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} (M)

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 $M \subset [a, b] \times ℝ^{n}$ of $ℝ^{n + 1}$ and any $x \in [a, b]$ the set

M_{x} ≔ {(y_{1}, \dots, y_{n}) \in ℝ^{n} | (x, y_{1}, \dots, y_{n}) \in M}

[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 $Q ≔ {[0, a]}^{2} \subset [0, a] \times ℝ$ are (constant) intervals as we have for any $x \in [0, a]$ :

Q_{x} = {y \in ℝ | (x, y) \in [0, a] \times [0, a]} = [0, a]

[1]

The sections in an ellipse $E ≔ {(x, y) \in ℝ^{2} | b^{2} x^{2} + a^{2} y^{2} \leq a^{2} b^{2}} \subset [- a, a] \times ℝ$ are (variable) intervals as we calculate for each $x \in [- a, a]$ :

\begin{array}{l} E_{x} & = {y \in ℝ | b^{2} x^{2} + a^{2} y^{2} \leq a^{2} b^{2}} \\ = {y \in ℝ | y^{2} \leq \frac{b^{2}}{a^{2}} (a^{2} - x^{2})} = [- \frac{b}{a} \sqrt{a^{2} - x^{2}}, \frac{b}{a} \sqrt{a^{2} - x^{2}}] \end{array}

[2]

Note that when $x^{2} = a$ the interval $E_{x}$ is a singleton: $[0, 0] = {0}$ .

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

\begin{array}{l} L_{x} & = {y \in ℝ | r^{2} - x^{2} \leq y^{2} \leq R^{2} - x^{2}} \\ = {y \in [- \sqrt{R^{2} - x^{2}}, \sqrt{R^{2} - x^{2}}] | y^{2} \geq r^{2} - x^{2}} \\ = {\begin{array}{l} [- \sqrt{R^{2} - x^{2}}, \sqrt{R^{2} - x^{2}}] if | x | \geq r \\ {y \in [- \sqrt{R^{2} - x^{2}}, \sqrt{R^{2} - x^{2}}] | | y | \geq \sqrt{r^{2} - x^{2}} if | x | < r} \end{array} \\ = {\begin{array}{l} [- \sqrt{R^{2} - x^{2}}, \sqrt{R^{2} - x^{2}}] if | x | \geq r \\ [- \sqrt{R^{2} - x^{2}}, - \sqrt{r^{2} - x^{2}}] \cup [\sqrt{r^{2} - x^{2}}, \sqrt{R^{2} - x^{2}}] if | x | < r \end{array} \end{array}

[3]

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

L_{x} = {\begin{array}{l} {0} if | x | = R \\ {- \sqrt{R^{2} - x^{2}}, \sqrt{R^{2} - x^{2}}} if | x | < R \end{array}

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

W_{x} = {(y, z) \in ℝ^{2} | (x, y, z) \in {[0, a]}^{3}} = {[0, a]}^{2}

[4]

Sections in a sphere

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

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

\begin{array}{l} S_{x} & = {(y, z) \in ℝ^{2} | x^{2} + y^{2} + z^{2} \leq r^{2}} \\ = {(y, z) \in ℝ^{2} | y^{2} + z^{2} \leq r^{2} - x^{2}} \end{array}

[5]

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 $M = [l_{1}, r_{1}] ⊍ \dots ⊍ [l_{k}, r_{k}]$ is a disjoint union
i

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of finitely many closed intervals, $l_{i} \leq r_{i}$ . In that case we call the number

V_{1} (M) ≔ \sum_{i = 1}^{k} r_{i} - l_{i}

[8.5.2]

the one-dimensional content of M. In addition we set: $V_{1} (\emptyset) ≔ 0$ .

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

V_{n + 1} (M) ≔ \int_{a}^{b} V_{n} (M_{X})

[8.5.3]

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

Consider:

Sometimes, especially in three-dimensional situations, the term volume is used instead of content.
A simple inductive argument shows that contents are positive: $V_{n} (M) \geq 0$ .
Due to the different add ons the content of any finite set is zero:

$V_{n} (\emptyset) = V_{n} ({x_{1}, \dots, x_{k}}) = 0$
[8.5.2] includes the common special case $k = 1$ . M then is a simple interval, $M = [l, r]$ , $l \leq r$ , and the content $V_{1}$ shortens to $V_{1} (M) = r - l$ .

[8.5.3] infact introduces the Cavalieri Principle:

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

V_{n} (M_{x}) = V_{n} (N_{x}) for all x \in [a, b] \Rightarrow V_{n + 1} (M) = V_{n + 1} (N)

[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 $f \in C^{0} ([a, b])$ the set

$M ≔ {(x, y) \in ℝ^{2} | a \leq x \leq b \land 0 \leq y \leq f (x)} \subset [a, b] \times ℝ$

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

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

$V_{2} (M) = \int_{a}^{b} V_{1} (M_{X}) = \int_{a}^{b} f$

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 $Q = {[0, a]}^{2} \subset [0, a] \times ℝ$ with edge length a is

V_{2} (Q) = a^{2}

[6]

Proof: According to [1] all sections in Q have a constant content:

V_{1} (Q_{x}) = V_{1} ([0, a]) = a

Thus $V_{1} (Q_{X})$ is integrable on $[0, a]$ so that Q has a content, namely

V_{2} (Q) = \int_{0}^{a} V_{1} (Q_{X}) = \int_{0}^{a} a = a^{2}

The content of an ellipse $E = {(x, y) \in ℝ^{2} | b^{2} x^{2} + a^{2} y^{2} \leq a^{2} b^{2}} \subset [- a, a] \times ℝ$ with semi axes a and b is

V_{2} (E) = a b π

[7]

Proof: From [2] we know that the sections in E are the intervals

E_{x} = [- \frac{b}{a} \sqrt{a^{2} - x^{2}}, \frac{b}{a} \sqrt{a^{2} - x^{2}}]

As the continuous function $V_{1} (E_{X}) = 2 \frac{b}{a} \sqrt{a^{2} - X^{2}}$ is integrable on $[- a, a]$ , E has a content that we calculate by using an integral from 8.4:

V_{2} (E) = \int_{- a}^{a} V_{1} (E_{X}) = 2 \frac{b}{a} \int_{- a}^{a} \sqrt{a^{2} - X^{2}} = \frac{b}{a} a^{2} π = a b π

The content of a circle (disc) $D = {(x, y) \in ℝ^{2} | x^{2} + y^{2} \leq r^{2}} \subset [- r, r] \times ℝ$ with radius r is

V_{2} (D) = r^{2} π

[8]

Proof: D is an ellipse with identical semi axes $a = b = r$ . Thus: $V_{2} (D) = a b π = r^{2} π$ .

The content of a perforated disc $L = {(x, y) \in ℝ^{2} | r^{2} \leq x^{2} + y^{2} \leq R^{2}} \subset [- R, R] \times ℝ$ with radii $r < R$ is

V_{2} (L) = (R^{2} - r^{2}) π

[9]

Proof: Due to [3] the sections in a perforated disc are calculated like this:

L_{x} = {\begin{array}{l} [- \sqrt{R^{2} - x^{2}}, \sqrt{R^{2} - x^{2}}] if | x | \geq r \\ [- \sqrt{R^{2} - x^{2}}, - \sqrt{r^{2} - x^{2}}] ⊍ [\sqrt{r^{2} - x^{2}}, \sqrt{R^{2} - x^{2}}] if | x | < r \end{array}

Thus their content (even in the special case $r = R$ ) is given by

V_{1} (L_{x}) = {\begin{array}{l} 2 \sqrt{R^{2} - x^{2}} if | x | \geq r \\ 2 (\sqrt{R^{2} - x^{2}} - \sqrt{r^{2} - x^{2}}) if | x | \leq r \end{array}

$V_{1} (L_{X})$ is representable as the concatenation of the restrictions $2 \sqrt{R^{2} - X^{2}} | [- R, - r]$ , $2 \sqrt{R^{2} - X^{2}} | [r, R]$ and $2 (\sqrt{R^{2} - X^{2}} - \sqrt{r^{2} - X^{2}}) | [- r, r]$ and thus integrable according to [8.1.14], so that the content of L could be calculated (use an integral from 8.4) to

\begin{array}{l} V_{2} (L) & = 2 \int_{- R}^{- r} \sqrt{R^{2} - X^{2}} + 2 (\int_{- r}^{r} \sqrt{R^{2} - X^{2}} - \sqrt{r^{2} - X^{2}}) + 2 \int_{r}^{R} \sqrt{R^{2} - X^{2}} \\ = 2 \int_{- R}^{- r} \sqrt{R^{2} - X^{2}} + 2 \int_{- r}^{r} \sqrt{R^{2} - X^{2}} + 2 \int_{r}^{R} \sqrt{R^{2} - X^{2}} - 2 \int_{- r}^{r} \sqrt{r^{2} - X^{2}} \\ = 2 \int_{- R}^{R} \sqrt{R^{2} - X^{2}} - 2 \int_{- r}^{r} \sqrt{r^{2} - X^{2}} \\ = R^{2} π - r^{2} π = (R^{2} - r^{2}) π \end{array}

For any function $f : [a, b] \to ℝ$ the content of the line f is

V_{2} (f) = 0

[10]

Proof: As $f = {(x, y) \in ℝ^{2} | x \in [a, b] \land y = f (x)} \subset [a, b] \times ℝ$ all the sections $f_{x}$ are singletons of content 0. Therefor we have

V_{2} (f) = \int_{a}^{b} f_{X} = \int_{a}^{b} 0 = 0

The content of a rectangle $R ≔ [0, a] \times [0, b]$ with edge lengths a and b is

V_{2} (R) = a b

[11]

Proof: ?

The content of a triangle $D ≔ {(x, y) \in ℝ^{2} | x \in [0, h] \land \frac{a}{h} \leq y \leq \frac{b}{h}}$ with base $g ≔ b - a$ and height h is

V_{2} (D) = \frac{1}{2} g h

[12]

Proof: ?

In the next example we calculate some of the classical three-dimensional volumia using our previous results [6] to [9].

Example:

Cube

Ellipsoid

Sphere

Cone

Pyramid

Torus

Cuboid

Cylinder

The content of a cube $W = {[0, a]}^{3}$ of edge length a is

$V_{3} (W) = a^{3}$

Proof: Due to [6] the sections $W_{x} = {[0, a]}^{2}$ (see [4]) in $W \subset [0, a] \times ℝ^{2}$ have the constant content $V_{2} (W_{x}) = a^{2}$ . Thus W has a content, namely

$V_{3} (W) = a^{2} \int_{0}^{a} 1 = a^{3}$

The content of an ellipsoid $E ≔ {(x, y, z) \in ℝ^{3} | b^{2} c^{2} x^{2} + a^{2} c^{2} y^{2} + a^{2} b^{2} z^{2} \leq a^{2} b^{2} c^{2}}$ with semi axes a, b and c is

$V_{3} (E) = \frac{4}{3} a b c π$

Proof: At first we see that $E \subset [- a, a] \times ℝ^{2}$ . If $x^{2} = a^{2}$ the inequality

$a^{2} c^{2} y^{2} + a^{2} b^{2} z^{2} \leq a^{2} b^{2} c^{2} - b^{2} c^{2} x^{2}$

only holds for $y = z = 0$ . If $x^{2} < a^{2}$ it is equivalent to

$\frac{c^{2} (a^{2} - x^{2})}{a^{2}} y^{2} + \frac{b^{2} (a^{2} - x^{2})}{a^{2}} z^{2} \leq \frac{b^{2} c^{2} (a^{2} - x^{2})^{2}}{a^{4}}$

Thus $E_{x}$ is either a point or an ellipse (see [7]) with semi axes $\frac{c^{2} (a^{2} - x^{2})}{a^{2}}$ and $\frac{b^{2} (a^{2} - x^{2})}{a^{2}}$ . In either case the content of $E_{x}$ equals to
$V_{2} (E_{x}) = \frac{c \sqrt{a^{2} - x^{2}}}{a} \cdot \frac{b \sqrt{a^{2} - x^{2}}}{a} π = \frac{b c (a^{2} - x^{2})}{a^{2}} π$
and as $V_{2} (E_{X})$ is integrable E has a content:

$V_{3} (E) = \frac{b c}{a^{2}} π \int_{- a}^{a} a^{2} - X^{2} = \frac{b c}{a^{2}} π (a^{3} - \frac{1}{3} a^{3}) \cdot 2 = \frac{4}{3} a b c π$

The content of a sphere $S = {(x, y, z) \in ℝ^{3} | x^{2} + y^{2} + z^{2} \leq r^{2}}$ of radius r is

$V_{3} (S) = \frac{4}{3} π r^{3}$

Proof: S is an ellipsoid with three identical semi axes: $r = a = b = c$ . The content of a sphere is thus a special case of the content for an ellipsoid.

For a separate proof without using the ellipsoid formula we consider the sections according to [5]:

$S_{x} = {(y, z) \in ℝ^{2} | y^{2} + z^{2} \leq r^{2} - x^{2}}$ .

$S_{x}$ is a circle with radius $\sqrt{r^{2} - x^{2}}$ (or a point if $x^{2} = r^{2}$ resp.). Due to [8] its content calculates to $V_{2} (S_{x}) = (r^{2} - x^{2}) π$ . As the continuous function $V_{2} (S_{X})$ is integrable on $[- r, r]$ , S has a content, namely:

$V_{3} (S) = π \int_{- r}^{r} r^{2} - X^{2} = π (r^{2} \cdot r - \frac{1}{3} r^{3}) \cdot 2 = \frac{4}{3} π r^{3}$

The content of a cone $C ≔ {(x, y, z) \in ℝ^{3} | x \in [0, h] \land y^{2} + z^{2} \leq (\frac{r}{h} x)^{2}}$ of height h and radius r is

$V_{3} (C) = \frac{1}{3} π r^{2} h$

Proof: We notice that $C \subset [0, h] \times ℝ^{2}$ and for any $x \in [0, h]$ we get

$C_{x} = {(x, y) \in ℝ^{2} | y^{2} + z^{2} \leq (\frac{r}{h} x)^{2}}$ .

Thus $C_{x}$ is a circle again (a point if $x = 0$ ). Due to [8] we find the function $V_{2} (C_{X}) = (\frac{r}{h} X)^{2} π$ which is integrable on $[0, h]$ and from that we eventually get the content

$V_{3} (C) = (\frac{r}{h})^{2} π \int_{0}^{h} X^{2} = (\frac{r}{h})^{2} π \cdot \frac{1}{3} h^{3} = \frac{1}{3} π r^{2} h$

The content of a pyramid $P ≔ {(x, y, z) \in ℝ^{3} | x \in [0, h] \land | y | \leq \frac{b}{2 h} x \land | z | \leq \frac{a}{2 h} x}$ with height h and rectangular base $a \times b$ is

$V_{3} (P) = \frac{1}{3} a b h$

Proof: $P \subset [0, h] \times ℝ^{2}$ and

$\begin{array}{l} P_{x} & = {(x, y) \in ℝ^{2} | | y | \leq \frac{b}{2 h} x \land | z | \leq \frac{a}{2 h} x} \\ = [- \frac{b}{2 h} x, \frac{b}{2 h} x] \times [- \frac{a}{2 h} x, \frac{a}{2 h} x], \end{array}$

for each $x \in [0, h]$ . Thus $P_{x}$ is a point if $x = 0$ and a rectangle with edge length $\frac{b}{2 h} x$ and $\frac{a}{2 h} x$ otherwise. The integrable function $V_{2} (P_{X}) = \frac{a b}{h^{2}} X^{2}$ now yields the content:

$V_{3} (P) = \frac{a b}{h^{2}} \int_{0}^{h} X^{2} = \frac{1}{3} a b h$

We abbreviate $r_{i} (x) ≔ R - \sqrt{r^{2} - x^{2}}$ and $r_{a} (x) ≔ R + \sqrt{r^{2} - x^{2}}$ and consider the set

T ≔ {(x, y, z) \in ℝ^{3} | x \in [- r, r] \land r_{i}^{2} (x) \leq y^{2} + z^{2} \leq r_{a}^{2} (x)}

T is a torus with radii r and R.
i

The sketch to the left shows a torus with radii r and R vertically halfed with respect to the z-axis.
It explaines how the inner ( $r_{i}$ ) and the outer ( $r_{a}$ ) radius of a section $T_{x}$ is calculated.

Its content is

V_{3} (T) = 2 R r^{2} π^{2}

Proof: We see that $T \subset [- r, r] \times ℝ^{2}$ and that the section

T_{x} = {(y, z) \in ℝ^{2} | r_{i}^{2} (x) \leq y^{2} + z^{2} \leq r_{a}^{2} (x)}

is a perforated disc for each $x \in [- r, r]$ . According to [9] the content of $T_{x}$ is given by

\begin{array}{l} V_{2} (T_{x}) & = (r_{a}^{2} (x) - r_{i}^{2} (x)) π \\ = ((R + \sqrt{r^{2} - x^{2}})^{2} - (R - \sqrt{r^{2} - x^{2}})^{2}) π \\ = 4 R \sqrt{r^{2} - x^{2}} π \end{array}

$V_{2} (T_{X})$ is integrable, thus T has a content, namely (the integral involved has been evaluated in 8.4)

V_{3} (T) = 4 R π \int_{- r}^{r} \sqrt{r^{2} - X^{2}} = 4 R π \cdot \frac{1}{2} r^{2} π = 2 R r^{2} π^{2}

The content of a cuboid $Q ≔ [0, a] \times [0, b] \times [0, c]$ of edge length a, b and c is

$V_{3} (Q) = a b c$

Proof: ?

The content of a cylinder $Z ≔ [0, h] \times {(y, z) \in ℝ^{2} | y^{2} + z^{2} \leq r^{2}}$ of height h and radius r is

$V_{3} (Z) = h r^{2} π$

Proof: ?

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 $f : [a, b] \to ℝ$ the set

R ≔ {(x, y, z) \in ℝ^{3} | x \in [a, b] \land y^{2} + z^{2} \leq {(f (x))}^{2}}

[8.5.5]

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

V_{3} (R) = π \int_{a}^{b} f^{2}

Proof: We note that $R \subset [a, b] \times ℝ^{2}$ and that $R_{x}$ is a circle of radius $| f (x) |$ for each $x \in [a, b]$ . So we have: $V_{2} (R_{X}) = π f^{2}$ , which is infact the assertion.

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

\begin{array}{l} V_{3} (R) & = π \int_{- 1}^{1} (X^{2} + 1)^{2} \\ = π \int_{- 1}^{1} X^{4} + 2 X^{2} + 1 \\ = π (\frac{1}{5} X^{5} + \frac{2}{3} X^{3} + X) |_{- 1}^{1} = \frac{56}{15} π \end{array}

Exercise: The content of the ellipsoid

E ≔ {(x, y, z) \in ℝ^{3} | x \in [- a, a] \land y^{2} + z^{2} \leq \frac{b^{2}}{a^{2}} \cdot (a^{2} - x^{2})}

calculates to $V_{3} (E) = \frac{4}{3} a b^{2} π$ .

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:

The content of the n-dimensional cube $W_{n} = {[0, a]}^{n}$ is

V_{n} (W_{n}) = a^{n}

[8.5.6]

The content of the n-dimensional sphere $S_{n} = {x \in ℝ^{n} | x_{1}^{2} + \dots + x_{n}^{2} \leq r^{2}}$ is

V_{n} (S_{n}) = {\begin{array}{l} \frac{1}{k!} π^{k} r^{n} if n = 2 k \\ \frac{2^{n} k!}{n!} π^{k} r^{n} if n = 2 k + 1 \end{array}

[8.5.7]

1. ► Proof by induction:

$n = 1$ : $W_{1} = [0, a]$ is a closed interval such that $V_{1} (W_{1}) = a = a^{1}$ .
$n \Rightarrow n + 1$ : Now let $W_{n + 1} = {[0, a]}^{n + 1} = [0, a] \times {[0, a]}^{n}$ be an $(n + 1)$ -dimensional cube. For any $x \in [0, a]$ we see that the section $W_{n + 1, x} = {[0, a]}^{n}$ is an n-dimensional cube, thus having a content due to the induction hypothesis. The constant function $V_{n} (W_{n + 1, X}) = a^{n}$ is integrable, so that $W_{n + 1}$ has a content as well, namely

$V_{n + 1} (W_{n + 1}) = a^{n} \int_{0}^{a} 1 = a^{n + 1}$

2. ► Proof by induction:

$n = 1$ : $S_{1} = {x \in ℝ | x^{2} \leq r^{2}} = [- r, r]$ is a closed interval such that

$V_{1} (S_{1}) = 2 r = \frac{2^{1} \cdot 0!}{1!} π^{0} r^{1}$

Note that $n = 2 \cdot 0 + 1$ in this case.
$n \Rightarrow n + 1$ : Next let $S_{n + 1} = {(x, y) \in ℝ^{n + 1} | x^{2} + y_{1}^{2} + \dots + y_{n}^{2} \leq r^{2}} \subset [- r, r] \times ℝ^{n}$ be an ( $n + 1$ )-dimensional sphere. The section $S_{n + 1, x} = {y \in ℝ^{n} | y_{1}^{2} + \dots + y_{n}^{2} \leq r^{2} - x^{2}}$ is an n-dimensional sphere for all $x \in [- r, r]$ , thus has a content according to the induction hypothesis. The function

$V_{n} (S_{n + 1, X}) = {\sqrt{r^{2} - X^{2}}}^{n} \cdot {\begin{array}{l} \frac{1}{k!} π^{k} if n = 2 k \\ \frac{2^{n} k!}{n!} π^{k} if n = 2 k + 1 \end{array}$

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

The integral (substituting $g = r \cdot \sin, g^{'} = r \cdot \cos$ , see [8.3.5])

$\begin{array}{l} \int_{- r}^{r} {\sqrt{r^{2} - X^{2}}}^{n} = \int_{r \cdot \sin (- \frac{π}{2})}^{r \cdot \sin (\frac{π}{2})} {\sqrt{r^{2} - X^{2}}}^{n} & = \int_{- \frac{π}{2}}^{\frac{π}{2}} {\sqrt{r^{2} - r^{2} \cdot \sin^{2}}}^{n} r \cdot \cos \\ = r^{n + 1} \int_{- \frac{π}{2}}^{\frac{π}{2}} {\sqrt{\underset{= \cos^{2}}{\underset{︸}{1 - \sin^{2}}}}}^{n} \cos \\ = r^{n + 1} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{n + 1} \\ \underset{[8.3.4]}{=} r^{n + 1} \cdot {\begin{array}{l} \frac{(n + 1)!}{(2^{k} k!)^{2}} \cdot π if n + 1 = 2 k \\ \frac{(2^{k} k!)^{2}}{(n + 1)!} \cdot 2 if n + 1 = 2 k + 1 \end{array} \end{array}$

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

$\begin{array}{l} V_{n + 1} (S_{n + 1}) & = \int_{- r}^{r} {\sqrt{r^{2} - X^{2}}}^{n} \cdot {\begin{array}{l} \frac{1}{k!} π^{k} if n = 2 k \\ \frac{2^{n} k!}{n!} π^{k} if n = 2 k + 1 \end{array} \\ = r^{n + 1} \cdot {\begin{array}{l} \frac{(2^{k} k!)^{2}}{(n + 1)!} \cdot 2 \cdot \frac{1}{k!} π^{k} if n = 2 k \Leftrightarrow n + 1 = 2 k + 1 \\ \frac{(n + 1)!}{(2^{k + 1} (k + 1)!)^{2}} \cdot π \cdot \frac{2^{n} k!}{n!} π^{k} if n = 2 k + 1 \Leftrightarrow n + 1 = 2 (k + 1) \end{array} \\ = {\begin{array}{l} \frac{2^{2 k + 1} {(k!)}^{2}}{(n + 1)! k!} π^{k} r^{n + 1} if n + 1 = 2 k + 1 \\ \frac{(n + 1) \cdot 2^{n}}{2^{2 k + 1} \cdot 2 (k + 1) (k + 1)!} π^{k + 1} r^{n + 1} if n + 1 = 2 (k + 1) \end{array} \\ = {\begin{array}{l} \frac{2^{n + 1} k!}{(n + 1)!} π^{k} r^{n + 1} if n + 1 = 2 k + 1 \\ \frac{1}{(k + 1)!} π^{k + 1} r^{n + 1} if n + 1 = 2 (k + 1) \end{array} \end{array}$

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

$Z ≔ [0, h] \times G$

[8.5.8]

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

$C ≔ {(x, \frac{x}{h} y_{1}, \dots \frac{x}{h} y_{n}) \in ℝ^{n + 1} | x \in [0, h] \land y \in G}$

[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 $Z = [0, h] \times G$ has a content as well. It calculates to

V_{n + 1} (Z) = V_{n} (G) \cdot h

[8.5.10]

As an example the cylinder $Z = [0, 2] \times {(x, y, z) \in ℝ^{3} | x^{2} + y^{2} + z^{2} \leq 9}$ , whose base is a sphere of radius 3 has the content $V_{4} (Z) = \frac{4}{3} \cdot 3^{3} π \cdot 2 = 72 π$ .

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 $d = (d_{0}, \dots, d_{n - 1})$ and $c = (c_{0}, \dots, c_{n - 1})$ , $c_{i} > 0$ , and an arbitrary subset $M \subset ℝ^{n}$ we set

c M + d ≔ {(c_{0} \cdot x + d_{0}, c_{1} \cdot y_{1} + d_{1}, \dots, c_{n - 1} \cdot y_{n - 1} + d_{n - 1}) | (x, y_{1}, \dots, y_{n - 1}) \in M}

If $d = (0, \dots,0)$ or $c = (1, \dots,1)$ we write $c M$ and $M + d$ respectively. $c M$ 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 $c M + d$ has a content. In this case we have:

V_{n} (c M + d) = c_{0} \cdot \dots \cdot c_{n - 1} \cdot V_{n} (M)

[8.5.11]

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

(\frac{1}{c_{0}}, \dots, \frac{1}{c_{n - 1}}) (c M + d) + (\frac{- d_{0}}{c_{0}}, \dots, \frac{- d_{n - 1}}{c_{n - 1}}) = M

We proceed now by induction.

$n = 1$ : If $M = \emptyset$ then $c M + d = \emptyset$ as well and both sets have the same content 0.
For $M = [l_{1}, r_{1}] ⊍ \dots ⊍ [l_{k}, r_{k}]$ we have

$c M + d = [c_{0} \cdot l_{1} + d_{0}, c_{0} \cdot r_{1} + d_{0}] ⊍ \dots ⊍ [c_{0} \cdot l_{k} + d_{0}, c_{0} \cdot r_{k} + d_{0}]$

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

$V_{1} (c M + d) = \sum_{i = 1}^{k} c_{0} \cdot r_{i} + d_{0} - (c_{0} \cdot l_{i} + d_{0}) = c_{0} \cdot \sum_{i = 1}^{k} r_{i} - l_{i} = c_{0} \cdot V_{1} (M)$
$n \Rightarrow n + 1$ : Now consider $M \subset [a, b] \times ℝ^{n}$ , i.e. $c M + d \subset [c_{0} \cdot a + d_{0}, c_{0} \cdot b + d_{0}] \times ℝ^{n}$ .
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 $a < b$ . At first we have

${(c M + d)}_{x} = (c_{1}, \dots, c_{n}) M_{\frac{x - d_{0}}{c_{0}}} + (d_{1}, \dots, d_{n})$ [+]

for all $x \in [c_{0} \cdot a + d_{0}, c_{0} \cdot b + d_{0}]$ , because:

$\begin{array}{l} (y_{1}, \dots, y_{n}) \in {(c M + d)}_{x} & \Leftrightarrow (x, y_{1}, \dots, y_{n}) \in c M + d \\ \Leftrightarrow (\frac{x - d_{0}}{c_{0}}, \frac{y_{1} - d_{1}}{c_{1}}, \dots, \frac{y_{n} - d_{n}}{c_{n}}) \in M \\ \Leftrightarrow (\frac{y_{1} - d_{1}}{c_{1}}, \dots, \frac{y_{n} - d_{n}}{c_{n}}) \in M_{\frac{x - d_{0}}{c_{0}}} \\ \Leftrightarrow (y_{1}, \dots, y_{n}) \in (c_{1}, \dots, c_{n}) M_{\frac{x - d_{0}}{c_{0}}} + (d_{1}, \dots, d_{n}) \end{array}$

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

$(c_{1}, \dots, c_{n}) M_{\frac{x - d_{0}}{c_{0}}} + (d_{1}, \dots, d_{n}) \underset{[+]}{=} {(c M + d)}_{x}$

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

$\begin{array}{l} V_{n + 1} (c M + d) & = \int_{c_{0} \cdot a + d_{0}}^{c_{0} \cdot b + d_{0}} V_{n} ({(c M + d)}_{X}) \\ = c_{0} \cdot c_{1} \cdot \dots \cdot c_{n} \int_{c_{0} \cdot a + d_{0}}^{c_{0} \cdot b + d_{0}} V_{n} (M_{\frac{X - d_{0}}{c_{0}}}) \cdot \frac{1}{c_{0}} \\ = c_{0} \cdot c_{1} \cdot \dots \cdot c_{n} \int_{a}^{b} V_{n} (M_{X}) \\ = c_{0} \cdot c_{1} \cdot \dots \cdot c_{n} \cdot V_{n + 1} (M) \end{array}$

There are some interesting implications of [8.5.11]:

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

V_{n} (M + d) = V_{n} (M)

[8.5.12]

Content is dilation-compatible, as for $d = (0, \dots,0)$ we have

V_{n} (c M) = c_{0} \cdot \dots c_{n - 1} \cdot V_{n} (M)

[8.5.13]

Content is shearing-proof: If $M \subset [a, b] \times ℝ^{n}$ has a content then for any shear vector $s = (s_{1}, \dots, s_{n}) \in ℝ^{n}$ the set

M^{s} ≔ {(x, \frac{s_{1}}{b - a} (x - a) + y_{1}, \dots, \frac{s_{n}}{b - a} (x - a) + y_{n}) | x \in [a, b] \land y \in M_{x}}

has a content of the same size:

V_{n + 1} (M^{s}) = V_{n + 1} (M)

[8.5.14]

Proof: For $x \in [a, b]$ we see that $M_{x}^{s} = M_{x} + (\frac{s_{1}}{b - a} (x - a), \dots, \frac{s_{n}}{b - a} (x - a))$ . According to [8.5.12] $M_{x}^{s}$ thus has the content $V_{n} (M_{x}^{s}) = V_{n} (M_{x})$ , 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 $C = {(x, \frac{x}{h} y_{1}, \dots \frac{x}{h} y_{n}) \in ℝ^{n + 1} | x \in [0, h] \land y \in G}$ has a content if its base $G \subset ℝ^{n}$ has a content. In this case it calculates to:

V_{n + 1} (C) = \frac{1}{n + 1} V_{n} (G) \cdot h

[8.5.15]

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

V_{n + 1} (C) = \int_{0}^{h} V_{n} (C_{X}) = \frac{1}{h^{n}} \cdot V_{n} (G) \int_{0}^{h} X^{n} = \frac{1}{h^{n}} \cdot V_{n} (G) \cdot \frac{h^{n + 1}}{n + 1} = \frac{1}{n + 1} V_{n} (G) \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.

8.4. 8.6.