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

Question

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 $C$.

Answer 1

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

Answer 2

Remark: I don't have a copy of Spivak's book at hand.

You don't need the full strength of the change of variables formula.

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.)