Spherical cap and dual slab scales

The cap lies in the closed hemisphere since ${\omega }_n\ge 1-{\delta }^2\ge 0$. For $\delta <1$ it lies in the open upper chart, and squaring $\sqrt{1-\mathrm{\mid }y{\mathrm{\mid }}^2}\ge 1-{\delta }^2$ gives $\mathrm{\mid }y{\mathrm{\mid }}^2\le 2{\delta }^2-{\delta }^4$, yielding the stated integral with [F1]. At $\delta =1$ the cap is the closed upper hemisphere. Its equator has surface measure zero: cover the equator by the other hemisphere charts; there ${\omega }_n$ is one parameter coordinate, and the equator is a coordinate hyperplane of Lebesgue measure zero by Tonelli (a singleton coordinate has zero length). The chart density is finite on its open domain, so integrating it over that null set gives zero. Thus the formula also holds at $\delta =1$, integrating over $\mathrm{\mid }y\mathrm{\mid }<1$; boundary values may be assigned arbitrarily.

The slab. $T_{\delta }$ is the box ${\prod }_{j<n}[-c{\delta }^{-1},c{\delta }^{-1}]\times [-c{\delta }^{-2},c{\delta }^{-2}]$, a product of $n-1$ intervals of length $2c{\delta }^{-1}$ and one of length $2c{\delta }^{-2}$. By [F2], ${\lambda }_n(T_{\delta })=(2c{\delta }^{-1}{)}^{n-1}(2c{\delta }^{-2})=(2c{)}^n{\delta }^{-(n-1)-2}=(2c{)}^n{\delta }^{-(n+1)}$.

Two-sided bounds for the cap. Lower bound: the ball $\mathrm{\mid }y\mathrm{\mid }\le \delta$ satisfies $\mathrm{\mid }y{\mathrm{\mid }}^2\le {\delta }^2\le 2{\delta }^2-{\delta }^4$ and on it the density is at least $1$; by [F2], $\sigma (C_{\delta })\ge {\omega }_{n-1}{\delta }^{n-1}$. Upper bound: on the cap $1-\mathrm{\mid }y{\mathrm{\mid }}^2\ge (1-{\delta }^2{)}^2$, so the density is at most $(1-{\delta }^2{)}^{-1}$; the domain is contained in the ball of radius $\sqrt{2{\delta }^2-{\delta }^4}\le \sqrt{2}\delta$, so for $\delta \le 1\mathrm{/}2$ the density factor is at most $4\mathrm{/}3$ and $\sigma (C_{\delta })\le (4\mathrm{/}3){\omega }_{n-1}{2}^{(n-1)\mathrm{/}2}{\delta }^{n-1}$. For $\delta \ge 1\mathrm{/}2$ one has $C_{\delta }\subseteq S^{n-1}$ and ${\delta }^{n-1}\ge 2^{-(n-1)}$, so $\sigma (C_{\delta })\le \sigma (S^{n-1})=n{\lambda }_n(B_1^n)\le n{\lambda }_n(B_1^n)2^{n-1}{\delta }^{n-1}$ by [F3]. Thus $c_n{\delta }^{n-1}\le \sigma (C_{\delta })\le C_n{\delta }^{n-1}$ with $c_n:={\omega }_{n-1}$ and $C_n:=\max \{(4\mathrm{/}3){\omega }_{n-1}{2}^{(n-1)\mathrm{/}2},n{\lambda }_n(B_1^n)2^{n-1}\}$.

The diameter. Let $\omega ,{\omega }^{\mathrm{\prime }}\in C_{\delta }$ and write $\omega =(u,{\omega }_n)$, ${\omega }^{\mathrm{\prime }}=(u^{\mathrm{\prime }},{\omega }_n^{\mathrm{\prime }})$ with $u,u^{\mathrm{\prime }}\in {\mathbb{R}}^{n-1}$. From step 1.1, $\mathrm{\mid }u{\mathrm{\mid }}^2\le 2{\delta }^2-{\delta }^4$ and $\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^2\le 2{\delta }^2-{\delta }^4$, while ${\omega }_n,{\omega }_n^{\mathrm{\prime }}\ge 1-{\delta }^2$. Hence $\omega \cdot {\omega }^{\mathrm{\prime }}=u\cdot u^{\mathrm{\prime }}+{\omega }_n{\omega }_n^{\mathrm{\prime }}\ge -\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }+(1-{\delta }^2{)}^2\ge -(2{\delta }^2-{\delta }^4)+(1-{\delta }^2{)}^2=1-4{\delta }^2+2{\delta }^4,$ so $\mathrm{\mid }\omega -{\omega }^{\mathrm{\prime }}{\mathrm{\mid }}^2=2-2\omega \cdot {\omega }^{\mathrm{\prime }}\le 8{\delta }^2-4{\delta }^4\le 8{\delta }^2$ and $\mathrm{diam}{C}_{\delta }\le 2\sqrt{2}\delta$.

Rotational reduction. For $v\in S^{n-1}$ choose an orthogonal map $R$ with $R{e}_n=v$; finite-dimensional orthogonal algebra supplies such an $R$ without choice. The cap $\{\omega :1-\omega \cdot v\le {\delta }^2\}=R[C_{\delta }]$ is the image of $C_{\delta }$ under an orthogonal transformation, which preserves $\sigma$ by [F1]; hence it has the same measure and the same diameter bound.

Conclusion. Step 1.1 rewrites the cap in the graph chart, step 2.1 gives the two-sided scale $c_n{\delta }^{n-1}\le \sigma (C_{\delta })\le C_n{\delta }^{n-1}$, step 1.2 computes ${\lambda }_n(T_{\delta })=(2c{)}^n{\delta }^{-(n+1)}$, step 2.2 gives $\mathrm{diam}{C}_{\delta }\le 2\sqrt{2}\delta$, and step 3.1 transfers the scales to arbitrary axis $v$ by orthogonal invariance.