Volume of a solid with specific cross sections

Question

We can use definite integrals to find volume of a solid with specific cross sections on the interval.

If the cross section is perpendicular to $x$-axis and its area is a function of $x$, say $A(x)$, then the volume $V$ of the solid on $[a, b]$ can be found using the formula

$V = \int_a^b A(x)dx$

Now in the proof of lecture notes it is assumed that all that cross sections are into the other like this(meaning they are not disjoint and they can't partially intersect)

For example take ellipsoid.

I understand why this is assumed because we are proving using darboux sums.But question is can you first bring some counterexample that this method does not work?And if I want to calculate volume with this method then how to check that all cross sections are into the other.Thank you.

Answer 1

$\newcommand{\d}{\,\mathrm{d}}$Cavalieri's principle is an intuitive statement relating the volumes of two solids if they have the same cross-sectional areas for every possible cross-section. An informal justification for $V=\int_a^bA(x)\d x$ is that the collection of cross-sections summed over $[a,b]$ will lead to the same volume, by Cavalieri's principle, so long as we account for every cross-section; thus taking limits leads to the integral from a Riemann sum. At no point do we have any requirements on the geometry of the solid other than it must have a meaningful volume, and the area function should be Riemann-integrable over $[a,b]$. The projections of the cross sections can theoretically be disjoint, but this would just lead to a rather bizarre shape that would be difficult to call a solid, but if there were finitely many places where the cross sections are disjoint you could think about the sum of a finite collection of "normal" solids. If there were infinitely many discontinuities in this sense, the solid might not be "measurable" and the volume would be ill-defined by most people's standards.

This is not really a "proof" however.

You may have heard of measure theory, a more general theory of integration. Henri Lebesgue developed the theory of the Lebesgue measure, $\mu$, on $\Bbb R^n$, which is the natural area/volume: the measure of a box is the product of its side lengths, and it is translationally invariant, and scales under linear maps by the determinant, etc. and if you study its construction a little more you'll see that $\mu(S)$ is truly the "volume" of $S\subset\Bbb R^n$ in its most natural sense. We can find the measure of a set by integrating along its characteristic function; $\chi_S$ is the function for which $\chi_S(x)=1$ if $x\in S$, and $\chi_S(x)=0$ otherwise:

$$\text{Volume of $S$}=\mu(S)=\int_{\Bbb R^n}\chi_S\d\mu=\int_{S}1\d\mu$$

This is just like how $b-a=\int_a^b1\d x$, the "volume" of a line in one-dimensional space, and is just a more general way of expressing this. Hopefully this is making sense, even if you've not studied measure theory.

A very important and famous theorem from measure theory is Fubini's theorem, one corollary of which is that I can split an integral over $\Bbb R^n$ into an integral over $\Bbb R^{n-1}$ and $\Bbb R$. This theorem is what powers the rigorous "reason" why these integrations work (but do think about how you want "volume" to be defined - under a different definition, i.e. a different measure, this might not work - Lebesgue's measure is just the "natural" measure of volume). From here on out I'll take $\Bbb R^3$ and split into $\Bbb R^2$ and $\Bbb R$ to give a direct analogy to your cross-sectional area problem.

So I can do this (where $\mu_2,\mu_1$ are two and one-dimensional measures, i.e. area and line length, or area and Riemann's "$\d x$"):

$$\begin{align}\text{Volume of $S$}=\mu(S)&=\int_{\Bbb R^3}\chi_S(x,y,z)\d\mu\\&=\int_{\Bbb R}\left(\int_{\Bbb R^2}\chi_S(x,y,z)\d\mu_2(y,z)\right)\d\mu_1(x)\\&=\int_{\Bbb R}A(x)\d x\\&=\int_a^b A(x)\d x\end{align}$$

Where $A(x)$ is $\int_{\Bbb R^2}\chi_S(x,y,z)\d\mu_2(y,z)$, i.e. the area (because $\mu_2,\Bbb R^2$) of the $2$-D $x$-slice of $S$, the cross-sectional area.

The measure theory there is just a very elegant way to justify it, but you don't need to know it all to make sense of this. It is just a fact (Fubini) that I can split volume integrals into $1$-D integrals of area and this is all justified and correlates to the intuitive definitions of area and volume - a powerful extension of $\int_a^b 1\d x=b-a$. Oh, and these fancy $\mu$-integrals are actually the same as Riemann integrals when the Riemann integral exists, but you can take a $\mu$-integral more often - here, we just need that $S$ is "measurable" - all this means is that it's a well-behaved patch of space; e.g. if I move it around, its volume won't change (although this is not the only requirement). Surprisingly, not all sets are measurable - see the Banach-Tarski paradox: Wikipedia, Banach-Tarski paradox: entertaining VSauce video and the Vitali set for more.

To re-iterate, this is only a proof in the sense that it shows that the natural definition of volume ($\mu$) can be related to these cross-sectional integrals. Had you used a different measure which splits differently, or a different notion of volume, or if you harbour objections to Lebesgue's measure being natural, then you can refute the above. This is just "the" definition that is overwhelmingly used.

About the torus:

Let $r,R$ be the minor and major radii, and put $a=R+r$, $b=R-r$. Then $a$ is the maximum distance from a point to the centre, and $b$ is the minimum. The area of an ellipse is $\pi\alpha\beta$, where $\alpha,\beta$ are the semi-minor/major axes. I will split the torus into two sections; the section where the cross-section is solid, and the inner section where there is a gap. In the image I provide I call these sections $A_1,A_2$. In $A_1$, if we consider $x$ to be the distance to the centre point, the semi-major axis of the cross-sectional ellipse about that point is given by pythagoras: $a^2=x^2+\alpha^2,\,\alpha=\sqrt{a^2-x^2}$. The semi-minor axis is always $r$ by construction of the torus ($r$ is how much the circular shape pokes up). Then, in $A_1$, $A(x)=\pi r\sqrt{a^2-x^2}$.

In the section $A_2$, the cross-section falls into two disjoint ellipses of equal area. The semi-minor axis is still $r$ for both, and the semi-major axis is given by (as explained geometrically in the image, taking $\alpha$ as the difference of the two horizontals) $\sqrt{a^2-x^2}-\sqrt{b^2-x^2}$. Then $A(x)=\pi r(\sqrt{a^2-x^2}-\sqrt{b^2-x^2})$ in this region.

Altogether:

$$\begin{align}V=\int_{-a}^aA(x)\d x&=\underset{A_1}{\underbrace{\int_{-a}^{-b}A(x)\d x+\int_b^aA(x)\d x}}+\underset{A_2}{\underbrace{\int_{-b}^bA(x)\d x}}\\&=2\pi r\int_b^a\sqrt{a^2-x^2}\d x+2\pi r\int_0^b\sqrt{a^2-x^2}\d x-2\pi r\int_0^b\sqrt{b^2-x^2}\d x\\&=2\pi r\left(\int_0^a\sqrt{a^2-x^2}\d x-\int_0^b\sqrt{b^2-x^2}\d x\right)\\&=2\pi r\left(a^2\color{red}{\int_0^1\sqrt{1-x^2}\d x}-b^2\color{red}{\int_0^1\sqrt{1-x^2}\d x}\right)\\&=2\pi r(a^2-b^2)\color{red}{\frac{\pi}{4}}\\&=2\pi r\cdot\frac{\pi}{4}(a-b)(a+b)\\&=2\pi r\cdot\frac{\pi}{4}(2r)(2R)\\&=\color{red}{2\pi}\cdot\color{green}{r\pi r}\cdot\color{red}{R}\\&=(2\pi R)(\pi r^2)\end{align}$$

Which is the correct answer, and so we see that even having disjoint cross-sections is fine if the solid is nice enough!