Radial integration over a bounded Euclidean shell

Fix a sphere of radius $r>0$. At least one coordinate of every point on it has absolute value at least $r\mathrm{/}\sqrt{3}$. On each of the six patches where a specified coordinate is largest in absolute value and has a specified sign, write that coordinate as $\pm \sqrt{r^2-u^2-v^2}$ over a bounded disk with $r^2-u^2-v^2\ge r^2\mathrm{/}3$. The graph gradient is uniformly bounded. A grid of squares of side $\delta$ covers the disk with $O({\delta }^{-2})$ squares; the graph over each square fits in a rectangular box of base area ${\delta }^2$ and height $O(\delta )$, so the total covering volume is $O(\delta )$. Thus the sphere has content zero. The sphere at radius zero is a point and is covered by one cube of arbitrarily small volume. Since $\mathrm{\partial }{A}_{a,b}$ is contained in the union of the spheres of radii $a$ and $b$, [F1] makes the bounded shell Jordan measurable.

If $a=b$, then $A_{a,b}$ is a sphere of Jordan content zero by step 1.1, so both integrals are zero. Now assume $0<a<b$. For $0<\epsilon <\min \{\pi \mathrm{/}4,(b-a)\mathrm{/}2\}$ put $K_{\epsilon }=[a,b]\times [\epsilon ,\pi -\epsilon ]\times [\epsilon ,2\pi -\epsilon ]$ and $\mathrm{\Psi }(r,\theta ,\varphi )=(r\sin \theta \cos \varphi ,r\sin \theta \sin \varphi ,r\cos \theta )$. On the open domain with $r>0$, $0<\theta <\pi$ and $0<\varphi <2\pi$, $\mathrm{\Psi }$ is injective and $C^1$ with determinant $r^2\sin \theta >0$; its inverse is given by radius, polar angle and the unique azimuth in $(0,2\pi )$. The compact set $K_{\epsilon }$ is Jordan, and [F2] makes $x\mapsto F(\mathrm{\mid }x\mathrm{\mid })$ integrable on $\mathrm{\Psi }(K_{\epsilon })$. Applying [F3] gives ${\int }_{\mathrm{\Psi }(K_{\epsilon })}F(\mathrm{\mid }x\mathrm{\mid })dx=({\int }_a^{b}F(r)r^{2}dr)({\int }_{\epsilon }^{\pi -\epsilon }\sin \theta d\theta )(2\pi -2\epsilon )\mathrm{.}$

The missing part of $A_{a,b}$ outside $\mathrm{\Psi }(K_{\epsilon })$ is contained in the union of the polar-cap box $[-b\sin \epsilon ,b\sin \epsilon {]}^2\times [-b,b]$ and the azimuth-seam box $[-b,b]\times [-b\sin \epsilon ,b\sin \epsilon ]\times [-b,b]$. Their total volumes are at most $8b^3({\epsilon }^2+\epsilon )$, which tends to zero. Since $F$ is bounded, the difference between its integral on the whole shell and on $\mathrm{\Psi }(K_{\epsilon })$ tends to zero. Letting $\epsilon \downarrow 0$ in step 2.1 gives the formula and the angular factor $2\cdot 2\pi =4\pi$. If $a=0$, first apply the result to $[\eta ,b]$ and then let $\eta \downarrow 0$; the omitted ball lies in $[-\eta ,\eta {]}^3$ and contributes at most $8{\eta }^3{\sup }_{[0,b]}\mathrm{\mid }F\mathrm{\mid }$, while the absolute value of the omitted radial integral is at most ${\sup }_{[0,b]}\mathrm{\mid }F\mathrm{\mid }{\eta }^3\mathrm{/}3$. Both tend to zero.