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

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    $\begingroup$ @Amzoti but what is the geometric identity? For example I already know Zeta(2)=$π^2$/6... But what geometrically proves that. $\endgroup$ Commented Jul 30, 2013 at 3:40
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    $\begingroup$ The area of a disk of radius $1$ is $\sqrt{\pi^2}$. With that, on a good day, I would get a great answer badge... $\endgroup$
    – Julien
    Commented Jul 30, 2013 at 5:17
  • $\begingroup$ @julien ^Made me laugh! Hahah! :) $\endgroup$ Commented Jul 30, 2013 at 7:02

2 Answers 2

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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 $4 \pi^2 R r$.

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    $\begingroup$ Yes, I guess this one is more geometrical than the 4D thing because we can see tori more easily than those weird 4D-spheres :P $\endgroup$ Commented Jul 30, 2013 at 4:46
  • $\begingroup$ In the same vein the volume of the "football" we get by revolving the cycloid $x(t)=R(t-\sin t),$ $y(t)=R(1-\cos t)$ about the $x$-axis is (a single arc, so $0\le t\le 2\pi$) $$V=5\pi^2R^3.$$ $\endgroup$ Commented Jul 30, 2013 at 11:19
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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 $$ x_1^2 + x_2^2 + x_3^2 + x_4^2 \leq r^2. $$


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 $$ x_1^2 + x_2^2 + x_3^2 + x_4^2+ x_5^2 \leq r^2. $$


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).

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    $\begingroup$ Thanks! Quick question though... I recognize the double struck "R" as the "real number set", correct? But what does it mean when you superscript a 4? (Also, are we talking about a sphere or a hypersphere, or the hyper volume of a 3D sphere if that's even a thing?) Thanks! :) $\endgroup$ Commented Jul 30, 2013 at 3:43
  • $\begingroup$ @AlbertRenshaw : If $3D$ has only a geometric meaning for you, the 3-dimensional space is usually modeled as $\mathbb R^3 = \{ (x,y,z) \, | \, x,y,z \in \mathbb R \}$ in mathematics. If you're in $\mathbb R^4$, then you can consider that you need to look at 4-coordinate elements such as (w,x,y,z) for instance. You need to make sense of what '4-dimensional volume' means though. The best way to see it is to visualize an hypercube with sidelength $a$ size to have "4-dimensional volume" $a^4$, in a similar way than a cube with sidelength $a$ has volume $a^3$. $\endgroup$ Commented Jul 30, 2013 at 3:49
  • $\begingroup$ @AlbertRenshaw, he meant the set of points $(x,y,z,w)$ such that $x, y, z, w$ are all real numbers. When he means the sphere in $\mathbb{R}^4$, he meant the set $\{(x,y,z,w)\in\mathbb{R}^4:x^2+y^2+z^2+w^2=r^2\}$. $\endgroup$
    – Marra
    Commented Jul 30, 2013 at 3:50
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    $\begingroup$ @WillJagy Got it! So just the hyper-volume of a 4-sphere! Thanks! :) $\endgroup$ Commented Jul 30, 2013 at 3:55
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    $\begingroup$ Woooow here calm down. Infinite squared is infinite in your context. And you need to watch out when trying to extend integer valued sequences in the way you tried it. What you just did is a pure guess, I am not convinced at all it makes sense. But what I can say is that a $4D$-cube indeed has '8 $3D$-cubes' in this animation : youtube.com/watch?v=CtSNStVW81M . I'll let you try figure out where they 'are'. $\endgroup$ Commented Jul 30, 2013 at 14:13

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