If the Greeks had been four dimensional, would they have been able to derive the pi squared coefficient for the hypersphere volume without calculus?

Question

I was reading about Archimedes' pre-calculus proof of the volume of the sphere and I realized that the trick he uses (volume of hemisphere + volume of cone = volume of cylinder) doesn't generalize to hyperspheres, so I was wondering if there's any way of proving the volume of a hypersphere is $\frac{π^2}{2}R^4$ with classical methods.

Answer 1

Your solution posted in the given link is so nice that it deserves to come alive here.

Consider a four-dimensional ball of radius $R$: $$B:=\bigl\{(x,y,u,v)\>\bigm|\>x^2+y^2+u^2+v^2\leq R^2\bigr\}\ ,$$ and at the same time a four-dimensional dicone $$C:=\bigl\{(x,y,u,v)\>\bigm|\>u^2+v^2\leq x^2+y^2\leq R^2\bigr\}\ .$$ Due to symmetry the dicone $C$ is just half of the dicylinder $$Z:=\bigl\{(x,y,u,v)\>\bigm|\>u^2+v^2\leq R^2, \ x^2+y^2\leq R^2\bigr\}\ ,$$ so that $${\rm vol}_4(C)={1\over2}{\rm vol}_4(Z)\ .\tag{1}$$ Now comes Archimedes. The following figure shows the projections of the three bodies onto the $(x,y)$-plane placed next to each other:

If we erect a two-dimensional ''stalk'' at $(x,y)\in B'$ this stalk will intersect $B$ in the $(u,v)$-disk $$\bigl\{(u,v)\>\bigm|\>u^2+v^2\leq R^2-x^2-y^2\bigr\}$$ of area $\pi(R^2-x^2-y^2)$, and the stalk erected at $(x,y)\in C'$ will intersect $C$ in the $(u,v)$-disk $$\bigl\{(u,v)\>\bigm|\>u^2+v^2\leq x^2+y^2\bigr\}$$ of area $\pi(x^2+y^2)$. The sum of these two areas is $=\pi R^2$ for all $(x,y)$, and is equal to the area that such a stalk cuts out of $Z$. By Cavalieri's principle we therefore can conclude that $${\rm vol}_4(B)+{\rm vol}_4(C)={\rm vol}_4(Z)\ .$$ Together with $(1)$ it follows that $${\rm vol}_4(B)={1\over2}{\rm vol}_4(Z)={\pi^2\over2}R^4\ .$$