Volumes of Elementary Solids and Solids of Revolution: Examples and Counterexamples

By [F1], the cylinder volume is $\pi \cdot 2^2\cdot 3=12\pi$.

By [F2], the cone volume is $\pi \cdot 2^2\cdot 3\mathrm{/}3=4\pi =(12\pi )\mathrm{/}3$.

At height $y\in [-r,r]$, put $q(y)=\sqrt{r^2-y^2}$. The outer and inner radii are $R+q(y)$ and $R-q(y)$; both are nonnegative because $R>r\ge q(y)$.

By [F1], the washer area is $\pi ((R+q{)}^2-(R-q{)}^2)=4\pi Rq$, so the torus volume is $4\pi R{\int }_{-r}^r\sqrt{r^2-y^2}dy$.

The integral in step 2.1 is the area under the upper semicircle of radius $r$, hence half the disc content [F2], namely $\pi r^2\mathrm{/}2$. Thus the volume is $2{\pi }^{2}R{r}^2$.

Put $f_1(x)=1$ and $f_2(x)=(x-1{)}^2$ on $[0,2]$; both are continuous and nonnegative and $f_2\le f_1$ there. By [F1] the solids $S_1$ and $S_2$ obtained by revolving $0\le y\le f_i(x)$ about the $y$-axis are compact and Jordan measurable with contents $2\pi {\int }_0^{2}xdx=4\pi$ and $2\pi {\int }_0^{2}x(x-1{)}^{2}dx=4\pi \mathrm{/}3$.

At height $0\le y\le 1$, the washer radii are $1+\sqrt{y}$ and $1-\sqrt{y}$. By [F2], its area is $4\pi \sqrt{y}$, and ${\int }_0^{1}4\pi \sqrt{y}dy=8\pi \mathrm{/}3$.

In cylindrical terms, writing $\rho$ for the distance to the $y$-axis, the solid $S$ of the Example is $\{f_2(\rho )\le y\le f_1(\rho ),\rho \le 2\}$, while $S_i=\{0\le y\le f_i(\rho ),\rho \le 2\}$. Hence $S\cup S_2=S_1$, and $S\cap S_2$ is the set $\{y=f_2(\rho )\}$, the graph of a continuous function on the closed disc of radius $2$, which has content zero by [F4]. The boundary of the bounded set $S$ lies in the union of the boundaries of $S_1$ and $S_2$, which have content zero by [F5] and step 1.1, so [F5] makes $S$ Jordan measurable.

By [F3] applied to $S$ and $S_2$, $\mathrm{cont}(S_1)=\mathrm{cont}(S)+\mathrm{cont}(S_2)$, so step 1.1 gives $\mathrm{cont}(S)=4\pi -4\pi \mathrm{/}3=8\pi \mathrm{/}3$; this is the shell value $2\pi {\int }_0^{2}x(1-(x-1{)}^2)dx$, the shell height being the difference of the two profiles.

The two independent descriptions therefore give the same volume $8\pi \mathrm{/}3$.

By [F1], the truncation has volume $\pi {\int }_1^R{x}^{-2}dx=\pi (1-1\mathrm{/}R)$.

By [F2] with $p=2$, the improper integral tends to $1$, so the truncated volumes tend to $\pi$.

Thus the horn has finite improper volume $\pi$. No assertion about its lateral surface area is used here.

Starting with $V_1=2$, [F1] and the upper-semicircle area give $V_2=\pi$, while ${\int }_{-1}^1(1-t^2)dt=4\mathrm{/}3$ gives $V_3=4\pi \mathrm{/}3$.

With $t=\sin \theta$, ${\int }_{-1}^1(1-t^2{)}^{3\mathrm{/}2}dt=2{\int }_0^{\pi \mathrm{/}2}{\cos }^4\theta d\theta =3\pi \mathrm{/}8$ by [F2], giving $V_4={\pi }^2\mathrm{/}2$. Also ${\int }_{-1}^1(1-t^2{)}^{2}dt=16\mathrm{/}15$, giving $V_5=8{\pi }^2\mathrm{/}15$.

Combining the preceding calculations gives the displayed table without using the Gamma closed form.

By [F3], the set $E$ is closed and bounded, hence compact, and $S$ has empty interior. Therefore $E$ has empty interior; being closed, it equals its boundary.

Consider any finite axis-parallel box cover of $E$. Partition $[0,1{]}^2$ at all endpoints of the last two coordinate intervals of those boxes. For the midpoint of each nondegenerate planar cell, the first-coordinate intervals of the boxes active there cover $S$, so [F1] makes their total length at least $1\mathrm{/}2$. Multiplying by the cell area and summing shows that the original boxes have total volume at least $1\mathrm{/}2$.

Thus $\mathrm{\partial }E=E$ does not have content zero. By [F2], the compact set $E$ is not Jordan measurable, refuting the claim.

By [F1], the set $S\times [0,1{]}^2$ is compact but not Jordan measurable, so its Jordan volume is not defined.

This single compact witness refutes the universal claim.

At each height $z\in [0,1]$, both boxes have a rectangular section of area $1$; outside that range both sections are empty. Thus [F1] gives equal volume.

The diameters are $\sqrt{3}$ for $E$ and $\sqrt{2^2+(1\mathrm{/}2{)}^2+1^2}=\sqrt{21}\mathrm{/}2$ for $F$. Since these are unequal and [F2] makes every isometry preserve diameter, the boxes are not congruent.

Hence equal parallel cross-sectional areas determine equal volume here but do not force congruence.

For each fixed $y$, the $x$-section is either $x\mapsto x$ or the zero function, and in either case its integral over $[-1,1]$ is $0$. Thus the $x$-first iterated integral exists and equals $0$.

For fixed $x\mathrm{\ne }0$, density from [F2] makes the lower and upper integrals of the $y$-section equal to $\min \{x,0\}$ and $\max \{x,0\}$. Integrating these envelopes over $[-1,1]$ gives $-1\mathrm{/}2$ and $1\mathrm{/}2$, which are unequal; [F1] therefore rules out multiple Riemann integrability.

The bounded function $f$ has the iterated integral from step 1.1 but is not Riemann integrable on the rectangle, so the claim is false.