Solving problem 3-29 in Spivak´s Calculus on Manifolds without using change of variables

Problem 3-29 (p. 61) in the section treating Fubini´s theorem reads:

Use Fubini´s theorem to derive an expression for the volume of a set of $\mathbb{R}^{3}$ obtained by revolving a Jordan-measurable set in the $yz$-plane about the $z$-axis.

By making some simplying assumptions about the plane region and changing variables to cylindrical coordinates we can obtain an expression for the volume. However, the material on the change of variables is treated two sections later (p. 66). Thus, is there is a way to solve the problem without using change of variables?

The definition of Jordan-measurable can be found on p. 56 (See also Theorem 3-9, p.55), but I sumarize it here: Spivak defines a bounded subset $C$ of $\mathbb{R}^{n}$ to be Jordan measurable if the topological boundary of $C$ has measure $0$, i.e. if for any $\varepsilon>0$ there is a cover $\{U_{i}\}$ of $\mathrm{Bd}(C)$ by closed $n$-cubes such that $\sum_{i}v(U_{i})<\varepsilon$ (where $v(U_{i})$ denotes the $n$-dimensional volume of the rectangle $U_{i}$). If $C$ is Jordan measurable it is contained inside some closed $n$-cube $A$ and the characteristic function $\chi_{C}$ is integrable on $A$. The integral $\int_{A}\chi_{C}$ is called the $n$-dimensional volume of $C$.

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The following argument in fact uses Fubini's theorem (twice!), but makes simplifying assumptions about the given domain $D$ in the meridian half-plane. I use $(r,z)$ instead of $(y,z)$ as coordinates in this "abstract" plane.

Assume that $D$ is given in the form $$D:=\{(r,z)\ |\ a\leq z\leq b,\ 0<c(z)\leq r\leq d(z)\}\ ,$$ with $a<b$ and $c(\cdot)$, $d(\cdot)$ integrable. Then the three-dimensional body $B$ whose volume we have to compute is given by $$B=\{(x,y,z)\ |\ a\leq z\leq b,\ c(z)\leq \sqrt{x^2+y^2}\leq d(z)\}\ .$$ For given $z\in[a,b]$ let
$$B_z:=\{(x,y)\ |\ c(z)\leq \sqrt{x^2+y^2}\leq d(z)\}$$ denote the intersection of $B$ with the horizontal plane at level $z$. The set $B_z$ is an annulus with inner radius $c(z)$ and outer radius $d(z)$. Therefore its area is given by $${\rm area}(B_z)=\pi\bigl(d^2(z)-c^2(z)\bigr)=2\pi\int_{c(z)}^{d(z)}r \ dr$$ (writing this area as an integral is a trick).

Now comes Fubini: \eqalign{{\rm vol}(B)&=\int_B 1\ {\rm d}(x,y,z)=\int_a^b \int_{B_z} 1\ {\rm d}(x,y)\ dz\cr &=\int_a^b {\rm area}(B_z)\ dz =2\pi\int_a^b \int_{c(z)}^{d(z)} r\ dr \ dz \cr &=2\pi \int_D r\ {\rm d}(r,z)\ .\cr}

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I may be mistaken, but isn´t this solution basically a change of variables to cylindrical coordinates? –  John Nov 19 '12 at 19:07
@John: Heavens; what have you expected, given that in the final formula there is a $2\pi$ in front and an extra factor $r$ in the integral? Using some elementary geometry about areas of circles it was possible to avoid Jacobians, etc. $--$ This comment is my last word on this matter. –  Christian Blatter Nov 19 '12 at 19:28
Assume that you are given a domain $D$ in the $(y,z)$-plane such that $y>0$ for all $(y,z)\in D$. Then a tiny rectangle $Q:=[y-u, y+u]\times[z-v,z+v]\subset D$ produces under rotation a cylindrical ring $R$ of exact volume $${\rm vol}(R)=2\pi y\cdot {\rm area}(Q)$$ (this is one of Guldin's rules).
Consider now approximations of $D$ by suitable unions of "almost disjoint" rectangles $Q_i$ $\ (1\leq i\leq N)$, and let $R_i$ be the rings generated by the $Q_i$. Then the volume of the considered body $B$ is approximatively given by $${\rm vol}(B)\doteq\sum_{i=1}^N{\rm vol}(R_i)=\sum_{i=1}^N 2\pi y_i\ {\rm area}(R_i)\doteq 2\pi \int_D y \ {\rm d}(y,z)\ ;\qquad(1)$$ whereby the $\doteq$'s can be made as exact as desired. As the outer terms in $(1)$ each have a fixed value these two values have to be equal, i.e., we obtain $${\rm vol}(B)= 2\pi \int_D y \ {\rm d}(y,z)\ .$$ (Depending on the analytical apparatus developed for the Jordan measure and integral you can replace $(1)$ by a squeeze argument or similar.)
@John: ${\rm d}(y,z)$ denotes the "area element" in ${\mathbb R}^2$. Spivak's integral notation seems to omit it. "Fubini" as a naked theorem has not been used. You have to bring in some elementary geometry, or the factor $2\pi$ wouldn't make its appearance in the final formula. –  Christian Blatter Nov 19 '12 at 13:18