Volumes of Elementary Solids and Solids of Revolution

If $A=\mathrm{\varnothing }$, the empty family covers $A\times [c,d]$. If $d=c$, use [F1] with tolerance $\min \{\epsilon ,1\mathrm{/}2\}$, obtaining a finite cube cover $Q_i$ with side lengths ${\mathrm{\ell }}_i\le 1$ and total base volume at most $\epsilon$; then the cubes $Q_i\times [c,c+{\mathrm{\ell }}_i]$ cover $A\times \{c\}$ and have total $(m+1)$-volume at most the base total.

Suppose $A\mathrm{\ne }\mathrm{\varnothing }$ and $L:=d-c>0$. Use [F1] with tolerance $\delta :=\min \{1\mathrm{/}2,\epsilon \mathrm{/}(2(L+1))\}$; enlarge any zero-side cubes slightly, using the unused half of this tolerance, so that the resulting finite cover has $0<{\mathrm{\ell }}_i\le 1$ and total base volume below $2\delta \le \epsilon \mathrm{/}(L+1)$. For each $i$, let $N_i:=1+\lfloor L\mathrm{/}{\mathrm{\ell }}_i\rfloor$. Fact [F2] gives $N_i{\mathrm{\ell }}_i>L$ and $N_i{\mathrm{\ell }}_i\le L+{\mathrm{\ell }}_i$, and the $N_i$ consecutive $(m+1)$-cubes of side ${\mathrm{\ell }}_i$ above $Q_i$ cover $Q_i\times [c,d]$.

The total volume of the cubes in step 1.2 is ${\sum }_i{N}_i{\mathrm{\ell }}_i^{m+1}\le {\sum }_i(L+{\mathrm{\ell }}_i){\mathrm{\ell }}_i^m\le (L+1){\sum }_i{\mathrm{\ell }}_i^m<\epsilon$, the first two inequalities being the bounds $N_i{\mathrm{\ell }}_i\le L+{\mathrm{\ell }}_i$ and ${\mathrm{\ell }}_i\le 1$ of step 1.2 and the last the strict bound on the base total. Together with step 1.1 this supplies an arbitrarily small finite cube cover in every case, so $A\times [c,d]$ has content zero.

If $C=\mathrm{\varnothing }$, then $G=\mathrm{\varnothing }$ and the empty finite cover proves the conclusion. If $m=0$, then ${\mathbb{R}}^0$ has exactly one point, so $C$ has at most one point and $G$ is a single point of ${\mathbb{R}}^1$, covered for every tolerance by one closed cube of small enough side; [F2] gives content zero. Hence assume $C\mathrm{\ne }\mathrm{\varnothing }$ and $m\ge 1$, so that the divisions by $\sqrt{m}$ below are defined.

By [F4], compactness places $C$ in a closed cube $Q$, and [F3] bounds $f(C)$ in an interval. By [F1], for every $\eta >0$ there is $\delta >0$ such that $\mathrm{\mid }f(u)-f(v)\mathrm{\mid }<\eta$ whenever $u,v\in C$ and $\mathrm{\parallel }u-v{\mathrm{\parallel }}_2<\delta$.

Partition a fixed cube slightly larger than $Q$ into coordinate cubes of side $h<\delta \mathrm{/}\sqrt{m}$. Keep only the base cells meeting $C$, and above each such cell keep the vertical $(m+1)$-grid cubes that meet $G$. Values of $f$ over one base cell differ by less than $\eta$, so its retained vertical stack has total height at most $\eta +2h$. If $B$ is the volume of the enlarged base cube, the retained cubes cover $G$ and have total volume at most $B(\eta +2h)$.

Choose $\eta$ and then $h$ so that $B(\eta +2h)<\epsilon$. Step 2.1 is then a finite closed-cube cover of $G$ with total volume below $\epsilon$; by [F2], $G$ has content zero.

If $D=\mathrm{\varnothing }$, then $K$ and its boundary are empty. Otherwise the extreme-value theorem bounds both functions in one interval $[c,d]$. Any point of $\mathrm{\partial }K$ either projects to $\mathrm{\partial }D$, or projects to the interior of $D$ and lies on $t=\alpha (u)$ or $t=\beta (u)$; hence $\mathrm{\partial }K\subseteq \mathrm{graph}(\alpha )\cup \mathrm{graph}(\beta )\cup (\mathrm{\partial }D\times [c,d])$. Since $D$ is Jordan measurable, [F3] makes $\mathrm{\partial }D$ content zero.

By [F2] the two graph pieces have content zero, and by [F1] the product $\mathrm{\partial }D\times [c,d]$ has content zero.

Given a positive tolerance, cover each of the three sets in step 2.1 with total cube volume below one third of it. Their union covers $\mathrm{\partial }K$, so the boundary has content zero.

If $D=\mathrm{\varnothing }$, every assertion and both integrals are zero. Otherwise [F4] bounds $\alpha$ and $\beta$; the defining weak inequalities make $K$ closed, so it is closed and bounded and therefore compact by [F5].

By [F1] the boundary of the bounded set $K$ has content zero, so the Jordan boundary criterion makes $K$ Jordan measurable.

By [F2], $H$ is integrable on $K$. For each $u\in D$, the vertical section is exactly $[\alpha (u),\beta (u)]$, and the restriction $t\mapsto H(u,t)$ is continuous and integrable; [F3] therefore gives the displayed iterated formula.

If $\alpha (u)=\beta (u)$, the corresponding section is a singleton and its integral is zero. Thus coincident graphs, whether at isolated points or everywhere, require no exceptional convention.

Apply [F1] with lower graph $0$, upper graph $g$, and integrand $H=1$. This includes the empty base and the identically zero graph.

With $H=1$ the left side of the formula in [F1] is ${\int }_{U_g}1=\mathrm{cont}(U_g)$ and the inner integral is ${\int }_0^{g(u)}1dt=g(u)$, so that formula becomes $\mathrm{cont}(U_g)={\int }_{D}g$.

First apply [F1] over the compact interval $[a,b]$ to the graphs $-f$ and $f$, obtaining the compact Jordan base $D=\{(x,y):a\le x\le b,\mathrm{\mid }y\mathrm{\mid }\le f(x)\}$. By [F4], $(x,y)\mapsto f(x{)}^2-y^2$ is continuous and nonnegative on $D$; [F5] and [F6] make its nonnegative square root $\rho$ continuous. A second application of [F1] to the graphs $-\rho$ and $\rho$ identifies their solid with $S_x(f)$, so $S_x(f)$ is compact and Jordan measurable.

For each $x\in [a,b]$, the section of $S_x(f)$ perpendicular to the $x$-axis is the closed disc $y^2+z^2\le f(x{)}^2$, and [F2] gives it content $\pi f(x{)}^2$, including when $f(x)=0$.

The sectional-content function $x\mapsto \pi f(x{)}^2$ is continuous, so [F3] gives $\mathrm{cont}(S_x(f))={\int }_a^b\pi f(x{)}^{2}dx$. If $a=b$ or $f$ is identically zero, the same formula gives zero.

The outer disc solid and the inner disc solid are compact Jordan sets by [F1]. The boundary of $W$, the outer solid with the interior of the inner solid removed, lies in the union of their content-zero boundaries; hence [F4] makes $W$ compact and Jordan measurable.

At coordinate $x$, the section is the annulus between radii $g(x)$ and $f(x)$. Splitting the outer disc into that annulus and the inner disc, whose overlap is a boundary circle of content zero, [F2] and [F3] give section content $\pi (f(x{)}^2-g(x{)}^2)$.

By [F5], integration of the continuous section-content function gives the displayed washer formula. It gives zero when $f=g$ and reduces to the disc formula when $g=0$. By [F6], coordinate permutations preserve the construction and volume, proving the coordinate-axis clause.

Apply [F1] to the constant profile $f(x)=R$ on $[0,h]$.

The integral of the constant $\pi R^2$ over an interval of length $h$ is $\pi R^{2}h$. This remains zero when $R=0$ or $h=0$.

If $h=0$ or $R=0$, the cone is degenerate and both sides are zero. Otherwise apply [F1] to $f(x)=Rx\mathrm{/}h$ on $[0,h]$.

The function $G(x)=\pi R^2{x}^3\mathrm{/}(3h^2)$ has derivative $\pi R^2{x}^2\mathrm{/}{h}^2$, so [F2] evaluates the volume as $G(h)-G(0)=\pi R^{2}h\mathrm{/}3$.

The positive-height computation and the zero-parameter case together prove the formula for all $R,h\ge 0$.

If $r=0$, the ball is the singleton $\{0\}$, which is covered by a cube of arbitrarily small volume; its content and the displayed formula are both zero.

Suppose $r>0$. The ball is the solid of revolution of $f(x)=\sqrt{r^2-x^2}$ on $[-r,r]$, so [F1] gives $\mathrm{cont}({\bar{B}}_3(0,r))=\pi {\int }_{-r}^r(r^2-x^2)dx$.

A primitive is $G(x)=\pi (r^{2}x-x^3\mathrm{/}3)$; by [F2], $G(r)-G(-r)=4\pi r^3\mathrm{/}3$.

Steps 1.1 and 2.1 cover respectively $r=0$ and $r>0$, so the formula holds for every $r\ge 0$.

Let $D\subseteq {\mathbb{R}}^2$ be the closed disc of radius $r$. It is Jordan measurable by [F2], and it is closed and bounded, hence compact by [F7]. The triangle inequality in [F11], applied in both orders, gives $\mathrm{\mid }\mathrm{\parallel }u{\mathrm{\parallel }}_2-\mathrm{\parallel }v{\mathrm{\parallel }}_2\mathrm{\mid }\le \mathrm{\parallel }u-v{\mathrm{\parallel }}_2$, so the norm is continuous. On $[0,r]$, the estimate $\mathrm{\mid }(r^2-s^2)-(r^2-t^2)\mathrm{\mid }=\mathrm{\mid }s-t\mathrm{\mid }\mathrm{\mid }s+t\mathrm{\mid }\le 2r\mathrm{\mid }s-t\mathrm{\mid }$ makes $s\mapsto r^2-s^2$ continuous; [F9] and [F10] make the nonnegative square root continuous, and [F8] then makes $\rho (u):=\sqrt{r^2-\parallel u{\parallel }_2^2}$ continuous on $D$. Thus [F6] identifies $B$ with the compact Jordan solid between $-\rho$ and $\rho$. Fact [F6] likewise makes the cylinder between the constant graphs $-r,r$, the upper and lower cones between $\parallel u{\parallel }_2,r$ and $-r,-\parallel u{\parallel }_2$, and the comparison solid $C$ between $-\parallel u{\parallel }_2,\parallel u{\parallel }_2$ compact Jordan sets.

At height $z\in [-r,r]$, [F2] gives the ball section area $\pi (r^2-z^2)$. The comparison section is the radius-$r$ disc with the open radius-$\mathrm{\mid }z\mathrm{\mid }$ disc removed. It is bounded and its boundary lies in the two disc boundary circles, so [F12] makes it Jordan measurable; its overlap with the closed inner disc is the inner boundary circle and has content zero. Facts [F2] and [F5] therefore give the same area $\pi (r^2-z^2)$. At $z=\pm r$ both areas are zero.

The cylinder is the union of $C$ and the two cones from step 1.1. Their pairwise overlaps lie in boundaries, which have content zero by [F12], so [F5], [F3], and [F4] give $\mathrm{cont}(C)=2\pi r^3-2(\pi r^3\mathrm{/}3)=4\pi r^3\mathrm{/}3$.

The bounded Jordan sets $B$ and $C$ have the equal Jordan sections of step 2.1, so [F1] gives $\mathrm{cont}(B)=\mathrm{cont}(C)=4\pi r^3\mathrm{/}3$. The construction and calculation include $r=0$.

Let $A:=\{(x,z):a\le \parallel (x,z){\parallel }_2\le b\}$ and $h(x,z):=f(\parallel (x,z){\parallel }_2)$. The annulus is closed and bounded, hence compact by [F12], and its boundary lies in the two circle boundaries of closed discs; [F2] and [F6] make those circles content zero and then make $A$ Jordan measurable. The triangle inequality in [F7], applied in both orders, gives $\mathrm{\mid }\mathrm{\parallel }u{\mathrm{\parallel }}_2-\mathrm{\parallel }v{\mathrm{\parallel }}_2\mathrm{\mid }\le \mathrm{\parallel }u-v{\mathrm{\parallel }}_2$, so the norm is continuous; [F8] then makes $h$ continuous. Fact [F1] applied between the graphs $0$ and $h$ identifies the resulting solid with $S_y(f)$ and its volume with ${\int }_{A}h$.

If $a=b$, the annulus $A$ is the boundary circle of the closed disc of radius $a$, so [F2] and [F6] give it content zero. Fact [F10] bounds $h$ on $A$, and the single-set upper bound in [F9], which transfers the rectangle monotonicity of [F5] to Jordan-set integrals, gives $0\le {\int }_{A}h\le ({\max }_{A}h)\mathrm{cont}(A)=0$. The integral $2\pi {\int }_a^{b}xf(x)dx$ is also zero, so the theorem holds in this case. Henceforth assume $a<b$.

For a partition $a=r_0<\cdots <r_N=b$, let $A_i$ be the closed subannulus with radii $r_i,r_{i+1}$, and let $m_i,M_i$ be the minimum and maximum of $f$ on $[r_i,r_{i+1}]$, which exist by [F10]. Fact [F2], the boundary criterion [F6], and additivity [F11] give $\mathrm{cont}(A_i)=\pi (r_{i+1}^2-r_i^2)$. The $A_i$ cover $A$ and have pairwise disjoint interiors, so [F9] bounds ${\int }_{A}h$ between ${\sum }_i{m}_i\pi (r_{i+1}^2-r_i^2)$ and ${\sum }_i{M}_i\pi (r_{i+1}^2-r_i^2)$.

Uniform continuity from [F3] makes ${\max }_i(M_i-m_i)$ tend to zero with the mesh. Hence the difference between the upper and lower annular sums in step 3.1 is at most $\pi (b^2-a^2){\max }_i(M_i-m_i)$ and tends to zero.

Since $r_{i+1}^2-r_i^2=(r_i+r_{i+1})(r_{i+1}-r_i)$, each annular sum differs by a vanishing mesh error from a tagged Riemann sum for $2\pi xf(x)$. By [F4], steps 3.1 and 4.1 therefore squeeze ${\int }_{A}h$ to $2\pi {\int }_a^{b}xf(x)dx$. Together with step 2.1, the argument permits $a=0$, zeros of $f$, and $a=b$.

For $n=1$, the ball is the closed bounded interval $[-r,r]$, hence compact by [F7], and it is Jordan measurable with content $2r$, including $r=0$.

Assume closed balls in dimension $n-1$ are compact and Jordan measurable. The triangle inequality in [F8], applied in both orders, gives $\mathrm{\mid }\mathrm{\parallel }u{\mathrm{\parallel }}_2-\mathrm{\parallel }v{\mathrm{\parallel }}_2\mathrm{\mid }\le \mathrm{\parallel }u-v{\mathrm{\parallel }}_2$, so the Euclidean norm is continuous. On $[0,r]$, the estimate $\mathrm{\mid }(r^2-s^2)-(r^2-t^2)\mathrm{\mid }=\mathrm{\mid }s-t\mathrm{\mid }\mathrm{\mid }s+t\mathrm{\mid }\le 2r\mathrm{\mid }s-t\mathrm{\mid }$ makes $s\mapsto r^2-s^2$ continuous; [F5] and [F6] make its nonnegative square root continuous, and [F4] makes the resulting composite continuous on the ball. Thus the $n$-ball is the solid between two continuous square-root graphs over ${\bar{B}}_{n-1}(0,r)$, and [F1] makes it compact and Jordan measurable.

For $\mathrm{\mid }t\mathrm{\mid }\le r$, the section at last coordinate $t$ is the $(n-1)$-ball of radius $\rho (t):=\sqrt{r^2-t^2}$. It is the image of the unit $(n-1)$-ball under scalar multiplication by $\rho (t)$, whose determinant has absolute value $\rho (t{)}^{n-1}$; [F2] gives section content $V_{n-1}(1)(r^2-t^2{)}^{(n-1)\mathrm{/}2}$. At $t=\pm r$ this is zero.

By [F3], integration of the section contents in step 3.1 gives the displayed recursion. Thus the induction proves Jordan measurability in every positive dimension and the recursion for every $n\ge 2$.