areageometryintegrationspheres
According to mathworld, let the sphere have radius $R$, then the surface area a spherical cap of height $h$ and base radius $a$ is given by
$$S=2\pi Rh=2\pi(a^2+h^2).$$
What is this value for an n-dimensional hypersphere?
If it helps simplify the problem we can assume $R=1$ and $a=0.5$.
Many thanks.
The Wikipedia article gives the volume of the 4-dimensional hypersphere, with radius $r$, to be $$V = \frac{\pi^2}{2}r^4$$
The surface area can be found by differentiating with respect to $r$:
$$A = \frac{\mathrm{d}V}{\mathrm{d}r} = 2\pi^2r^3$$
The article also gives a proof of how to calculate the volume, and hence surface area.
The following is due to Archimedes of Syracuse (287 BC - 212 BC).
First, remeber that the lateral surface area of a right conical frustum is $$\pi(r_1+r_2)S=2\pi mS\tag 1$$ where $r_1$ is radius of the base, $r_2$ is the radius of the top, $m=\frac{r_1+r_2}2$, the "mid-radius", and $S$ is the lateral (slant) height.
Now, consider the following figure.
Here a horizontal slice of the black sphere has been replaced by a red frustum. Also there is a blue cylinder containing the sphere. Consider the blue slice of the cylinder matching the frustum in its height and matching the sphere in its radius. We want to show that the lateral surface area of the blue cylindrical slice equals the lateral surface area of the red frustum.
The lateral surface area of the blue cylindrical slice is$$2\pi rd.$$
How come that $2\pi rd=2\pi mS$? The red filled triangle is similar to the triangle $MNC$. So $$\frac{m}{r'}=\frac d S,$$ from where $$d=\frac{mS}{r'}.$$ Substituting this into the formula for the lateral cylindrical surface, we get $$2\pi \frac{r}{r'}mS.$$ Compare this to $(1).$
Then think: If the spherical slice is very thin then $r'$ is very close to $r$ and the lateral surface area of the frustum is very close to the lateral surface area of the spherical slice... This is all independent from the actual position of the slice we chose.
Then slice the whole spherical cap. The lateral surface area of the slices will equal the corresponding lateral surface area of the cylinder.
Calculus is needed only if the argumentation above seems to be shaky. But the genius did not need calculus.
Best Answer
The solution requires special classes of functions, namely regularized incomplete beta functions $I_{\sin^2 \Phi} (\frac{n-1}{2}, \frac{1}{2})$ and Gamma functions. The result and its derivation can e.g. be found in the very nice article by S. Li here. Let $\Phi$ be the angle of the cap, so $a = R\sin \Phi$.
Then the surface area of the cap is $$ A(\Phi) = R^{n-1} \frac{\pi^{n/2}}{\Gamma(n/2)} I_{\sin^2\Phi} (\frac{n-1}{2}, \frac{1}{2}) $$