I have already proved (without integration) that the volume of a hypersphere in $R^{4}$is given by
$$V=2\pi^{2}R^{3},$$
with $R$ being the radius. According to Wikipedia, the hypervolume is then given by:
$$\textbf{V}=\frac{1}{2}\pi^{2}R^{4}=\frac{1}{4}VR.$$
I was wondering whether anyone could give me any pointers as to how to derive this formula without integrals. Thanks in advance.
Let $AB$ be a quadrant of a great circle. Two hyperplanes perpendicular to the radius $OB$ cut the hypersphere in two spheres, whose interiors are the bases of a frustum of a hypercone inscribed in the hypersphere. Let $H'$ be the slant height of this frustum, and $r_{1}$ and $r_{2}$ the radii of the bases. $H'$ is the length of a chord of the great circle. The lateral volume of the frustum is given by the formula
$$\frac{4}{3}\pi H'(r_{1}^{2}+r_{1}r_{2}+r_{2}^{2}).$$
Let $H$ be the altitude of the frustum, and $K$ the distance from to the middle point of the chord which represents $H'$. From similar triangles we have
$$\frac{H'}{K}=\frac{H}{\frac{1}{2}(r_{1}+r_{2})}$$
or
$$r_{1}+r_{2}=\frac{2KH}{H'}.$$
Also from right triangles
$$H'^{2}=H^{2}+(r_{1}+r_{2})^{2}$$
and
$$R^{2}=K^{2}+(\frac{1}{2}H')^{2}$$
Therefore
\begin{align*}\frac{4}{3}(r_{1}^{2}+r_{1}r_{2}+r_{2}^{2})&=(r_{1}+r_{2})^{2}+\frac{1}{3}(r_{1}-r_{2})^{2}\\
\qquad&=(4R^{2}-H'^{2})\frac{H^{2}}{H'^{2}}-\frac{1}{3}(H'^{2}-H^{2})\\
\qquad&=4R^{2}\frac{H^{2}}{H'^{2}}+\frac{1}{3}H'^{2}-\frac{4}{3}H^{2}.
\end{align*}
For an inscribed hypercone with vertex at $B$ and $r_{1}$ the radius of the base, we have only to make $r_{2}=\circ$ in all of these expressions. If we take two arcs symmetrically situated on the arc $AB$ with respect to $P$, the middle point of this arc, we shall have two right triangles with hypothenuse equal to $H'$ symmetrically situated with respect to the radius $OP$, and therefore equal. But the legs denoted by $H$ are non-homologous sides in the two triangles, and so the sum of their squares is $H'^{2}$. Therefore, if we write down the formula given above for each of two divisions of $AB$ symmetrically situated with respect to $P$, the sum of the two expressions will be
$$4R^{2}-\frac{2}{3}H'^{2};$$
and the sum of the volumes of the frustums corresponding to the two arcs will be
$$4\pi R^{2}H'-\frac{2}{3}\pi H'^{3}.$$
Now if we divide the arc $AB$ into $2n$ equal parts, we shall have a hypercone and frustum, and $n-I$ pairs of frustums, all inscribed in a half-hypersphere; and when $n$ is increased indefinitely the sum of their lateral volumes will have for limit the volume of the half-hypersphere. The sum is
$$4n\pi R^{2}H'-\frac{2}{3}n\pi H'^{3}.$$
But
$$\lim_{n=\infty}nH'=\text{arc}{AP}=\frac{\pi R}{4},$$
and
$$\lim_{n=\infty}nH'^{3}=\lim_{n=\infty}\frac{(nH')^{3}}{n^{2}}=\circ$$
Thus the volume of the half-hypersphere is
$$4\pi R^{2}\cdot\frac{\pi R}{4}$$
and the volume of the hypersphere is
$$2\pi^{2}R^{3}.$$
Best Answer
Basic approach. One way to do what I think you might want is to observe that you can dissect a hypercube into four congruent parts by drawing hyperpyramids from any vertex $V$ to the four hyperfaces that do not contain $V$. This is by analogy of dividing a square into two congruent right triangles from any vertex to the two sides that don't contain that vertex, or of dividing a cube into three congruent pyramids from any vertex to the three faces that don't contain that vertex.
This demonstrates that a hyperpyramid has a volume of one-fourth the base volume times the altitude. By appropriate scaling and skewing, one can demonstrate the same formula for any hyperpyramid.
One can then apply this formula to the hypersphere, whose (hypersurface) volume you have determined to be $2\pi^2R^3$. By analogy with determining the area of a circle as the limit of a sequence of triangles of ever more finely divided base, the hypervolume of the hypersphere should be one fourth the common altitude of the hyperpyramids, $R$, times the combined base volume, $2\pi^2R^3$, or
$$ \frac12 \pi^2 R^4 $$
I'm not exactly sure how one would do this rigorously without essentially re-inventing calculus, but perhaps the somewhat intuitive argument here is sufficient for your needs.