8.5. Calculating Content

We are now going to develop a general concept for calculating content. We will do this by assigning adequate values ${V}_{n+1}\left(M\right)$ to suitable subsets M of ${ℝ}^{n+1}$ that, in the two-dimensional case coincide with the areas introduced in 8.4 and meet our image of a volume in the three-dimensional one.

These values will be recursively designed, as suggested by the interpretation of areas at the end of 8.4: The (n + 1)-dimensional content of a set M will result from integrating the n-dimensional contents of its sections ${M}_{x}$.

Definition:  For a subset $M\subset \left[a,b\right]×{ℝ}^{n}$ of ${ℝ}^{n+1}$ and any $x\in \left[a,b\right]$ the set

 ${M}_{x}≔\left\{\left({y}_{1},\dots ,{y}_{n}\right)\in {ℝ}^{n}|\left(x,{y}_{1},\dots ,{y}_{n}\right)\in M\right\}$ [8.5.1]

is called a section in M.

The following applet visualises a section in a sugar loaf at $x=0$ (display by JavaView i JavaView is an interactive 3D geometry viewer. Display is controlled by the left-mouse, for example: press "o" and drag to rotate (preset) press "s" and drag to scale press "t" and drag to translate Go to www.javaview.de/jars/shortcuts.html for a complete list of options. The right-mouse launches an extensive context menu.
).

As a start we calculate sections in some manageable subsets of ${ℝ}^{2}$ and of ${ℝ}^{3}$. All the parameters in this and the subsequent examples are strictly positive.

Example:

• All sections in a Square  $Q≔{\left[0,a\right]}^{2}\subset \left[0,a\right]×ℝ$ are (constant) intervals as we have for any $x\in \left[0,a\right]$:

 ${Q}_{x}=\left\{y\in ℝ|\left(x,y\right)\in \left[0,a\right]×\left[0,a\right]\right\}=\left[0,a\right]$ 
• The sections in an ellipse  $E≔\left\{\left(x,y\right)\in {ℝ}^{2}|{b}^{2}{x}^{2}+{a}^{2}{y}^{2}\le {a}^{2}{b}^{2}\right\}\subset \left[-a,a\right]×ℝ$ are (variable) intervals as we calculate for each $x\in \left[-a,a\right]$:

 $\begin{array}{ll}{E}_{x}\hfill & =\left\{y\in ℝ|{b}^{2}{x}^{2}+{a}^{2}{y}^{2}\le {a}^{2}{b}^{2}\right\}\hfill \\ \hfill & =\left\{y\in ℝ|{y}^{2}\le \frac{{b}^{2}}{{a}^{2}}\left({a}^{2}-{x}^{2}\right)\right\}=\left[-\frac{b}{a}\sqrt{{a}^{2}-{x}^{2}},\frac{b}{a}\sqrt{{a}^{2}-{x}^{2}}\right]\hfill \end{array}$ 

Note that when ${x}^{2}=a$ the interval ${E}_{x}$ is a singleton: $\left[0,0\right]=\left\{0\right\}$.

• Sections in a perforated disc  $L≔\left\{\left(x,y\right)\in {ℝ}^{2}|{r}^{2}\le {x}^{2}+{y}^{2}\le {R}^{2}\right\}\subset \left[-R,R\right]×ℝ$ are a bit laborious to calculate. Let us just start with an $x\in \left[-R,R\right]$:

 

If $r=R$ the sections in L are one- or two-element sets:

• Sections in a cube  $W≔{\left[0,a\right]}^{3}\subset \left[0,a\right]×{ℝ}^{2}$ are (constant) squares because we have for $x\in \left[0,a\right]$:

 ${W}_{x}=\left\{\left(y,z\right)\in {ℝ}^{2}|\left(x,y,z\right)\in {\left[0,a\right]}^{3}\right\}={\left[0,a\right]}^{2}$ 
• Sections in a sphere

$S≔\left\{\left(x,y,z\right)\in {ℝ}^{3}|{x}^{2}+{y}^{2}+{z}^{2}\le {r}^{2}\right\}\subset \left[-r,r\right]×{ℝ}^{2}$

are circles (with varying radii) as we have for every $x\in \left[-r,r\right]$:

 $\begin{array}{ll}{S}_{x}\hfill & =\left\{\left(y,z\right)\in {ℝ}^{2}|{x}^{2}+{y}^{2}+{z}^{2}\le {r}^{2}\right\}\hfill \\ \hfill & =\left\{\left(y,z\right)\in {ℝ}^{2}|{y}^{2}+{z}^{2}\le {r}^{2}-{x}^{2}\right\}\hfill \end{array}$ 

If ${x}^{2}={r}^{2}$ the section ${S}_{x}$ is a circle with radius 0, i.e. a point.

Now we define by recursion which sets are suitable for measuring content and how to calculate it. The remarks at the end of the previous section serve as a guideline.

Definition:

• A subset $M\subset ℝ$ has a one-dimensional content if $M=\left[{l}_{1},{r}_{1}\right]⊍\dots ⊍\left[{l}_{k},{r}_{k}\right]$ is a disjoint union i The union $A\cup B$ of two sets A and B is called disjoint if $A\cap B=\varnothing$. In that case the symbol $\cup$ is often replaced by $⊍$. Display problems should be solved by downloading the font Lucida Bright Math Symbol.
of finitely many closed intervals, ${l}_{i}\le {r}_{i}$. In that case we call the number

 ${V}_{1}\left(M\right)≔\sum _{i=1}^{k}{r}_{i}-{l}_{i}$ [8.5.2]

the one-dimensional content of M. In addition we set: ${V}_{1}\left(\varnothing \right)≔0$.

• A subset $M\subset \left[a,b\right]×{ℝ}^{n}$, $a, has an ($n+1$)-dimensional content if every section ${M}_{x}$, $x\in \left[a,b\right]$, has an n-dimensional content ${V}_{n}\left({M}_{x}\right)$ such that the funtion ${V}_{n}\left({M}_{\mathrm{X}}\right):\left[a,b\right]\to ℝ$ is integrable on $\left[a,b\right]$. In this case the number

 ${V}_{n+1}\left(M\right)≔\underset{a}{\overset{b}{\int }}{V}_{n}\left({M}_{\mathrm{X}}\right)$ [8.5.3]

is called the ($n+1$)-dimensional content of M. If $M\subset \left[a,a\right]×{ℝ}^{n}$ we additionally set: ${V}_{n+1}\left(M\right)≔0$.

Consider:

• Sometimes, especially in three-dimensional situations, the term volume is used instead of content.

• A simple inductive argument shows that contents are positive:  ${V}_{n}\left(M\right)\ge 0$.

• Due to the different add ons the content of any finite set is zero:

${V}_{n}\left(\varnothing \right)={V}_{n}\left(\left\{{x}_{1},\dots ,{x}_{k}\right\}\right)=0$

• [8.5.2] includes the common special case $k=1$. M then is a simple interval, $M=\left[l,r\right]$, $l\le r$, and the content ${V}_{1}$ shortens to ${V}_{1}\left(M\right)=r-l$.

• [8.5.3] infact introduces the Cavalieri Principle:

If any two sets $M,N\subset \left[a,b\right]×{ℝ}^{n}$ have a content such that all of their sections have the same content then they coincide in their own content as well:

 [8.5.4]

• [8.5.2] carries forward the area concept of 8.4 as the areas calculated there now reappear as two-dimensional contents: For any positive function $f\in {\mathcal{C}}^{0}\left(\left[a,b\right]\right)$ the set

$M≔\left\{\left(x,y\right)\in {ℝ}^{2}|a\le x\le b\text{\hspace{0.28em}}\wedge \text{\hspace{0.28em}}0\le y\le f\left(x\right)\right\}\subset \left[a,b\right]×ℝ$

has a two-dimensional content ${V}_{2}\left(M\right)$ that is exactly the area of the region generated by f and the x-axis on $\left[a,b\right]$.

Proof:  For $x\in \left[a,b\right]$ we have: ${M}_{x}=\left\{y\in ℝ|0\le y\le f\left(x\right)\right\}=\left[0,f\left(x\right)\right]$, so that ${V}_{1}\left({M}_{x}\right)=f\left(x\right)$. Therefor the content of M is calculated to

${V}_{2}\left(M\right)=\underset{a}{\overset{b}{\int }}{V}_{1}\left({M}_{\mathrm{X}}\right)=\underset{a}{\overset{b}{\int }}f$

In a first example we recalculate the content of some well known two-dimensional objects.

 Example: Square Ellipse Circle Perforated Disc Line Rectangle Triangle
• The content of a square $Q={\left[0,a\right]}^{2}\subset \left[0,a\right]×ℝ$ with edge length a is

 ${V}_{2}\left(Q\right)={a}^{2}$ 

Proof:  According to  all sections in Q have a constant content:

${V}_{1}\left({Q}_{x}\right)={V}_{1}\left(\left[0,a\right]\right)=a$

Thus ${V}_{1}\left({Q}_{\mathrm{X}}\right)$ is integrable on $\left[0,a\right]$ so that Q has a content, namely

${V}_{2}\left(Q\right)=\underset{0}{\overset{a}{\int }}{V}_{1}\left({Q}_{\mathrm{X}}\right)=\underset{0}{\overset{a}{\int }}a={a}^{2}$

In the next example we calculate some of the classical three-dimensional volumia using our previous results  to .

 Example: Cube Ellipsoid Sphere Cone Pyramid Torus Cuboid Cylinder
• The content of a cube $W={\left[0,a\right]}^{3}$ of edge length a is

${V}_{3}\left(W\right)={a}^{3}$

Proof:  Due to  the sections ${W}_{x}={\left[0,a\right]}^{2}$ (see ) in $W\subset \left[0,a\right]×{ℝ}^{2}$ have the constant content ${V}_{2}\left({W}_{x}\right)={a}^{2}$. Thus W has a content, namely

${V}_{3}\left(W\right)={a}^{2}\underset{0}{\overset{a}{\int }}1={a}^{3}$

Some of the examples above (for instance the sphere and the cone) represent a general class of subsets of ${ℝ}^{3}$ that have a content, the so-called solids of revolution. They are turn-shaped and are generated by a function rotating around the x-axis.

Notation and Proposition:  For any continuous function $f:\left[a,b\right]\to ℝ$ the set

 $R≔\left\{\left(x,y,z\right)\in {ℝ}^{3}|x\in \left[a,b\right]\text{\hspace{0.28em}}\wedge \text{\hspace{0.28em}}{y}^{2}+{z}^{2}\le {\left(f\left(x\right)\right)}^{2}\right\}$ [8.5.5]

is called the solid of revolution generated (on $\left[a,b\right]$) by the casing function f. R has the content

${V}_{3}\left(R\right)=\pi \underset{a}{\overset{b}{\int }}{f}^{2}$

Proof:  We note that $R\subset \left[a,b\right]×{ℝ}^{2}$ and that ${R}_{x}$ is a circle of radius $|f\left(x\right)|$ for each $x\in \left[a,b\right]$. So we have: ${V}_{2}\left({R}_{\mathrm{X}}\right)=\pi {f}^{2}$, which is infact the assertion.

As an example we calculate the content of the solid of revolution R (pulley) generated on $\left[-1,1\right]$ by the casing function ${\mathrm{X}}^{2}+1$:

$\begin{array}{ll}{V}_{3}\left(R\right)\hfill & =\pi \underset{-1}{\overset{1}{\int }}\left({\mathrm{X}}^{2}+1{\right)}^{2}\hfill \\ \hfill & =\pi \underset{-1}{\overset{1}{\int }}{\mathrm{X}}^{4}+2{\mathrm{X}}^{2}+1\hfill \\ \hfill & =\pi \left(\frac{1}{5}{\mathrm{X}}^{5}+\frac{2}{3}{\mathrm{X}}^{3}+\mathrm{X}\phantom{\phantom{\rule{0pt}{12pt}}}\right){|}_{-1}^{1}=\frac{56}{15}\pi \hfill \end{array}$

Exercise:  The content of the ellipsoid

$E≔\left\{\left(x,y,z\right)\in {ℝ}^{3}|x\in \left[-a,a\right]\text{\hspace{0.28em}}\wedge \text{\hspace{0.28em}}{y}^{2}+{z}^{2}\le \frac{{b}^{2}}{{a}^{2}}\cdot \left({a}^{2}-{x}^{2}\right)\right\}$ i The ellipsoid in this exercise is characterised by two identical semi axes, namely b which is the semi axis in y and z direction as well. This is due to the fact that E is generated as rotational solid by the upper semi ellipse $\left\{\left(x,y\right)\in {ℝ}^{2}|\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\right\}$. semi ellipse with a = 2 and b = 1.2 For the radius $y=f\left(x\right)$ that is required to calculate the sections ${E}_{x}$ we look at the following consideration: $\begin{array}{ll}\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\hfill & ⇔{b}^{2}{x}^{2}+{a}^{2}{y}^{2}={a}^{2}{b}^{2}\hfill \\ \hfill & ⇔{y}^{2}=\frac{{a}^{2}{b}^{2}-{b}^{2}{x}^{2}}{{a}^{2}}\hfill \\ \hfill & ⇔y=\frac{b}{a}\sqrt{{a}^{2}-{x}^{2}}\hfill \end{array}$

calculates to  ${V}_{3}\left(E\right)=\frac{4}{3}a{b}^{2}\pi$.

Proof: ?

If we turn to n-dimensional content, employment of the induction principle is unavoidable. As a first example we calculate the content of an n-dimensional cube and an n-dimensional sphere respectively.

Proposition:

1. The content of the n-dimensional cube ${W}_{n}={\left[0,a\right]}^{n}$ is

 ${V}_{n}\left({W}_{n}\right)={a}^{n}$ [8.5.6]
1. The content of the n-dimensional sphere ${S}_{n}=\left\{x\in {ℝ}^{n}|{x}_{1}^{2}+\dots +{x}_{n}^{2}\le {r}^{2}\right\}$ is

 [8.5.7]

1.   Proof by induction:

• $n=1$:  ${W}_{1}=\left[0,a\right]$ is a closed interval such that ${V}_{1}\left({W}_{1}\right)=a={a}^{1}$.

• $n\text{\hspace{0.28em}}⇒\text{\hspace{0.17em}}n+1$:  Now let ${W}_{n+1}={\left[0,a\right]}^{n+1}=\left[0,a\right]×{\left[0,a\right]}^{n}$ be an $\left(n+1\right)$-dimensional cube. For any $x\in \left[0,a\right]$ we see that the section ${W}_{n+1,x}={\left[0,a\right]}^{n}$ is an n-dimensional cube, thus having a content due to the induction hypothesis. The constant function ${V}_{n}\left({W}_{n+1,\mathrm{X}}\right)={a}^{n}$ is integrable, so that ${W}_{n+1}$ has a content as well, namely

${V}_{n+1}\left({W}_{n+1}\right)={a}^{n}\underset{0}{\overset{a}{\int }}1={a}^{n+1}$

2.   Proof by induction:

• $n=1$:  ${S}_{1}=\left\{x\in ℝ|{x}^{2}\le {r}^{2}\right\}=\left[-r,r\right]$ is a closed interval such that

${V}_{1}\left({S}_{1}\right)=2r=\frac{{2}^{1}\cdot 0!}{1!}{\pi }^{0}{r}^{1}$

Note that $n=2\cdot 0+1$ in this case.

• $n\text{\hspace{0.28em}}⇒\text{\hspace{0.28em}}n+1$:  Next let ${S}_{n+1}=\left\{\left(x,y\right)\in {ℝ}^{n+1}|{x}^{2}+{y}_{1}^{2}+\dots +{y}_{n}^{2}\le {r}^{2}\right\}\subset \left[-r,r\right]×{ℝ}^{n}$ be an ($n+1$)-dimensional sphere. The section ${S}_{n+1,x}=\left\{y\in {ℝ}^{n}|{y}_{1}^{2}+\dots +{y}_{n}^{2}\le {r}^{2}-{x}^{2}\right\}$ is an n-dimensional sphere for all $x\in \left[-r,r\right]$, thus has a content according to the induction hypothesis. The function

is continuous, and that means integrable as well, so that ${S}_{n+1}$ has a content.

The integral (substituting $g=r\cdot \mathrm{sin},\text{\hspace{0.28em}}{g}^{\prime }=r\cdot \mathrm{cos}$, see [8.3.5])

now allows to calculate the ($n+1$)-dimensionale sphere's content:

We follow up our subject and consider now generalised cones and cylinders. For an arbitrary non-empty subset $\varnothing \ne G\subset {ℝ}^{n}$ of ${ℝ}^{n}$ we call the set

 $Z≔\left[0,h\right]×G$ [8.5.8]

a (generalised) cylinder with base G and height h.

 $C≔\left\{\left(x,\frac{x}{h}{y}_{1},\dots \frac{x}{h}{y}_{n}\right)\in {ℝ}^{n+1}|x\in \left[0,h\right]\text{\hspace{0.28em}}\wedge \text{\hspace{0.28em}}y\in G\right\}$ [8.5.9]

a (generalised) cone with base G and height h.

Consider:

• Only if $n=2$ the base is an area in the usual sense, of course.

• A (generalised) pyramide is a cone whose base is a polytope.

Exercise:  If G has a content then the cylinder $Z=\left[0,h\right]×G$ has a content as well. It calculates to

 ${V}_{n+1}\left(Z\right)={V}_{n}\left(G\right)\cdot h$ [8.5.10]

As an example the cylinder $Z=\left[0,2\right]×\left\{\left(x,y,z\right)\in {ℝ}^{3}|{x}^{2}+{y}^{2}+{z}^{2}\le 9\right\}$, whose base is a sphere of radius 3 has the content  ${V}_{4}\left(Z\right)=\frac{4}{3}\cdot {3}^{3}\pi \cdot 2=72\pi$.

Proof: ?

Obviously [8.5.10] confirms the well-known formula "base times hight" for the content of a cylinder. In order to get a similar result for a cone we need some technical preparations: For any two vectors $d=\left({d}_{0},\dots ,{d}_{n-1}\right)$ and $c=\left({c}_{0},\dots ,{c}_{n-1}\right)$, ${c}_{i}>0$, and an arbitrary subset $M\subset {ℝ}^{n}$ we set

$cM+d≔\left\{\left({c}_{0}\cdot x+{d}_{0},{c}_{1}\cdot {y}_{1}+{d}_{1},\dots ,{c}_{n-1}\cdot {y}_{n-1}+{d}_{n-1}\right)|\left(x,{y}_{1},\dots ,{y}_{n-1}\right)\in M\right\}$

If $d=\left(0,\dots ,0\right)$ or $c=\left(1,\dots ,1\right)$ we write $cM$ and $M+d$ respectively. $cM$ is generated from M by dilation with the dilation vector c whereas $M+d$ is generated by translation with the translation vector d.

The following proposition will not only provide the desired formula for cones but will yield further important results as well.

Proposition:  A subset $M\subset {ℝ}^{n}$ has a content if and only if $cM+d$ has a content. In this case we have:

 ${V}_{n}\left(cM+d\right)={c}_{0}\cdot \dots \cdot {c}_{n-1}\cdot {V}_{n}\left(M\right)$ [8.5.11]

Proof:  It is sufficient to prove only the "$⇒$" direction because it includes the reverse direction already due to the identity

$\left(\frac{1}{{c}_{0}},\dots ,\frac{1}{{c}_{n-1}}\right)\left(cM+d\right)+\left(\frac{-{d}_{0}}{{c}_{0}},\dots ,\frac{-{d}_{n-1}}{{c}_{n-1}}\right)=M$

We proceed now by induction.

• $n=1$:  If $M=\varnothing$ then $cM+d=\varnothing$ as well and both sets have the same content 0.
For $M=\left[{l}_{1},{r}_{1}\right]⊍\dots ⊍\left[{l}_{k},{r}_{k}\right]$ we have

$cM+d=\left[{c}_{0}\cdot {l}_{1}+{d}_{0},{c}_{0}\cdot {r}_{1}+{d}_{0}\right]⊍\dots ⊍\left[{c}_{0}\cdot {l}_{k}+{d}_{0},{c}_{0}\cdot {r}_{k}+{d}_{0}\right]$

which is again a disjoint union of closed intervals, having thus a content, namely:

${V}_{1}\left(cM+d\right)=\sum _{i=1}^{k}{c}_{0}\cdot {r}_{i}+{d}_{0}-\left({c}_{0}\cdot {l}_{i}+{d}_{0}\right)={c}_{0}\cdot \sum _{i=1}^{k}{r}_{i}-{l}_{i}={c}_{0}\cdot {V}_{1}\left(M\right)$

• $n\text{\hspace{0.28em}}⇒\text{\hspace{0.28em}}n+1$:  Now consider $M\subset \left[a,b\right]×{ℝ}^{n}$, i.e. $cM+d\subset \left[{c}_{0}\cdot a+{d}_{0},{c}_{0}\cdot b+{d}_{0}\right]×{ℝ}^{n}$.

If $a=b$ then ${c}_{0}\cdot a+{d}_{0}={c}_{0}\cdot b+d$ and the content of either set is 0. Assume now $a. At first we have

${\left(cM+d\right)}_{x}=\left({c}_{1},\dots ,{c}_{n}\right){M}_{\frac{x-{d}_{0}}{{c}_{0}}}+\left({d}_{1},\dots ,{d}_{n}\right)$[+]

for all $x\in \left[{c}_{0}\cdot a+{d}_{0},{c}_{0}\cdot b+{d}_{0}\right]$, because:

$\begin{array}{ll}\left({y}_{1},\dots ,{y}_{n}\right)\in {\left(cM+d\right)}_{x}\hfill & \text{ }⇔\text{ }\left(x,{y}_{1},\dots ,{y}_{n}\right)\in cM+d\hfill \\ \hfill & \text{ }⇔\text{ }\left(\frac{x-{d}_{0}}{{c}_{0}},\frac{{y}_{1}-{d}_{1}}{{c}_{1}},\dots ,\frac{{y}_{n}-{d}_{n}}{{c}_{n}}\right)\in M\hfill \\ \hfill & \text{ }⇔\text{ }\left(\frac{{y}_{1}-{d}_{1}}{{c}_{1}},\dots ,\frac{{y}_{n}-{d}_{n}}{{c}_{n}}\right)\in {M}_{\frac{x-{d}_{0}}{{c}_{0}}}\hfill \\ \hfill & \text{ }⇔\text{ }\left({y}_{1},\dots ,{y}_{n}\right)\in \left({c}_{1},\dots ,{c}_{n}\right){M}_{\frac{x-{d}_{0}}{{c}_{0}}}+\left({d}_{1},\dots ,{d}_{n}\right)\hfill \end{array}$

Now, if M has a content so does every section ${M}_{x}$ and ${V}_{n}\left({M}_{\mathrm{X}}\right)$ is integrable on $\left[a,b\right]$. Due to the induction hypothesis

$\left({c}_{1},\dots ,{c}_{n}\right){M}_{\frac{x-{d}_{0}}{{c}_{0}}}+\left({d}_{1},\dots ,{d}_{n}\right)\underset{{\left[+\right]}}{=}{\left(cM+d\right)}_{x}$

has the content ${V}_{n}\left({\left(cM+d\right)}_{x}\right)={c}_{1}\cdot \dots \cdot {c}_{n}\cdot {V}_{n}\left({M}_{\frac{x-{d}_{0}}{{c}_{0}}}\right)$ for each $x\in \left[{c}_{0}\cdot a+{d}_{0},{c}_{0}\cdot b+{d}_{0}\right]$. According to the substitution formula [8.3.5] the function ${V}_{n}\left(\left(cM+d{\right)}_{\mathrm{X}}\right)$ integrable and the content of $cM+d$ is calculated like this:

$\begin{array}{ll}{V}_{n+1}\left(cM+d\right)\hfill & =\underset{{c}_{0}\cdot a+{d}_{0}}{\overset{{c}_{0}\cdot b+{d}_{0}}{\int }}{V}_{n}\left({\left(cM+d\right)}_{\mathrm{X}}\right)\hfill \\ \hfill & ={c}_{0}\cdot {c}_{1}\cdot \dots \cdot {c}_{n}\underset{{c}_{0}\cdot a+{d}_{0}}{\overset{{c}_{0}\cdot b+{d}_{0}}{\int }}{V}_{n}\left({M}_{\frac{\mathrm{X}-{d}_{0}}{{c}_{0}}}\right)\cdot \frac{1}{{c}_{0}}\hfill \\ \hfill & ={c}_{0}\cdot {c}_{1}\cdot \dots \cdot {c}_{n}\underset{a}{\overset{b}{\int }}{V}_{n}\left({M}_{\mathrm{X}}\right)\hfill \\ \hfill & ={c}_{0}\cdot {c}_{1}\cdot \dots \cdot {c}_{n}\cdot {V}_{n+1}\left(M\right)\hfill \end{array}$

There are some interesting implications of [8.5.11]:

1. Content is translation-proof, as for $c=\left(1,\dots ,1\right)$ we have

 ${V}_{n}\left(M+d\right)={V}_{n}\left(M\right)$ [8.5.12]
2. Content is dilation-compatible, as for $d=\left(0,\dots ,0\right)$ we have

 ${V}_{n}\left(cM\right)={c}_{0}\cdot \dots {c}_{n-1}\cdot {V}_{n}\left(M\right)$ [8.5.13]
3. Content is shearing-proof: If $M\subset \left[a,b\right]×{ℝ}^{n}$ has a content then for any shear vector $s=\left({s}_{1},\dots ,{s}_{n}\right)\in {ℝ}^{n}$ the set

${M}^{s}≔\left\{\left(x,\frac{{s}_{1}}{b-a}\left(x-a\right)+{y}_{1},\dots ,\frac{{s}_{n}}{b-a}\left(x-a\right)+{y}_{n}\right)|x\in \left[a,b\right]\text{\hspace{0.28em}}\wedge \text{\hspace{0.28em}}y\in {M}_{x}\right\}$

has a content of the same size:

 ${V}_{n+1}\left({M}^{s}\right)={V}_{n+1}\left(M\right)$ [8.5.14]

Proof:  For $x\in \left[a,b\right]$ we see that ${M}_{x}^{s}={M}_{x}+\left(\frac{{s}_{1}}{b-a}\left(x-a\right),\dots ,\frac{{s}_{n}}{b-a}\left(x-a\right)\right)$. According to [8.5.12]  ${M}_{x}^{s}$ thus has the content ${V}_{n}\left({M}_{x}^{s}\right)={V}_{n}\left({M}_{x}\right)$, so that the Cavalieri Principle [8.5.4] yields the assertion.

Now that content is compatible with dilation we are able to compute the content of a cone.

Proposition:  The cone $C=\left\{\left(x,\frac{x}{h}{y}_{1},\dots \frac{x}{h}{y}_{n}\right)\in {ℝ}^{n+1}|x\in \left[0,h\right]\text{\hspace{0.28em}}\wedge \text{\hspace{0.28em}}y\in G\right\}$ has a content if its base $G\subset {ℝ}^{n}$ has a content. In this case it calculates to:

 ${V}_{n+1}\left(C\right)=\frac{1}{n+1}{V}_{n}\left(G\right)\cdot h$ [8.5.15]

Proof:  For any $x\in \left[0,h\right]$ we have ${C}_{x}=\left\{\left(\frac{x}{h}{y}_{1},\dots ,\frac{x}{h}{y}_{n}\right)|y\in G\right\}=\left(\frac{x}{h},\dots ,\frac{x}{h}\right)\cdot G$. Along with G the section ${C}_{x}$ has a content as well (see [8.5.13]), namely ${V}_{n}\left({C}_{x}\right)=\left(\frac{x}{h}{\right)}^{n}\cdot {V}_{n}\left(G\right)$. Being a multiple of ${\mathrm{X}}^{n}$ the function ${V}_{n}\left({C}_{\mathrm{X}}\right)$ is integrable, i.e. C has the content

${V}_{n+1}\left(C\right)=\underset{0}{\overset{h}{\int }}{V}_{n}\left({C}_{\mathrm{X}}\right)=\frac{1}{{h}^{n}}\cdot {V}_{n}\left(G\right)\underset{0}{\overset{h}{\int }}{\mathrm{X}}^{n}=\frac{1}{{h}^{n}}\cdot {V}_{n}\left(G\right)\cdot \frac{{h}^{n+1}}{n+1}=\frac{1}{n+1}{V}_{n}\left(G\right)\cdot h$

Consider:

• The content formula [8.5.15] of course comprises the previously calculated formulas for triangles, three-dimensional circular cones and pyramids.

• The height of a cone might be perpendicular to the base, but does not need to. Its position has no impact on the content due to [8.5.14]. All three triangles below for instance have the same content of 1.5.    