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Start with a circular region, and build a right-circular cylinder of any height. Next construct the right-triangular cone with apex the center of one of the faces of the cylinder, and between them construct the half-ellipsoid with a vertex at that same apex. I just learned a well-known fact that the ratios of the volumes of these three solids will be 1 : 2 : 3.

A cone, half-circle, and cylinder, from Marsden and Tromba's Vector Calculus book

If you go down a dimension, and place a triangle in a half-ellipse in a rectangle you get a ratio of areas of ${1:\frac{\pi}{2}:2}$. Generalizing this to $n$ dimensional space, starting with an $(n-1)$-ball as your base, the ratio of the volumes of the cone and cylinder will be $1$ to $n$. But how does the hyper-volume of the higher-dimensional ellipsoid fit into this ratio?

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Considering Cavalieri's principle we only need to think about the case of a ball (instead of a general ellipsiod). After seeing the Wikipedia page for the $n$-ball and the closed-form and recursive formulas for a hyper-volume of an $n$-ball of radius $r$, the volume $V_n$ is given by $$ V_n = \frac{\sqrt{\pi^n}}{\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}r^n \quad\text{ and }\quad V_n = \sqrt{\pi}\frac{\operatorname{\Gamma}\left(\frac{n+1}{2}\right)}{\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}r V_{n-1} \,. $$ where $\Gamma$ is the gamma function. The ratio of these three $n$-dimensional volumes will then be $$ \frac{1}{n}r V_{n-1} \;:\; \frac{1}{2}V_n \;:\; rV_{n-1} \\[3ex]\text{or}\\[1ex] 1 \;:\; \frac{\sqrt{\pi}\operatorname{\Gamma}\left(\frac{n+1}{2}\right)}{2\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}n \;:\; n \,. $$ For the first few values of $n$, the middle number $B_n$ in this ratio is $$ \begin{array}{c|cc} n & 2&3&4&5&6&7&8&\dotsb&\\\hline B_n & \frac{1}{2}\pi & 2 & \frac{3}{4}\pi & \frac{8}{3} & \frac{15}{16}\pi &\frac{16}{5} & \frac{35}{16}\pi & \dotsb\\ \end{array} \,. $$ You get the factor of $\pi$ only in the even numbered terms because $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$, so they will have an additional $\sqrt{\pi}$ in the numerator, whereas odd numbered terms will have a $\sqrt{\pi}$ in the denominator to cancel with the $\sqrt{\pi}$ already there.

Now let's consider what happens to this ratio of volumes as $n \to \infty$. Using Gautschi's inequality and the squeeze theorem we have, for $s=\frac{1}{2}$ and $x=\frac{n}{2}$ that $$ x^{1-s} < \frac{\Gamma(x+1)}{\Gamma(x+s)} < (x+1)^{1-s}\\[2ex] \sqrt{\frac{2}{n}}n> \frac{\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}n> \sqrt{\frac{2}{n+2}}n \\[2ex] \infty> \lim_{n\to \infty}\frac{\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}n> \infty $$ So as $n \to \infty$, the ratio of the volume of the half-ball to cone approaches infinity! Now considering the volume of the cylinder compared to the volume of the half-ball (so we divide out the factor of $n$): $$ \sqrt{\frac{2}{n}}> \frac{\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}> \sqrt{\frac{2}{n+2}} \\[2ex] 0> \lim_{n\to \infty}\frac{\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}> 0 $$ So the ratio of the volume of the half-ball to the cylinder approaches zero as $n \to \infty$. This is weird, but it's not new: this oddity is basically the same oddity as the known oddity that the volume of the $n$-ball of a fixed radius approaches zero as $n \to \infty$.

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