= Solution
Normalize the <surface area> of the <spherical cap> by writing
$$
q(\rho)=\frac{\operatorname{Vol}_{n-1}(D_\rho)}{s_{n-1}},\qquad c_n=\frac{v_{n-1}}{nv_n}.
$$
Two elementary descriptions of a <spherical cap> are useful. The <spherical polar coordinates> formula gives
$$
q'(\rho)=(n-1)c_n\sin^{n-2}\rho,\qquad q(\pi/2)=\frac12.
$$
Projecting orthogonally onto the last $n-1$ coordinates gives
$$
q(\rho)\leq c_n\frac{\sin^{n-1}\rho}{\cos\rho}.
$$
Indeed, the upper <hemisphere> is the graph $x_1=\sqrt{1-\lVert x'\rVert^2}$, whose <surface area> factor is $(1-\lVert x'\rVert^2)^{-1/2}\leq1/\cos\rho$ on $\lVert x'\rVert\leq\sin\rho$. This establishes the upper projection bound directly.
Set $F(\rho)=\sin^n\rho-q(\rho)$. The supplied ratio bound implies $c_n<\sqrt{(n+1)/(2\pi)}/n\leq1/2$ for $n\geq2$. Consequently, at $\rho=\pi/4$,
$$
q(\pi/4)\leq2c_n\,2^{-n/2}\leq2^{-n/2},\qquad F(\pi/4)\geq0.
$$
Also,
$$
F'(\rho)=\sin^{n-2}\rho\bigl(n\sin\rho\cos\rho-(n-1)c_n\bigr).
$$
On $[\pi/4,\pi/2]$, the expression in parentheses is strictly decreasing; it starts positive and ends negative. Thus $F$ first increases and then decreases, so its minimum on this interval is at an endpoint. Since $F(\pi/2)=1/2$, both endpoint values are nonnegative. \b[The <spherical cap area upper bound> follows:]
$$
\boxed{\operatorname{Vol}_{n-1}(D_\rho)\leq\sin^n\rho\,s_{n-1}\qquad(\pi/4\leq\rho<\pi/2).}
$$
Back to article page
