volume) $$ \frac{8 \pi^2 \; r^5}{15} $$ Here, we refer to the set $$ x_1^2 + x_2^2 + x_3^2 + x_4^2+ x_5^2 \leq r^2. $$

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EDIT. My impression is that most students are not aware of the easy way to do volumes of the higher dimensional spheres. Let $V_m$ be the volume of the sphere of radius $1$ in $\mathbb R^m.$ So $$V_1=2, \; V_2 = \pi, \; V_3 = \frac{4 \pi}{3}.$$ In order to find $V_{m+2},$ we integrate $V_m R^m$ over the unit disc in the plane, where this radius $R= \sqrt {1 - x^2 - y^2}.$ Switching to polar coordinates, we get $$ V_{m+2} = \int_0^{2 \pi} \int_0^1 V_m (1-r^2)^{m/2} \, r \, dr \, d \theta = 2 \pi V_m \int_0^1 (1-r^2)^{m/2} \, r \, dr = \frac{2 \, \pi \, V_m}{m+2}. $$ So $$ V_4 = \frac{2 \, \pi \, V_2}{4} = \frac{\pi^2}{2} $$ and $$ V_5 = \frac{2 \, \pi \, V_3}{5} = \frac{2 \pi \cdot 4 \pi}{3 \cdot 5} = \frac{8\pi^2}{15} $$

By induction, $$ V_n = \frac{\pi^{n/2}}{(n/2)!}, $$ where in the case of odd $n$ we need to understand $$ (n/2)! = \Gamma \left( 1 + \frac{n}{2} \right) $$ which is always a rational multiple of $\sqrt \pi$ (for odd $n$ it is).

Answer 2

Volume of a torus is $2 \pi^2 R^2 r$, where $r$ is the inner radius and $R$ is the outer radius. Its surface area is $4 \pi^2 R r$.