1. Mensuration, or the art of measuring, involves the construction of measures, the methods of using them, and the investigation of rules by which magnitudes, which it may be difficult or impossible to measure directly, are calculated from the ascertained value of some associated magnitude. It is usual, however, to employ the term mensuration in the last of these senses; and we may therefore define it to be that department of mathematical science by which the various dimensions of bodies are calculated from the simplest possible measurements.

The determination of the lengths and directions of straight lines, including what are familiarly known as problems in heights and distances, generally depends on the solution of triangles, and will be discussed in the articles Trigonometry and Surveying. The remaining portions of the subject, which will form the subject of the present article, are the determinations of the lengths of curves, the areas of plane or other figures, and the volumes and surfaces of solids. Even thus restricted, the science of mensuration is obviously of very great extent, and our space will only permit us to discuss some of its more interesting and important problems.

The reader will find tables of the numbers most frequently required in mensuration at the end of this article, and a complete account of measures in the article Weights and Measures.

2. On the Numerical Expression of the Length of a Line, the Area of a Surface, and the Volume of a Solid.—If any line be chosen as the unit of length, the square and cube described upon it will be the units of surface and of volume, or the "square" and "cubic" units. Thus, if a foot be the linear unit, a square whose side is a foot, and a cube whose edge is a foot, are the square and cubic units. The length of a line, the area of a surface, and the volume of a solid, are then expressed by the numbers of units of length, surface, and volume which they respectively contain.

3. From this it follows (1st) that if \( l \) be the linear unit, the length of a line which contains \( a \) units is \( al \); or simply \( a \), for \( l \) is reckoned as unity.

4. (2d) If the length and breadth of a rectangle \( AD \) be divided into linear units, and lines be drawn through the points of section, parallel to its sides, it is evident that the rectangle will be divided into as many square units \( AE \), as there are linear units in its length \( AB \), repeated as often as there are linear units in its breadth \( AC \); or the number of square units contained by the figure will be obtained by multiplying its length by its breadth. Hence if the length \( AB \) contain \( a \), and the breadth \( AC \) contain \( b \) units, the surface \( AD \) will contain \( ab \) square units.

5. (3d) If the length, breadth, and height of a parallelopiped \( BCD \) be divided into linear units by planes drawn parallel to its faces, it will obviously be divided into cubic units such as \( EF \); and if \( AB, AC, AD \) contain \( a, b, c \) units respectively, it appears from the diagram that the portion of the solid between the face \( BC \), and the plane passing through \( EG, EH \), contains \( ob \) cubic units. Similarly, each of the sections of the solid between two contiguous planes passing through the divisions of \( AD \), contains \( ab \) cubic units; and since \( AD \) is divided into \( c \) parts, the whole solid contains \( abc \) units of volume.

The demonstrations of the preceding propositions might be extended to include the cases where the linear dimensions are fractional, or incommensurable with the linear unit.

SECTION I.—PLANE FIGURES CONTAINED BY STRAIGHT LINES.

6. To find the Area of a Rectangle.—It has been proved in article 4, that if \( a \) and \( b \) be the sides of a rectangle,

\[ \text{the area} = ab; \]

or the area of a rectangle is equal to its length multiplied by its breadth.

Whence, also, if \( a \) be the side of a square,

\[ \text{the area of the square} = a^2. \]

Example 1.—Find the area of a square BD, whose side is 15 chains 40 links.

Here the side of the square \( = 1540 \) links; therefore

\[ \text{the area} = (1540)^2 = 2371600 \text{ square links} \]

\[ = 23 \text{ a. } 2 \text{ r. } 34\frac{1}{2} \text{ p.} \]

Example 2.—Find the area of a rectangle \( EG \), whose sides are 12 feet 6 inches and 9 inches.

The area \( = 12\frac{1}{2} \times 9 = 93\frac{1}{2} \text{ square feet.} \)

7. To find the Area of a Parallelogram.—(i.) The area in terms of the base and perpendicular breadth.

The parallelogram \( ABCD \) is equal to the rectangle \( BEFC \) (Geom., sec. iv., theor. 1; or, Euc. i., 35).

Wherefore the area \( ABCD = BC \cdot BE = AD \cdot BE \); and putting \( AD = b \), and \( BE = h \),

\[ \text{the area} = bh; \]

or the area of a parallelogram is equal to its length, multiplied by its perpendicular breadth.

Example.—Find the area of a parallelogram whose length is 37 and its breadth 5\(\frac{1}{2}\) feet.

The area \( = 37 \times 5\frac{1}{2} = 194\frac{1}{2} \text{ square feet.} \)

8. (ii.) The area in terms of the sides and the included angle.—Let \( AB = a, AD = b, \) angle \( BAD = A. \)

Then \( BE = a \sin A \); and since \( ABCD = BE \cdot AD, \)

\[ \text{the area} = ab \sin A. \]

Whence the area is obtained by multiplying together the two sides of the parallelogram and the sine of the contained angle.

Example.—The sides of a parallelogram are 36 and 25\(\frac{1}{2}\), and the contained angle 58°.

The area \( = 36 \times 25\frac{1}{2} \times \sin 58^\circ = 36 \times 25\frac{1}{2} \times 0.8480181 \)

\[ = 775\frac{1}{2}; \]

or log area = \log a + \log b + L \sin A - 10^8

\begin{align*} a &= 36 & \log a &= 1.5568025 \\ b &= 25.5 & \log b &= 1.4065402 \\ A &= 58^\circ & L \sin A &= 9.9284205 \end{align*}

Area = 778.5 \quad \text{log area} = 2.8912632

9. (iii.) The area in terms of the diagonals and their contained angle.

If \( m, n \) are the diagonals, and \( I \) the contained angle, it follows, from article 17, that the

area = \(\frac{1}{2}mn \sin I\).

Example.—Find the area of a parallelogram whose diagonals are 30 and 25, which make with each other an angle of 60°.

\[ \text{Area} = \frac{1}{2} \times 30 \times 25 \times \sin 60^\circ = 15 \times 25 \times \sqrt{3} = 324.76; \]

or \(\log \text{area} = \log m + \log n + L \sin I - \log_2 - 10\).

\begin{align*} m &= 30 & \log m &= 1.4771213 \\ n &= 25 & \log n &= 1.3979400 \\ I &= 60^\circ & L \sin I &= 9.9695306 \\ & & \log_2 &= 9.6989700 \end{align*}

Area = 324.76 \quad \text{log area} = 2.5115619

If the parallelogram is equilateral, or a rhombus, the diagonals intersect at right angles, and \(\sin I = \sin 90^\circ = 1\).

Whence the area = \(\frac{1}{2}mn\).

Example.—Find the area of a rhombus whose diagonals are 30 and 20 chains.

The area = \(\frac{1}{2} \times 30 \times 20 = 300\) chains = 30 acres.

10. To find the Area of a Triangle—(i.) When the base AC, and height BD are given.—Let AC = \(a\), BD = \(h\). Then, since a triangle is half a rectangle of the same base and altitude (GEOM. p. 520), the area of the triangle

\[ \text{area} = \frac{1}{2}ah; \]

or the area is equal to one-half the product of one of the sides multiplied by the perpendicular let fall upon it from the opposite angle of the triangle.

Example.—A side of a triangle is 40, and the perpendicular on it from the opposite angle is 14.52 chains.

The area = \(\frac{1}{2} \times 40 \times 14.52 = 290.4\) sq. chains.

\(= 29\) a. 0 r. 64 p.

11. (ii.) When two sides and the included angle are given.

Let AC = \(a\), AB = \(c\).

Since ABC = \(\frac{1}{2}\)ACBD, and BD = \(c \sin A\),

the area of the triangle = \(\frac{1}{2}bc \sin A\);

or the area is equal to one-half the continued product of the two sides, and the sine of the contained angle.

Example.—The sides of a triangle are 30 and 40, and the contained angle is 28° 57'.

The area = \(\frac{1}{2} \times 30 \times 40 \times \sin 28^\circ 57' = 600 \times 0.4840462 = 290.43\);

or, by logarithms—

Log area = \(\log b + \log c + L \sin A + \log_2 - 10\)

\begin{align*} b &= 30 & \log b &= 1.4760913 \\ c &= 40 & \log c &= 1.6020600 \\ A &= 28^\circ 57' & L \sin A &= 9.6848868 \end{align*}

Area = 290.43 \quad \text{log area} = 2.4630381

12. When the three sides of the triangle are given.—Let BC = \(a\), AC = \(b\), AB = \(c\); and put BD = \(x\), and AD = \(y\).

Then CD = \(b - y\);

\[ \therefore c^2 = x^2 + y^2, \quad \text{and} \quad a^2 = x^2 + (b - y)^2. \]

Whence \(y = \frac{b^2 + c^2 - a^2}{2b}\).

But \(x^2 = c^2 - y^2 = c^2 - \left(\frac{b^2 + c^2 - a^2}{2b}\right)^2\)

\[ = \frac{(a + b + c)(b + c - a)(a + c - b)(a + b - c)}{4b^2}. \]

Whence, since the square of the area of the triangle is \((\frac{1}{2}ACDB)^2 = \frac{1}{4}b^2x^2\), we have

(area)\(^2 = \frac{1}{4}(a + b + c)(b + c - a)(a + c - b)(a + b - c)\).

Now if we put \(s = \frac{1}{2}(a + b + c)\), then \(\frac{1}{2}(b + c - a) = s - a\), etc. Therefore substituting

The area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\).

Otherwise, since \(\sin A = \frac{2}{bc} \sqrt{(s-a)(s-b)(s-c)}\),

the triangle = \(\frac{bc}{2} \sin A = \sqrt{(s-a)(s-b)(s-c)}\).

Whence, to obtain the area of a triangle, from half the sum of the three sides, subtract each side separately, multiply the half sum by the three remainders successively, and extract the square root.

Example 1.—The sides of a triangle are 3, 4, and 5 feet.

Find its area.

\begin{align*} a &= 5 & b &= 4 & c &= 3 \\ s &= \frac{1}{2}(5 + 4 + 3) = 6; & s - a &= 1; & s - b &= 2; & s - c &= 3. \end{align*}

The area = \(\sqrt{6 \times 1 \times 2 \times 3} = \sqrt{36} = 6\) square feet.

Example 2.—The sides of a triangle are 221, 255, and 238.

\begin{align*} a &= 255 & b &= 238 & c &= 221 \\ s &= \frac{1}{2}(255 + 238 + 221) = 371.5 \\ s - a &= 122 & s - b &= 113 & s - c &= 154 \end{align*}

Area = 24276 \quad \text{log area} = 4.3851772

13. When two angles and the adjacent side are given.

Let the angles \(A, B\) and the side \(c\) be given.

The area = \(\frac{1}{2}bc \sin A\).

But \(b = \frac{c \sin B}{\sin C} = \frac{c \sin B}{\sin(A+B)} \quad \because \sin C = \sin(A+B)\).

Therefore substituting,

\[ \text{area} = \frac{c^2 \sin A \sin B}{2 \sin(A+B)} \]

\[ = \frac{1}{2}c^2 \sin A \sin B \cos(A+B); \]

and \(\log \text{area} = 2 \log c + L \sin A + L \sin B + L \cos(A+B)\)

\[ + \log_2 - 2 - 10. \]

Example.—The side of a triangle is 2405 feet, and the angles at its ends 77° 54', and 87° 40'. Find its area in square miles.

\begin{align*} c &= 2405 & \log c &= 3.3822302 \\ A &= 77^\circ 54' & L \sin A &= 9.9502426 \\ B &= 87^\circ 40' & L \sin B &= 9.9996398 \\ A + B &= 165^\circ 34' & L \cos(A+B) &= 10.6033590 \\ & & \log_2 &= 9.6989700 \end{align*}

\[ \log \text{area in ft.} = 7.0544416 \]

\[ 2 \log 5280 = 7.4452678 \]

Area = 0.40661 sq. miles. \quad \text{log area in miles} = 1.6091788

14. To find the Area of a Trapezoid.—Let \(AB = a\), \(CD = b\), \(CE = h\). Then \(ABCD = ABC + ACD = \frac{1}{2}ah + \frac{1}{2}bh\)

(art. 10).

Throughout this article we shall denote the logarithm of a number to the base 10, by \(\log_{10}\) or simply \(\log\); and the tabular logarithm of the sine, cosine, etc., of an angle \(A\) by \(L \sin A, L \cos A\), etc. Since the logarithmic sines, cosines, etc., in the tables, are all increased by 10, we have \(L \sin A = \log_{10} \sin A + 10\), etc. We shall also denote the arithmetical complement of \(\log n\), or \(10 - \log n\), by \(\log_2 n\). Hence the trapezoid \(ABCD = \frac{1}{2}(a+b)h\); or the area is equal to half the sum of the parallel sides multiplied by the perpendicular breadth.

**Example.** — Find the area of a trapezoid whose parallel sides are 12'25 and 7'5 chains, and its breadth 15'4.

The area \(= \frac{1}{2} \times (12'25 + 7'5) \times 15'4 = 7'7 \times 19'75\)

\(= 152'075\) square chains.

\(= 15\) a. o r. 32'2 p.

15. To find the Area of a Quadrilateral Figure—(i.) When a diagonal AC and perpendiculars DF, BE are given.—Let AC = a, BE = m, and DF = n. Then \(ABCD = ABC + ADC = \frac{1}{2}am + \frac{1}{2}an\) (art. 10); or,

the quadrilateral \(= \frac{1}{2}a(m+n)\).

Hence the area is obtained by multiplying the diagonal by half the sum of the perpendiculars.

**Example.** — In a quadrilateral the diagonal is 42, and the perpendiculars 16 and 18.

The area \(= \frac{1}{2} \times 42 \times (16+18) = 21 \times 34 = 714\).

16. (ii.) When a side AB and the perpendiculars DF, CE are given.

The area \(= ADF + CDEF + BCE\).

\(= \frac{1}{2}AF \cdot DF + \frac{1}{2}(DF + CE)EF + \frac{1}{2}BE \cdot CE\) (arts. 10, 14)

\(= \frac{1}{2}(AE \cdot DF + BF \cdot CE)\).

**Example.** — Let \(AB = 12\), \(AF = 5\), \(BE = 3\), \(DF = 8\), and \(CE = 6\). Then \(AE = 12 - 3 = 15\) or 9.

BF \(= 12 - 5 = 7\)

The area \(= \frac{1}{2}(9 \times 8 + 7 \times 6) = 4 \times 9 + 3 \times 7 = 57\); or,

\(= \frac{1}{2}(15 \times 8 + 7 \times 6) = 4 \times 15 + 3 \times 7 = 81\).

17. (iii.) When the diagonals and the included angle are given.—Let AC = m, BD = n, AEB = I;

\(\therefore\) BEC \(= 180^\circ - AEB\);

\(\therefore\) Sin BEC \(= \sin I\).

ABC \(= ABE + BCE\)

\(= \frac{1}{2}AEB \cdot BE \cdot \sin AEB\)

\(+ \frac{1}{2}CE \cdot BE \cdot \sin BEC\) (art.11)

\(= \frac{1}{2}AC \cdot BE \cdot \sin AEB\).

Similarly \(ADC = \frac{1}{2}AC \cdot DE \cdot \sin AEB\).

Whence \(ABCD = \frac{1}{2}AC \cdot BD \cdot \sin AEB\); or,

the area \(= \frac{1}{2}mn \cdot \sin I\).

Therefore the area of a quadrilateral is equal to half the product of the diagonals, and the sine of their contained angle.

This rule obviously applies to parallelograms.

**Example.** — The diagonals of a quadrilateral are 30 and 40, and the included angle 48°.

The area \(= \frac{1}{2} \times 30 \times 40 \times \sin 48^\circ\)

\(= 600 \times 7'341448\)

\(= 435'887\).

18. (iv.) When the sides and the angle between the Mensuration diagonals are given.—In the figure of article 17, let \(AB = a\), BC = b, CD = c, AD = d, BEC = I, AE = x, BE = y, EC = z, ED = w.

Then \(x^2 + y^2 + 2xy \cos I = a^2\)

\(y^2 + z^2 - 2yz \cos I = b^2\)

\(z^2 + w^2 + 2zw \cos I = c^2\)

\(w^2 + x^2 - 2wx \cos I = d^2\)

Whence \(2(x+z)(y+w) \cos I = a^2 - b^2 + c^2 - d^2\).

But \(2(x+z)(y+w) \cos I = \frac{1}{2}AC \cdot BD \cdot \sin I \times 4 \cot I\),

\(= \text{area } ABCD \times 4 \cot I\) (art. 17).

Therefore the area \(= \frac{1}{2}(a^2 - b^2 + c^2 - d^2) \tan I\).

**Example.** — The sides of a quadrilateral in order are 10, 9, 8, and 7, and the angle between the diagonals is 80°. Find the area.

\(a^2 - b^2 + c^2 - d^2 = 100 - 81 + 64 - 49 = 34\).

\(\therefore\) log area \(= \log 34 + L \tan 80^\circ + \colog 4 - 10\)

\(\log 34 = 1'5314789\)

L tan 80° = 10'7536812

\(\colog 4 = 9'3979400\)

Area \(= 48'206\), log area \(= 1'6831001\).

19. (v.) When the quadrilateral is inscribed in a circle.

—In the figure of article 17, let \(AB = a\), BC = b, CD = c, AD = d. Then, since the quadrilateral is inscribed in a circle, \(D = 180^\circ - B\);

and \(a^2 + b^2 - 2ab \cos B = AC^2 = c^2 + d^2 + 2cd \cos B\).

From which \(\cos B = \frac{a^2 + b^2 - c^2 - d^2}{2(ab + cd)}\).

But \((\text{area})^2 = \frac{1}{4}(ab + cd) \sin B\)

\(= \frac{1}{2}(b+c+d-a) \frac{1}{2}(a+c+d-b) \frac{1}{2}(a+d+c-b) \frac{1}{2}(a+b+c-d)\).

Whence putting \(S = \frac{1}{2}(a+b+c+d)\),

the area \(= \sqrt{(s-a)(s-b)(s-c)(s-d)}\).

**Example.** — In a quadrilateral, two sides are 120 and 104, and the contained angle 59° 29' 23"; and the remaining sides are 78 and 50, and the contained angle 120° 30' 37".

Since the sum of the opposite angles \(= 180^\circ\), the figure may be inscribed in a circle, and the formula applies.

| a | 120 | |---|-----| | b | 104 | | c | 78 | | d | 50 |

\(\frac{352}{s} = 176\)

\(s-a = 56\)

\(\log = 1'7481880\)

\(s-b = 72\)

\(\log = 1'8573325\)

\(s-c = 98\)

\(\log = 1'9912261\)

\(s-d = 126\)

\(\log = 2'003705\)

\(2'76971171\)

Area \(= 7056\)

log area \(= 3'8485586\)

20. (vi.) When the four sides of the quadrilateral and a pair of opposite angles are given.—Since the figure (art. 17) may be divided into two triangles whose areas can be found by article 11; if \(a, b, c,\) and \(d\) be the sides, we have evidently

the area \(= \frac{1}{2}(ab \sin B + cd \sin D)\).

**Example.** — To verify the calculation of the area in the example of last article.

| a | 129 | |---|-----| | b | 20791812 | | c | 78 | | d | 18929946 | | e | 19989700 | | f | 9352745 | | G | 120°29'37" | | L | 9-352745 | | 2ABC = 10752 | | 2ADC = 3360 | | 3'5283391 | | \(\therefore\) 2ABCD = 14112, and ABCD = 7056. |

21. To find the Area of an Irregular Polygon.—The figure may be divided into triangles or trapezia; To find the area of the hexagon by logarithms.

The logarithm for a hexagon = .4146519

Area of hexagon = 1039230

24. To find the Area of a Regular Polygon inscribed in a Circle.—Let AB (fig. art. 22) be a side of the polygon; the radius AC = r, and the number of sides = n.

Then ACD = \(\frac{180^\circ}{n}\);

∴ AD = \(r \sin \frac{180^\circ}{n}\), and CD = \(r \cos \frac{180^\circ}{n}\);

and since the polygon = nABC = nAD · DC,

its area = \(nr^2 \sin \frac{180^\circ}{n} \cos \frac{180^\circ}{n} = \frac{1}{2}nr^2 \sin \frac{360^\circ}{n}\).

Example.—The radius of a circle being 10, find the area of an inscribed hexagon.

The area = \(\frac{1}{2} \times 6 \times 100 \sin 60^\circ = 300 \times \frac{\sqrt{3}}{2}\)

= 150 \times 1.7320508 = 259.81.

25. To find the Area of a Regular Polygon described about a Circle.—In fig. to article 22, if FG be a side of the polygon of n sides; since FG = CE · tan ECF = r tan \(\frac{180^\circ}{n}\), and the polygon = nCFG = nCE · EF;

therefore area = \(nr^2 \tan \frac{180^\circ}{n}\).

Example.—Find the area of a regular hexagon described about a circle whose radius is 10 feet.

Area = 6 \times 100 \times \tan 30^\circ = 600 \times \frac{1}{\sqrt{3}} = 200 \sqrt{3}

= 200 \times 1.7320508 = 346.41.

SECTION II.—SOLIDS CONTAINED BY PLANES.

26. To find the Volume of a Rectangular Parallelopiped.

If the parallelopiped be rectangular, and its edges be a, b, and c, it has been proved in article 5, that its volume = abc;

or the volume is equal to the length multiplied by the breadth and by the height.

Example.—The volume of a parallelopiped whose length, breadth, and height are 5, 4, and 3,

= 5 \times 4 \times 3 = 60.

27. To find the Surface of a Rectangular Parallelopiped.

The surface of the parallelopiped (fig. art. 5) is evidently

= 2BC + 2BD + 2CD;

∴ surface = 2(ab + ac + bc.)

Hence the surface is double the sum obtained by adding together the length multiplied by the breadth, the length multiplied by the height, and the breadth multiplied by the height.

Example.—The surface of the rectangular parallelopiped, whose length, breadth, and thickness are 5, 4, and 3,

= 2(5 \times 4 + 5 \times 3 + 4 \times 3) = 2(20 + 15 + 12) = 94.

28. To find the Volume of any Parallelopiped.—(i.) When the length AC of the base BC (fig. art. 29), its perpendicular breadth CK, and the altitude DE of the parallelopiped are given.—It is proved (Geom. p. 336; Euc. xi. 34), that a parallelopiped is equal to any other parallelopiped having an equal base and the same altitude; and as this other parallelopiped may be rectangular, its volume = base \times height (art. 26). Hence the volume of the parallelopiped AM = AC · CK · DE; or if AB = l, KL = m, and DE = n, the volume = lmn.

Example.—If the length of the base be 57, its perpendicular breadth 5\(\frac{1}{4}\), and the height of the parallelopiped 13, its volume \(=57 \times 525 \times 13 = 2525\).

29. (ii) Given the edges of a parallelopiped, and their inclinations, to find its volume.—Let AM be the parallelopiped, DE a perpendicular from D upon BC; and let a spherical surface whose centre is A, cut the faces of the parallelopiped in the arcs FH, FG, GH, and the plane ADE in the arc HI.

Let \(AB = a\), \(AC = b\), \(AD = c\), and the angles \(BAC = \alpha\), \(BAD = \beta\), \(CAD = \gamma\). In the spherical triangle FGH,

\[ \sin F = \frac{2}{\sin \alpha \sin \beta} [\sin s \sin (\alpha - \beta) \sin (\beta - \gamma)], \]

where \(s = \frac{1}{2}(\alpha + \beta + \gamma)\),

and \(HI = \sin \beta \sin F\);

also \(DE = AD \sin DAE = c \sin HI = e \sin \beta \sin F\);

and the parallelopiped

\(= \text{parallelogram } BC \times DE = DE \times ab \sin \alpha\) (art. 8);

therefore the volume

\[ = 2abc \sqrt{\sin s \sin (\alpha - \beta) \sin (\beta - \gamma)}. \]

Example.—The edges of a parallelopiped are 4, 5, and 6, and their inclination \(40^\circ\), \(50^\circ\), and \(60^\circ\). Find the volume.

\[ \begin{align*} a &= 40^\circ \\ \beta &= 50^\circ \\ \gamma &= 60^\circ \\ S &= 75 \log \sin 1.9849438 \\ s - \alpha &= 35 \ldots = 1.7585913 \\ s - \beta &= 25 \ldots = 1.6259483 \\ s - \gamma &= 15 \ldots = 1.4129962 \\ &= 212.7824796 \\ &= 1.3912398 \\ a &= 4 \log a = .6020600 \\ b &= 5 \log b = .6989700 \\ c &= 6 \log c = .7781513 \\ \log 2 &= .3010300 \\ \end{align*} \]

Volume \(= 59.0814 \log \text{volume} = 1.7714511\)

30. The Surface of the Parallelopiped.—From art. 8 the surface is evidently

\[ = 2(ab \sin \alpha + ac \sin \beta + bc \sin \gamma). \]

Example.—Find the surface of the parallelopiped in example to art. 29.

\[ \begin{align*} a &= 4, b = 5, c = 6, \alpha = 40^\circ, \beta = 50^\circ, \gamma = 60^\circ \\ \log a &= .6020600 \quad \therefore BC = 12.856 \\ \log b &= .6989700 \quad \text{similarly BD} = 18.385 \\ 4 \sin \alpha &= 9.8080675 \quad \text{and CD} = 25.981 \\ \log \text{area BC} &= 1.1090975 \\ &= 57.222 \\ \therefore \text{Surface} &= 114.444 \end{align*} \]

31. To find the Volume of a Prism.—If the parallelopiped AG be cut by a plane BEFC, it will be divided into two equal prisms (GEOMETRY, p. 536, theor. iv.; Eucl. xi. 28). Therefore the prism AEF \(= \frac{1}{2} AG\)

\(= \frac{1}{2} ABHC \times DI\) (art. 28) \(= ABC \times DI\).

If the prism be polygonal, it may be divided into triangular prisms, as shown in the figure; and its volume is equal to the sum of these prisms.

\[ = ABC \times FL + ACD \times FL + ADE \times FL = ABCDE \times FL. \]

Hence, in any prism, if \(m^2\) = area of base, and \(h\) = perpendicular height,

the volume \(= m^2 h\);

or the volume of any prism is equal to its base multiplied by its height.

Example.—The sides of the base of a triangular prism are 3, 4, and 5 feet, and its height 7 feet. What is its volume?

The area of the base \(= 6\) cubic feet (ex. 1, art. 12);

Therefore the volume \(= 6 \times 7 = 42\) cubic feet.

32. To find the Surface of a Prism.—The surface will be obtained by adding the areas of the triangles or polygons which form the ends of the solid, and of the parallelograms which form its sides.

If the edges are perpendicular to the base, it is evident (art. 6) that if \(p\) = the perimeter of the base, and \(h\) = the height, the lateral surface \(= ph\);

and \(m^2\) being the area of the end, the whole surface

\[ = ph + 2m^2. \]

Example.—Find the volume and surface of a prism whose base is a regular polygon of 15 sides, a side of the base 23 feet, and the height 75 feet.

The area of the base \(= 9332.8\) square feet (ex. art. 22);

Therefore the volume \(= 75 \times 9332.8 = 699960\) cubic feet.

Also if the edges are perpendicular to the base, the surface

\[ = 2 \times 9332.8 + 15 \times 23 \times 75 = 445406 \text{ square feet}. \]

33. To find the Volume of a Wedge.—The base AC of the wedge is a parallelogram, and the edge EF is parallel to AB or CD. If the plane EH be parallel to ADF, the wedge will be equal to the sum or difference of the prism AEHG, and the pyramid CEG.

Therefore, if \(AB = a\), \(EF = a_1\) a perpendicular from A upon CD \(= b\), and a perpendicular from F upon the plane AC \(= h\), the wedge

\[ = \frac{1}{2} AG \times bh = \frac{1}{2} BG \times bh \quad (\text{arts. 31 and 35}) \]

\[ = \frac{1}{2} a_1 bh = \frac{1}{2} (a_1 a_2) bh = bh \left( \frac{a_1 + a_2}{2} \right) \]

\[ = \frac{1}{2} bh (2a_1 + a_2). \]

Example.—The length and breadth of the base of a wedge are 32 and 9, and the height 28 feet.

The volume \(= \frac{1}{2} \times 9 \times 28 (64 + 21) = 3570\).

34. To find the Volume of a Rectangular Prismoid.—The faces BD and GE are rectangles, whose planes and sides are parallel; hence the figure is evidently a frustum of a pyramid.

Let \(AB = a\), \(AD = b\), \(EF = a_1\), \(FG = b_1\),

and the perpendicular from F upon the plane DB \(= h\). The prismoid = wedge AED + wedge DEH

\[= \frac{1}{3} h (2a_1 + a_2) + \frac{1}{3} h (2b_1 + b_2)\]

But if the figure were cut by a plane parallel to, and equidistant from the planes AC, EG, and if \(a_1, b_1\) be the length and breadth of this plane reckoned respectively parallel to AD, AB,

then \(a_2 = \frac{a_1 + a_2}{2}; b_2 = \frac{b_1 + b_2}{2}\).

Hence substituting

the prismoid = \(\frac{1}{3} h (a_1 b_1 + a_2 b_2 + a_3 b_3)\).

Since \(a_1 b_1, a_2 b_2, a_3 b_3\) are the areas of the ends, and what may be termed the middle section of the figure parallel to the ends, the volume of the prismoid is obtained by adding together the areas of its ends, and four times the area of its middle section, and multiplying the sum by one-sixth of the height.

Example.—The sides of the base of a rectangular prismoid are 12 and 8; the sides of the top respectively parallel to those of the base are 6 and 4; and the height is 5.

Here \(a_1 = 12, a_2 = 6, b_1 = 8, b_2 = 4, h = 5\),

\[a_2 = \frac{a_1 + a_2}{2} = 9; \quad b_2 = \frac{b_1 + b_2}{2} = 6.\]

The volume = \(\frac{1}{3} (12 \times 8 + 4 \times 9 \times 6 + 4 \times 6) = 280\).

35. To find the Volume of any Pyramid.—The volume of a pyramid is one-third of that of a prism of the same base and altitude (Geom. p. 539, theor. 17, cor. 1; Euclid xii. 7, cor. 1). It therefore follows (art. 31) that if \(a^2\) is the base, and \(h\) is the height,

the volume = \(\frac{1}{3} a^2 h\);

or the volume is one-third of the product of the base multiplied by the height.

36. To find the Volume of a Right Pyramid.—The base AD is supposed to be a regular polygon; F is the centre of a circle described about the polygon AD, and EF is perpendicular to the plane AD. EI, perpendicular to AB, is the "slant height."

Let \(AB = a, EF = h, EI = l, n =\) the number of sides of the polygon.

The area \(AD = \frac{1}{2} na^2 \cot \frac{180^\circ}{n}\) (art. 22),

and hence, since (art. 35) the pyramid = \(\frac{1}{3} AD, EF\),

its volume = \(\frac{1}{3} na^2 h \cot \frac{180^\circ}{n}\).

37. To find the Surface of a Right Pyramid.—The lateral surface of the pyramid = \(nABE = nAI - EI = \frac{1}{2} nal\).

Hence the whole surface = \(\frac{1}{2} na \left(1 + \frac{1}{2} a \cot \frac{180^\circ}{n}\right)\).

In the triangle EFI,

\[EI^2 = EF^2 + FI^2 = EF^2 + IB^2 \cot^2 BFI,\]

by which, according to the data given for calculating the surface and volume of the pyramid, we may obtain \(l\) from \(h\), or \(h\) from \(l\).

Example.—Find the volume and surface of an hexagonal pyramid, each side of the base being 2 feet 6 inches, and the length of the axis 10 feet.

\[a = 2.5, h = 10, n = 6; \quad \frac{180^\circ}{n} = 30^\circ.\]

\[\log \frac{1}{2} a = -0.969100 \quad \text{hence} \quad \frac{1}{2} a^2 \cot^2 \frac{180^\circ}{n} = 4.687\]

\[L \cot 30^\circ = 10^{-2385606} \quad \text{hence} \quad k^2 = 1000000\]

\[\log \frac{1}{2} a^2 \cot^2 \frac{180^\circ}{n} = 6709412\]

Lateral surface = \(3 \times 2.5 \times 10.23 = 76.725\) sq. feet

\[\log a = 3.979400 \quad \text{area of base} = 16238\]

\[\log n = 7.781513 \quad \text{whole surf.} = 92963\] sq. feet

\[L \cot 30^\circ = 10^{-2385606} \quad \text{log base} = 12105319\]

\[\log 3 = 9.228787\]

\[\log 4 = 9.3979400 \quad \text{log volume} = 17334106\]

Log area of base = 12105319 volume = 54126 cubic feet.

38. To find the Volume of the Frustum of any Pyramid.

—Let the pyramid ACF be cut by a plane \(abc\) parallel to the base \(ABC\), and let \(AB = a, ab = a_1\); the areas \(ACB = m_1, acb = m_2\); the heights \(FG = h_1, FHI = h_2, GHI = h\). Then the frustum AcB

\[= \text{pyramid ACF} - \text{pyramid mid abc}\]

\[= \frac{1}{3} m_1 h_1 - \frac{1}{3} m_2 h_2 (\text{art. 35}).\]

But \(\frac{h_1}{h} = \frac{AF}{aF} = \frac{a_1}{a_2};\)

\[\therefore \frac{h_1}{h} = \frac{a_1}{a_2} = \frac{m_1}{m_2}.\]

Hence \(h_1 = \frac{m_1 h}{m_1 - m_2}\)

and \(h_2 = \frac{m_2 h}{m_1 - m_2}.\)

Therefore the volume of the frustum

\[= \frac{1}{3} h \left(m_1^2 - m_2^2\right) = \frac{1}{3} h \left(m_1^2 + m_2^2 + m_3^2\right);\]

a formula which applies to the frusta of all pyramids whether right or oblique.

39. The formula of article 38 may be put into a somewhat different form.—If \(AB = a, ab = a_1\); area \(AC = m_1\), area \(ac = m_2\); and if \(a_2\) be the side and area of the middle section of the frustum formed by a plane parallel to, and equidistant from, the planes AC, ac, we shall evidently have (Eucl. vi. 20, cor. 3)—

\[m_1^2 = c^2 a_1^2, m_2^2 = c^2 a_2^2, m_3^2 = c^2 a_3^2;\]

where \(c\) depends on the nature of the polygon AD.

Also \(a_2 = \frac{1}{2} (a_1 + a_3);\)

therefore \(4m_3^2 = c^2 (a_1 + a_3)^2 = c^2 a_1^2 + 2c^2 a_1 a_3 + c^2 a_3^2\)

\[= m_1^2 + 2m_1 m_2 + m_2^2.\]

Now the volume of the frustum

\[= \frac{1}{3} h (2m_1^2 + 2m_1 m_2 + m_2^2) (\text{art. 37}).\]

Whence substituting

the volume = \(\frac{1}{3} h \left(m_1^2 + 4m_2^2 + m_3^2\right);\)

or the volume of the frustum of a pyramid is obtained by adding the areas of the ends to four times the area of the middle section, and multiplying the sum by one-sixth of the height.

Example.—Find the volume of the frustum of an hexagonal pyramid, the sides of whose terminating polygons are 4 and 3 feet, and its height 9 feet.

(i.) By the formula of article 38.—By art. 22 we find \( m^2 = 41569 \), and \( m^2 = 23383 \).

Whence \( m_2 = 31177 \); and the volume \( = 3(41569 + 31177 + 23383) = 28839 \) cubic feet.

(ii.) By the formula of article 39.—The side of the middle section, \( a_2 = \frac{1}{2}(4+3) = 3.5 \).

Whence (art. 22) \( m_2 = 31826 \); and the volume \( = \frac{3}{2}(41569 + 127304 + 23383) = 28839 \) cubic feet.

40. To find the Surface of the Frustum of a Right Pyramid.—In fig. art. 38 let the perimeter of AC = \( p_1 \), and that of AE = \( p_2 \); and the slant heights DF = \( l_1 \), EF = \( l_2 \), DE = \( l_3 \), AB = \( a_1 \), \( a_2 = a_3 \); then the lateral surface of the frustum is equal to the difference of the lateral surfaces of the pyramids AFC, \( a/c = \frac{1}{2}p_1l_1 - \frac{1}{2}p_2l_2 \).

But \( \frac{l_1}{l_2} = \frac{a_1}{a_2} \), \( \frac{l_2}{l_3} = \frac{a_2}{a_3} \), \( \frac{p_1}{p_2} = \frac{a_1}{a_2} \), \( \frac{p_2}{p_3} = \frac{a_2}{a_3} \).

Hence \( l_1 = \frac{p_1l_2}{p_1-p_2} \), and \( l_2 = \frac{p_2l_3}{p_2-p_3} \);

Therefore the lateral surface of the frustum

\[ = \frac{1}{2}l_1^2 - \frac{1}{2}l_2^2 = \frac{1}{2}l(p_1+p_2). \]

Since \( h_1 = \frac{a_1}{a_2} \), \( h_2 = \frac{a_2}{a_3} \),

Also, \( l_1 = h_1^2 + \frac{1}{2}a_1^2 \cot^2 \frac{180^\circ}{n} \) (art. 37);

and \( l = \frac{a_1-a_2}{a_1}l_1 \).

Hence we may find \( l \) when \( h \) is given.

Example.—Find the lateral surface of the frustum in last example.

\( a_1 = 4, a_2 = 3, h = 9 \).

Whence \( h_1 = \frac{4}{3} \times 9 = 36 \);

and since \( \frac{1}{2}a_1^2 \cot^2 \frac{180^\circ}{n} = 4687 \), we have

\[ l_1 = 36-07, \text{ and } l = \frac{(4-3) \times 36-07}{4} = 90175. \]

Hence lateral surface \( = 3(4+3) \times 90175 = 18937 \) square feet.

41. To find the Volume and Surface of a Regular Polyhedron.—Regular polyhedrons are solids contained by planes, which are equal, similar, and regular polygons; each solid angle of the polyhedron is contained by the same number of planes, having the same inclinations, and the polyhedron admits of having a sphere inscribed within it, or described about it. There are only five regular polyhedrons, which are enumerated in Table II.

Let ABE be one of the faces of the polyhedron, O the centre of a circumscribed sphere, and OC, OD perpendiculars to the plane ABE and to the line AB; then it is evident that AO and OC are the radii of spheres described about the polyhedron, and inscribed within it; that AB is bisected in D, and that C is the centre of the polygon ABE.

If \( l \) is the number of faces in the polyhedron,

\( m \) is the number of faces in each of its solid angles,

\( n \) is the number of sides in each of its faces, and

\( a = AB \), the length of each of those sides;

and if we suppose the planes ACO, ADQ, CDQ, to meet the surface of a sphere whose centre is O in the arcs pq, pr, qr,

the angles, \( p = \frac{\pi}{m}, q = \frac{\pi}{n}, r = \frac{\pi}{2} \);

and in the spherical triangle pqr,

\[ \cos AOC = \cot p \cot q = \cot \frac{\pi}{m} \cot \frac{\pi}{n} \]

whence \( \cot AOC = \frac{\cos \frac{\pi}{m} \cos \frac{\pi}{n}}{\left\{ -\cos \left( \frac{\pi}{m} + \frac{\pi}{n} \right) \cos \left( \frac{\pi}{m} - \frac{\pi}{n} \right) \right\}^{\frac{1}{2}}} \)

Also CO = AC cot AOC = AD cosec ACD cot AOC,

\[ = \frac{1}{2}a \cos \frac{\pi}{n} \cot AOC; \]

and the polygon ABE \( = \frac{n a^2}{4} \cot \frac{\pi}{n} \) (art. 22).

Now, as the polyhedron is made up of \( l \) equal pyramids, whose bases and altitudes are respectively the same as the polygon ABE and the line CO,

its volume \( = \frac{1}{2}l \cdot CO \cdot AEB \) (art. 35),

\[ = \frac{nla^2}{24} \cdot \frac{\cos \frac{\pi}{m} \cdot \cot \frac{\pi}{n}}{\left\{ -\cos \left( \frac{\pi}{m} + \frac{\pi}{n} \right) \cos \left( \frac{\pi}{m} - \frac{\pi}{n} \right) \right\}^{\frac{1}{2}}} \]

a form apparently impossible, but not really so; for in every polyhedron \( \cos \left( \frac{\pi}{m} + \frac{\pi}{n} \right) \) is negative, the angle \( \frac{\pi}{m} + \frac{\pi}{n} \) being \( > 90^\circ \) and \( < 180^\circ \).

It is evident (art. 22) that the surface

\[ = \frac{nla^2}{4} \cdot \cot \frac{\pi}{n}. \]

Example.—Find the volume of a tetrahedron whose edge is 1.

The tetrahedron is contained by 4 equilateral triangles and each of its solid angles has 3 faces; whence

\[ a = 1, m = 3, n = 3, l = 4 \]

\[ \frac{\pi}{m} = \frac{\pi}{n} = 60^\circ. \]

Hence the volume \( = \frac{4 \times 3}{24} \cdot \cos 60^\circ \cdot \cot^2 60^\circ \)

\[ = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot 1178513; \]

and the surface \( = \frac{3}{4} \cdot \cot 60^\circ \)

\[ = \sqrt{3} = 1.7320508. \]

42. It obviously follows from the formulae of last article, that the surface and volume of a polyhedron whose edge is \( a \), may be obtained by multiplying the surface and volume of a similar polyhedron, whose edge is 1, by \( a^2 \) and \( a^3 \) respectively. The surfaces and volumes of the five regular polyhedrons, whose edges = 1, are given in table II.

Example.—Find the surface and volume of a regular dodecahedron whose edge is 5.

The tabular surface and volume from table II. are 206457788 and 76631189.

The surface \( = 25 \times 206457788 = 51614447. \)

The volume \( = 125 \times 76631189 = 95788986. \) SECTION III.—THE CIRCLE.

The Circumference of a Circle.

43. If \( CP = r \), \( CN = x \), \( MP = y \), the equation to the circle is \( x^2 + y^2 = r^2 \);

\[ \frac{dy}{dx} = -\frac{x}{y} \]

and the length of the arc \( AP \)

\[ = \int_0^x \left( 1 + \left( \frac{dy}{dx} \right)^2 \right)^{\frac{1}{2}} dx \]

\[ = r \int_0^x \left( \frac{r^2}{(r^2 - x^2)^{\frac{3}{2}}} \right) dx \]

\[ = r \left( \frac{x}{r} + \frac{1}{2} \cdot \frac{x^3}{3r^2} + \frac{1}{2} \cdot \frac{x^5}{5r^4} + \cdots \right) \]

Putting \( x = r \), the length of the quadrant \( AP \)

\[ = r \left( 1 + \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{6} \cdot \frac{1}{7} \cdot \cdots \right) \]

44. A very rapidly converging series for the circumference of a circle is obtained by the following process. It is shown in Trigonometry, that

\[ \tan^{-1} m + \tan^{-1} n = \tan^{-1} \frac{m-n}{1+mn}; \]

\[ \tan^{-1} 1 - \tan^{-1} \frac{1}{5} = \tan^{-1} \frac{1}{3} \]

\[ \tan^{-1} \frac{2}{3} - \tan^{-1} \frac{1}{5} = \tan^{-1} \frac{1}{7} \]

\[ \tan^{-1} \frac{7}{17} - \tan^{-1} \frac{1}{5} = \tan^{-1} \frac{1}{9} \]

\[ \tan^{-1} \frac{9}{46} - \tan^{-1} \frac{1}{5} = \tan^{-1} \frac{1}{239} \]

Whence, adding,

\[ \tan^{-1} 1 - 4 \tan^{-1} \frac{1}{3} = \tan^{-1} \frac{1}{239} \]

or

\[ \pi = 4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239}; \]

for \( \tan^{-1} 1 = \frac{\pi}{4} \); where \( \pi \) is the circumference of a circle whose diameter is unity.

Now, it is proved in Trigonometry that

\[ \tan^{-1} x = x - \frac{1}{3} x^3 + \frac{1}{5} x^5 - \frac{1}{7} x^7 + \cdots \]

Therefore,

\[ \frac{\pi}{4} = 4 \left( \frac{1}{5} - \frac{1}{3} \cdot \frac{1}{5^3} + \frac{1}{5^5} - \frac{1}{5^7} + \cdots \right) \]

a rapidly converging series, from which the value of \( \pi \) is calculated as follows:

\[ \begin{align*} \frac{1}{5} &= 0.20000000 \\ \frac{1}{5^3} &= 0.00064000 \\ \frac{1}{5^5} &= 0.00000057 \\ \frac{1}{5^7} &= 0.000000024 \\ \frac{1}{5^9} &= 0.000000004 \\ \end{align*} \]

\[ \pi = 3.141592704, \text{ which is correct to 7 places.} \]

The value of \( \pi \) to 20 places of decimals is

\[ 3.14159265358979323846. \]

The functions of \( \pi \), which are most frequently useful in Mensuration, will be found in Table III.

45. Since \( \pi \) is the circumference of a circle whose diameter is unity; and since the circumference of circles are proportional to their radii, or to their diameters. (See Trigonometry.)

therefore if \( r \) is the radius; \( d \) is the diameter;

the circumference of the circle \( = \pi d = 2 \pi r \).

Whence the following rules:

(i.) To find the circumference of a circle, multiply the diameter, or twice the radius, by 3.14159..., or 3.1416.

Example 1.—The circumference of a circle whose diameter is 5 feet

\[ = 5 \times 3.1415927 = 15.7079635 \text{ feet.} \]

Example 2.—Find the circumference of the earth at the equator, the equatorial diameter being 79256 miles.

\[ \log 79256 = 3.8990321 \]

\[ \log \pi = 4.971499 \]

Circumference \( = 24890.0 \text{ miles.} \)

46. If we convert the value of \( \pi \) into a continued fraction we obtain

\[ \frac{314159}{100000} = 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1 + \frac{1}{25 + \cdots}}}} \]

Of which the successive convergents are

\[ \frac{22}{7}, \frac{333}{106}, \frac{355}{113}, \cdots \]

Of these

\[ \frac{355}{113} = 3.1415929, \]

which differs from the accurate value of \( \pi \) only in the 7th place of decimals. Whence the following rules:

(ii.) To find the circumference of a circle roughly, multiply the diameter by 22, and divide by 7.

Example.—The circumference of a circle 5 feet in diameter is about

\[ 5 \times \frac{22}{7} = 15.71; \]

which is correct only in one place of decimals.

(iii.) To find the circumference of a circle very nearly, multiply the diameter by 355, and divide by 113.

Example.—The circumference of a circle whose diameter is 5 is

\[ \frac{5 \times 355}{113} = 15.7079645; \]

a result correct in 5 places of decimals.

(iv.) To multiply by \( \pi \). Multiply by 22 and divide by 7, and from the result subtract \( \frac{1}{22} \)th of \( \frac{1}{22} \)th of the multiplicand.—The result is too great by about its 200,000th part.

Example.—Find the circumference in last example.

\[ \begin{array}{c|c} \frac{22}{7} & 100 \\ \hline 110 & 89.05 \\ \end{array} \]

Circumference \( = 15.708035 \)

47. To find the diameter of a Circle when the Circumference is given.

(i.) It follows from article 45, that to find the diameter we must divide the circumference by 3.14159, or multiply by 3.18309866184. Or if \( c \) is the circumference of the circle,

\[ d = \frac{c}{\pi} ; \quad r = \frac{c}{2\pi}. \]

**Example.**—The diameter of the circle whose circumference is 1570796

\[ = 1570796 \times 3183098 = 4999998 = 5 \text{ nearly}. \]

(ii.) To find the diameter roughly, multiply the circumference by 7, and divide by 22.

(iii.) To find the diameter nearly, multiply the circumference by 113, and divide by 355.

(iv.) To divide by \( \pi \), multiply by 7, divide by 11 and by 2, and to the result add \( \frac{1}{3} \) of \( \frac{1}{3} \) of the dividend. The result is too small by about its 100,000th part.

**Example.**—Find the diameter in last example.

\[ \begin{align*} 15707964 & \\ 7 & \\ 1109955748 & \\ 299995977 & \\ 4999998 & \\ \end{align*} \]

\( = 4999998 \) is the diameter.

**Arcs of Circles.**

48. Since angles at the centres of circles are proportional to the arcs on which they stand (Geom. sec. iv., theor. 31; or Eucl. vi., 38),

if \( A \) is the number of degrees in the angle at the centre;

\( r \) is the radius,

\( a \) is the arc which subtends the angle \( A \),

\[ \pi r : a :: 180^\circ : A; \]

\[ A = \frac{180^\circ}{\pi} \cdot \frac{a}{r}; \]

or \( A = 57.2957795 \times \frac{a}{r} \).

From this equation the following rules are obtained:

49. (i.) To find the number of degrees in an arc when its length and the radius of the circle are given, multiply the length of the arc by 57.295778, and divide by the radius.

**Example.**—If the radius of a circle be 25 and the arc 30, the number of degrees in the arc

\[ = 57.29578 \times \frac{30}{25} = 68^\circ 75494 = 68^\circ 45' 17'' 8; \]

or, by logarithms,

\[ \begin{align*} \log \frac{180}{\pi} & = 1.7581226 \\ \log a & = 1.4771213 \\ \log r & = 1.3979400 \\ \log A & = 1.8373039 \\ & = 68^\circ 45' 17'' 8. \end{align*} \]

50. (ii.) To find the radius of the circle when the length of the arc and the number of degrees which it contains are given, multiply the length of the arc by 57.29578, and divide by the number of degrees in the angle.

For, by the formula (art. 48),

\[ r = 57.2957795 \times \frac{a}{A}. \]

**Example.**—The length of the arc is 30 feet, and the angle, which it subtends at the centre, 68° 45' 17'' 8. Find the radius of the circle.

\[ \begin{align*} a = 30 & \\ \log \frac{180}{\pi} & = 1.7581226 \\ A = 68^\circ 45' 17'' 8 & \\ \log a & = 1.4771213 \\ \log A & = 1.8373039 \\ \log r & = 1.3979400 \\ \end{align*} \]

Hence the radius \( = 25 \) feet.

51. (iii.) To find the arc when the radius and the angle subtended at the centre are given, multiply the number of degrees in the angle by the radius and by 0.174533. For, by the formula of art. 48,

\[ a = \frac{\pi}{180} \cdot Ar = 0.174533 \times Ar. \]

**Example.**—Given the radius 25 feet, and the angle at the centre 68° 45' 17'' 8, to find the length of the subtending arc.

\[ \begin{align*} r & = 25 \\ \log r & = 1.3979400 \\ A & = 68^\circ 75494 \\ \log A & = 1.8373039 \\ \end{align*} \]

From Table III. \( \log \frac{\pi}{180} = 2.2418774 \)

\[ \text{Arc} = 30 \text{ feet}. \quad \log a = 1.4771213 \]

Since the length of an arc \( = \frac{\pi A}{180} \times \text{radius}; \) if \( \frac{\pi A}{180} \) (or the length of an arc to radius = 1) be computed for the various values of \( A \), and arranged in a table (See Table IV.), the length of any arc may be found by multiplying the proper tabular number by the radius of the arc.

**Example.**—To find the length of an arc whose radius is 25, and angle 68° 45' 17'' 8.

From Table IV. arc of 60° = 1.0471976

\[ \begin{align*} 8^\circ & = 1.3962663 \\ 40^\circ & = 1.16355 \\ 5^\circ & = 1.45444 \\ 10^\circ & = 1.485 \\ 7^\circ & = 1.339 \\ 0^\circ 8 & = 1.39 \end{align*} \]

\[ \therefore \text{Arc to radius } 25 = 1.2 \times 25 = 30. \]

A great variety of problems in finding the lengths of arcs may be proposed. The following seem to be the most important:

52. (iv.) To find the length of an arc whose chord and radius are given.—In fig. art. 22 we have given \( AC = r \), \( AB = 2c \).

To find the length of the arc \( AEB \). Put the angle \( ACB = A \), then \( \sin \frac{A}{2} = \frac{c}{r} \); whence \( A \) is known, and the length of the arc is found by art. 51.

**Example.**—The chord of an arc is 28-23214 feet, and the radius 25 feet. Find the length of the arc.

Here \( c = 14.11607 \)

\[ \begin{align*} \log c & = 1.1497136 \\ \log r & = 1.3979400 \\ \end{align*} \]

\[ \therefore \frac{A}{2} = 34^\circ 22' 38'' 9 \quad L \sin \frac{A}{2} = 9.7517736 \]

and \( A = 68^\circ 45' 17'' 8 \)

Having ascertained \( A \), we find, by art. 51, that the length of the arc is 30 feet.

53. To find the Length of an Arc when its Chord, and Height, or Versed Sine, are given.—In fig. art. 22, putting \( DE = h \);

\[ AC^2 = AD^2 + CD^2; \quad \therefore r^2 = c^2 + (r - h)^2: \]

Hence \( r = \frac{1}{2} \left( h + \frac{c^2}{h} \right) \)

The radius \( r \) being known, the length of the arc is found by art. 52.

**Example.**—Given the chord of an arc 28-23214 feet, and its versed sine 4-36661, to find its length.

\[ \begin{align*} c & = 14.11607 \\ \log c & = 1.1497136 \\ \log h & = 2.2994272 \\ \log h & = 6.401444 \\ \log \frac{c^2}{h} & = 1.6592828 \\ \end{align*} \]

The radius \( r = 25.0 \). We have now \( \sin A = \frac{c}{r} \) (art. 52),

whence \( A = 68^\circ 75'49'' \);

and finally, by art. 52, the arc is found to be 30 feet.

The Area of a Circle.

54. In the figure to article 43, if \( CP = r^2 \), \( CM = x \), and \( MP = y \), the equation to the circle is \( x^2 + y^2 = r^2 \), and the area of the quadrant ABC.

\[ \int_0^r y dx = \int_0^r (r^2 - x^2)^{\frac{1}{2}} dx = \frac{\pi r^2}{4} \]

Whence the area of the circle \( = \pi r^2 \). See also article 59.

55. A more elementary demonstration may be obtained as follows:

The area of a circle is greater than that of any inscribed, and less than that of any circumscribed polygon; and by continually increasing the number of sides of the polygons, their areas will obviously approach to equality with each other, and with that of the circle.

Let \( AB, FG \) (fig. art. 22) be the sides of regular inscribed and circumscribed polygons,

\( n = \) the number of sides in each, \( P = \) the area of the circumscribed polygon, \( p = \) the area of the inscribed polygon.

Then putting \( \theta \) for the angle \( ACB \),

\[ \theta = \frac{n}{n}; \]

and, by continually increasing \( n \), \( \theta \) diminishes indefinitely, and ultimately

\[ \tan \theta = \frac{\sin \theta}{\theta} = 1. \quad (\text{See Trigonometry.}) \]

Now, \( P = nr^2 \tan \frac{\pi}{n} = \pi r^2 \cdot \frac{\tan \frac{\pi}{n}}{\pi} \) (art. 25).

\[ = \pi r^2 \cdot \frac{\tan \theta}{\theta} = \pi r^2 \text{ ultimately}; \]

for \( \frac{\tan \theta}{\theta} = 1 \) ultimately.

Also \( p = nr^2 \sin \frac{\pi}{n} \cos \frac{\pi}{n} = \pi r^2 \cdot \frac{\sin \frac{\pi}{n}}{\pi} \cos \frac{\pi}{n} \) (art. 24).

\[ = \pi r^2 \cdot \frac{\sin \theta}{\theta} \cdot \cos \theta = \pi r^2 \text{ ultimately}; \]

for \( \frac{\sin \theta}{\theta} = 1 \) and \( \cos \theta = 1 \) ultimately.

Therefore, since we have always

\( P > \) area of circle \( > p \);

and since \( P \) and \( p \) both ultimately differ from \( \pi r^2 \) by less than any assignable quantity, it follows that

the area of the circle \( = \pi r^2 \).

56. If \( d \) be the diameter of the circle,

\( d = 2r \), and area \( = \frac{\pi}{4} d^2 \);

where \( \frac{\pi}{4} = 0.785398163397 \),

again the area \( = \pi r^2 \cdot r \).

But if \( c = \) the circumference, \( c = 2\pi r \),

therefore the area \( = \frac{1}{2} cr = \frac{1}{2} c \times \frac{1}{2} d \).

Lastly, the area \( = \pi r^2 = \frac{4}{4} \frac{\pi^2}{4} \),

where \( \frac{1}{4} = 0.7857747155 \).

Whence the following rules:

(i.) The area of a circle is obtained by multiplying the square of the radius by 3.14159.

Example.—The area of a circle whose radius is 3.5,

\((3.5)^2 \times 3.14159 = 12.25 \times 3.14159 = 38.4845.\)

(ii.) The area of a circle is equal to the square of the diameter multiplied by 7.85398.

Example.—The area of a circle whose diameter is 7,

\[ = 49 \times 7.85398 = 38.4845. \]

(iii.) The area of a circle is equal to half the diameter multiplied by half the circumference.

Example.—The diameter of a circle is 7; find the area.

By article 45, the circumference \( = 21.9911486 \);

therefore the area \( = 3.5 \times 10.9955743 = 38.4845 \).

(iv.) The area of a circle is equal to the square of the circumference multiplied by 0.785398.

Example.—The circumference being 21.9911486;

the area \( = (21.9911486)^2 \times 0.785398 = 38.4845 \).

57. In the formulae of articles 55 and 56, if we put

\( C^2 = \) the area of the circle,

\[ r = \frac{1}{\sqrt{\pi}} \cdot C \quad \text{where} \quad \frac{1}{\sqrt{\pi}} = 0.5641896 \]

\( d = \frac{2}{\sqrt{\pi}} \cdot C \quad \text{where} \quad \frac{2}{\sqrt{\pi}} = 1.283792 \)

\( c = 2\sqrt{\pi} \cdot C \quad \text{where} \quad 2\sqrt{\pi} = 3.5449077 \).

Whence, to find the radius, diameter, or circumference of a circle when the area is given, multiply the square root of the area by 0.5641896, 1.283792, or 3.5449077 respectively.

Example.—The area of a circle is 38.4845.

\[ \sqrt{38.4845} = 6.2036 \]

\[ \therefore \text{radius} = 6.2036 \times 0.5641896 = 3.5 \]

\[ \text{diameter} = 6.2036 \times 1.283792 = 7.9 \]

\[ \text{circumference} = 6.2036 \times 3.5449077 = 21.9911. \]

58. To find the Area of a Circular Ring.—The ring is a plane surface bounded by two circles, described one within the other, but not necessarily concentric. If \( r_1, r_2 \) be the outer and inner radii of the ring, its area

\[ = \pi (r_1^2 - r_2^2) = \pi (r_1 + r_2)(r_1 - r_2). \]

Example.—Find the area of a ring whose outer and inner diameters are 10 and 6.

\( r_1 = 5 \) and \( r_2 = 3 \);

\[ \therefore \text{area} = 8 \times 2 \times 3.14159 = 50.2655. \]

59. To find the Area of a Sector of a Circle.—If \( r \) be the radius of the circle (fig. art. 43), and \( \theta \) the angle ACP, then the area of the sector ACP

\[ = \frac{1}{2} r^2 \int_0^\theta d\theta = \frac{1}{2} \theta r^2 = \frac{1}{2} \text{arc} \times \text{radius}. \]

For the whole circle \( \theta = 2\pi \);

\[ \therefore \text{circle} = \pi r^2, \]

as was already proved in articles 54 and 55.

Or, since sectors have the same ratio as the arcs on which they stand (Euclid 33, vi.), if \( A \) = the angle of the sector, in degrees,

\[ 360^\circ : A :: \text{area of circle} : \text{area of sector}; \]

\[ \therefore \text{sector} = \frac{A}{360} \pi r^2. \] Again, since \( \frac{A}{360} = \frac{a}{2\pi r} \) (art. 48),

the sector \(= \frac{1}{2} ar\);

and, finally, since \( \theta = \frac{a}{r} \),

the sector \(= \frac{1}{2} \theta r^2\).

We have therefore the following rules:

(i.) The area of a sector is equal to the area of the whole circle, multiplied by the number of degrees in the arc of the sector, and divided by 360.

Example.—The radius of a sector is 25 feet, and its angle 68° 45' 17" 8'. Required its area.

Here \( r = 25, A = 68^\circ 75^\circ 494 \);

\[ \text{area of sector} = \frac{(25)^2 \times 314159 \times 68^\circ 75^\circ 494}{360} = 375 \text{ sq. ft.} \]

Or, by logarithms,

\[ \log r = 1.3979400 \\ \log \pi = 0.4971499 \\ \log A = 1.8373039 \\ \colog 360 = 7.4436975 \\ \log \text{area} = 2.5740313 \]

Area \(= 375\).

Since, by art. 63, a sector

\[ \frac{A}{360} \pi r^2 = \frac{1}{2} \times \frac{A\pi}{180} \times r^2; \]

where \( \frac{A\pi}{180} \) is the length of an arc of \( A^\circ \) to radius 1, its area will be found by ascertaining the value of \( \frac{A\pi}{180} \) from Table IV., and multiplying by the square of the radius.

Example.—To find the sector whose radius is 25 and angle 68° 45' 17" 8'.

From Table II. (see art. 51).—The length of an arc of 68° 45' 17" 8' to radius 1 = 1.2;

\[ \text{area of sector} = 1.2 \times (25)^2 = 1.2 \times 625 = 375. \]

(ii.) The area of a sector is equal to one-half of the length of its arc multiplied by its radius.

Example.—If the radius of a sector be 25 feet, and the length of its arc 30 feet,

its area \(= \frac{1}{2} \times 30 \times 25 = 375 \text{ square feet.}\)

(iii.) If the radius AC and chord AB (fig. art. 22) be given, putting AC \(= r, AB = 2c,\)

we have \( \sin \frac{A}{2} = \frac{c}{r}, \)

whence A is known; and the area of the sector is found by Rule (i.).

Example.—If the radius of a sector is 25, and the chord of its arc 28° 23' 24", what is its area?

Here A is ascertained to be 68° 45' 17" 8' (see art. 52); and by Rule (i), the area of the sector is found to be 375.

60. To find the Area of a Segment of a Circle.—Let C be the centre of the circle, CD \(= r, DE = a, MP = b,\)

CE \(= c, EM = h, AE = l.\)

Then, if EM \(= x,\) and MP \(= y,\)

\[ (y + c)^2 + x^2 = r^2, \]

and the area DEMP

\[ = \int_0^h ydx = \int_0^h (r^2 - h^2)dx, \]

\[ = \frac{r^2}{2} \sin^{-1} \frac{h}{r} + \frac{h}{2} (r^2 - h^2) - ch. \]

Also the area ADB \(= \int_{-l}^{+l} ydx\)

\(= r^2 \sin^{-1} \frac{l}{r} - cl = \text{sector ACBD} - \text{triangle ABC}.\)

61. We may also express the areas DEMP, ADBE, in terms of AP, DE, and MP.

For, from the equation to the circle,

\[ (b + c)^2 + h^2 = r^2 (a + c)^2. \]

Whence \( e = \frac{b^2 + h^2 - a^2}{2(a-b)}, \quad r = \frac{(a-b)^2 + h^2}{2(a-b)}, \quad \text{and} \quad (r^2 - h^2) = \frac{b+c}{2(a-b)}.\)

Therefore, substituting area DEMP

\[ = \frac{1}{4(a-b)} \left( \frac{(a-b)^2 + h^2}{2(a-b)} \right)^2 \sin^{-1} \frac{2al}{a^2 + l^2} - \frac{(l^2 - a^2)}{2(a-b)}. \]

When \( h = l, b = o; \)

\[ \therefore \text{area ABD} = \frac{1}{2a} \left( \frac{(a^2 + l^2)^2}{2a} \right) \sin^{-1} \frac{2al}{a^2 + l^2} - \frac{(l^2 - a^2)}{2(a-b)}. \]

62. Again, since the area of the segment ADB \(= \text{sector ADBC} - \text{triangle ABC};\) putting angle ACB \(= \theta,\) and AC \(= r,\)

the segment \(= \frac{1}{2} \theta r^2 - \frac{1}{2} r^2 \sin \theta \) (arts. 59 and 11)

\(= \frac{1}{2} r^2 (\theta - \sin \theta).\)

Whence (i.) to find the area of a segment of a circle, find the area of the corresponding sector, and subtract the triangle contained by the radii of the sector and the base of the segment.

Among various particular cases of this problem, the following may be noted:

(ii.) Given the radius AC \(= r,\) and the angle at the centre ACB \(= A^\circ.\)

The sector ADBC \(= \frac{A}{360} \pi r^2 \) (art. 59);

the triangle ABC \(= \frac{1}{2} r^2 \sin A;\)

\[ \therefore \text{area of segment ABD} = \frac{1}{2} r^2 \left( \frac{\pi A}{180} - \sin A \right). \]

Example.—Find the area of the segment whose arc has a radius of 25 feet, and which subtends an angle of 68° 45' 17" 8' at the centre.

\( r = 25 \log r = 1.3979400 \) In Ex. Rule (I.) (art. 59), the area of the sector has been found

\[ \log r = 2.7958800 \text{ to be } = 375.000 \]

\( A = 68^\circ 45' 17" 8' \) \( \sin A = 0.9994343 \) the triangle \(= 291.262\)

\[ \colog 2 = 9.6289700 \quad \therefore \text{segment } = 83.738 \]

Area of triangle \(= 291.262.\)

(iii.) Given the Chord and Radius of the Segment.—Let \(2c\) \(= \) the chord, \(r\) \(= \) the radius, \(A\) \(= \) angle at centre.

Then \( \sin \frac{A}{2} = \frac{c}{r}; \)

and A being known, the area of the segment is found by Rule (ii.).

Example.—Find the area of the segment whose radius is 25, and chord 28° 23' 24".

\( c = 14^\circ 11607,\) from which \( A = 68^\circ 45' 17" 8' \) (see Ex. art. 52);

and the area is then found, by Rule (ii.), to be 83.738.

(iv.) Given the chord and height, or versed sine of the segment.—Let \(2c\) \(= \) the chord, \(h\) \(= \) the height, \(r\) \(= \) the radius.

\[ r = \frac{1}{2}(h + \frac{c^2}{h}) \quad \text{(art. 63).} \]

Then A is found as in Rule (iii.), and finally the area is obtained by Rule (ii.)

Example.—Given the chord of a segment 28° 23' 24", and its height 4° 36661, to find its area.

\( c = 14^\circ 11607 \) and \( h = 4^\circ 36661,\) from which \( r = 25 \) (see art. 53).

64. To find the Surface of a Right Cone.—Putting \( AB = l \), the lateral surface of the pyramid described about the cone,

\[ = n' \cdot \text{BED} = n' \cdot \text{AB} \cdot \text{AD} = nlr \tan \frac{\pi}{n} = \pi lr \cdot \frac{\tan \frac{\pi}{n}}{\pi}. \]

But by increasing \( n \) indefinitely,

\[ \tan \frac{\pi}{n} = 1; \]

and the surface of the pyramid coincides with that of the cone;

\[ \therefore \text{surface of cone} = \pi lr. \]

Also, if \( p \) is the circumference of the base,

\[ \therefore \pi r^2 + \pi lr = \pi r(r + l); \]

Hence the convex surface of a cone is equal to half the perimeter of the base multiplied by the slant height.

If we add the surface of the base, the whole surface,

\[ = \pi r^2 + \pi lr = \pi r(r + l). \]

Again, if \( h \) is the vertical height,

\[ l = (r^2 + h^2)^{\frac{1}{2}}; \]

\[ \therefore \text{surface} = \pi r[r + (r^2 + h^2)^{\frac{1}{2}}]. \]

Example.—Find the surface of a cone whose height is 13, and the radius of its base 3:5.

The slant height \( l = (169 + 1225)^{\frac{1}{2}} = 13:46 \) nearly.

The convex surface \( = 3:14159 \times 3:5 \times 13:46 = 148:0 \).

The surface of the base (art. 56) \( = 38:5 \).

The whole surface \( = 186:5 \).

65. To find the Volume of the Frustum of a right Cone.—The frustum is cut off by a plane parallel to the base.

Let the radii of the ends \( BC = r_1, BC = r_2 \), and the height \( Ce = h \). Also, let \( AC = h, AC = a_2 \).

Then \( \frac{h}{r_1} = \frac{r_1}{r_2} \); \( \frac{h}{r_2} = \frac{r_2}{r_1 - r_2} \).

From which \( h_1 = \frac{r_1 h}{r_1 - r_2} \), and \( h_2 = \frac{r_2 h}{r_1 - r_2} \).

Since the frustum = difference of cones ABC, \( A_b c \),

its volume \( = \frac{1}{3} \pi r^2 h_1 - \frac{1}{3} \pi r^2 h_2 \) (art. 63)

\[ = \frac{1}{3} \pi h \left( \frac{r_1^2 - r_2^2}{r_1 - r_2} \right) = \frac{1}{3} \pi h \left( r_1^2 + r_1 r_2 + r_2^2 \right). \]

Or if \( a_1^2, a_2^2 \) be the areas of the ends of the frustum,

the volume \( = \frac{1}{3} h(a_1^2 + a_2^2 + a_3^2) \).

66. It may also be shown, as in art. 39, that the volume

\[ = \frac{1}{3} h(a_1^2 + a_2^2 + a_3^2). \]

Where \( a_1^2, a_2^2 \) are the areas of the terminating planes of the frustum, and \( a_3^2 \) the area of a section parallel to these planes, and equidistant from them.

---

Mensuration.

From \( r \) and \( c \) we have

\[ A = 68^\circ 45' 17''8 \text{ by Rule (iii)}. \]

Finally, from \( r \) and \( A \) we find by Rule (ii) that

the area \( = 83:738 \).

(v.) Given the height and radius or diameter of the segment.—Let \( r \) be the radius, \( h \) the height, then, by (ii),

the area \( = \frac{1}{2} \left( \frac{\pi A}{180} - \sin A \right) \times r^2 \);

and we have

\[ \text{versin } A = \frac{h}{r}. \]

Whence \( A \) is known, and the area may be calculated as in (ii).

If the radius be divided into \( n \) equal parts, and \( h \) be taken successively equal to \( \frac{r}{n}, \frac{2r}{n}, \ldots \), the corresponding values of \( \frac{h}{r} \), and of the factor \( \frac{\pi A}{180} - \sin A \) may be calculated and arranged in a table (see Table V.). The area of any segment may then be ascertained by dividing its height by the radius of its arc, so as to obtain the tabular versed sine and the corresponding tabular number, and then multiplying the tabular number by the square of the radius of the segment.

Example.—To find the area of a segment whose height is 4:36661, and whose radius is 25.

Here \( h = 4:36661, d = 50 \).

Tabular versin \( = \frac{h}{r} = \frac{4:36661}{25} = 1:75 \) nearly.

The tabular number for 1:75, obtained by taking the mean of those for 1:7 and 1:8, is 1:34396;

\[ \therefore \text{area of segment} = 1:34396 \times (25)^2 = 84 \text{ nearly}. \]

The Cone.

63. To find the Volume of a Right Cone.—The right cone is generated by the revolution of a right-angled triangle \( ABC \) about its side \( BC \).

Let \( BED \) be the side of a regular pyramid of \( n \) sides described about the cone, and put \( AC = r, BC = h \).

Then the volume of the pyramid (art. 36)

\[ = \frac{1}{3} n \cdot CED \cdot BC = \frac{1}{3} n \cdot AC \cdot AD \cdot BC \]

\[ = \frac{1}{3} \pi r^2 h \tan \frac{\pi}{n} = \frac{1}{3} \pi r^2 h \cdot \frac{\tan \frac{\pi}{n}}{\pi}. \]

But by increasing \( n \) indefinitely,

\[ \tan \frac{\pi}{n} = 1 \text{ ultimately}, \]

and the pyramid becomes ultimately equal to the cone.

Therefore the volume of the cone

\[ = \frac{1}{3} \pi r^2 h, \text{ or } \frac{1}{3} m^2 h; \]

where \( m^2 \) is put for the area of the base of the cone.

Hence the volume of a cone is equal to one-third the product of the area of its base multiplied by its height.

Example.—Find the volume of the cone whose height is 13, and the radius of its base 3:5.

The area of the base \( = 38:4845 \) (Ex. (i.) art. 56);

Therefore the volume \( = \frac{1}{3} \times 13 \times 38:4845 = 166:77 \).

VOL. XIV. Example.—Find the volume of the frustum of a cone, the radii of whose ends are 3 and 4 feet, and its height 5 feet.

\[ \log_2 = -971218 \quad \log_4 = -6020600 \]

\[ \log(3)^2 = -9542425 \quad \log(4)^2 = 1-2041200 \]

\[ \log \pi = -4971499 \quad \log \pi^2 = -9942998 \]

\[ \log a_1^2 = 1-4513924 \quad \log a_2^2 = 1-7012699 \]

\[ a_1^2 = 50-266 \quad a_2^2 = 28-274 \]

\[ a_1^2 = 37-699 \quad a_2^2 = 1-5763312 \]

Volume \(= 193-73\). log volume \(= 2-2872006\)

67. To find the Surface of the Frustum of a Cone.—If the slant heights (fig. art. 65) \(AB, AB_b, BC, BC_b\) be \(l_1, l_2, l_3\) respectively, it may be shown, as in last article, that

\[ l_1 = \frac{r_1}{r_1 - r_2}, \quad l_2 = \frac{r_2}{r_1 - r_2}; \]

and since the lateral surface of the frustum is equal to the difference of the lateral surfaces of the cones;

\[ \text{lateral surface} = \pi(r_1^2 - r_2^2) \quad (\text{art. 64}); \]

\[ = \pi(r_1^2 + r_2^2). \]

Again, because the surface

\[ = \frac{1}{2}(2\pi r_1 + 2\pi r_2), \]

if \(p_1, p_2\) be put for the perimeters of the ends,

the lateral surface \(= \frac{1}{2}(p_1 + p_2)\).

The whole surface is obtained by adding to the lateral surface the areas of the circles which form the ends of the frustum.

If the perpendicular height \(h\) be given, we have evidently

\[ l = \left(h + (r_1 - r_2)\right)^{\frac{1}{2}}. \]

Example.—Find the surface of the frustum of a right cone whose height is 5 feet, and the radii of its ends 3 and 4 feet.

Here \(h = 5, r_1 = 4, r_2 = 3\);

\[ l = (25 + 1)^{\frac{1}{2}} = 5-059 \]

\[ p_1 = 2 \times 4 \times 3-14159 = 25-133 \]

\[ p_2 = 2 \times 3 \times 3-14159 = 18-850 \]

\[ p_1 + p_2 = 44-983 \]

Convex surface \(= \frac{1}{2} \times 5-059 \times 43-983 = 112-13\)

The areas of ends \(= 3-14159(9 + 16) = 78-84\) (art. 56)

Whole surface \(= 190-67\)

The Cylinder.

68. To find the Volume of a Right Cylinder.—The cylinder is generated by the revolution of a rectangle \(ACCA\) about its side \(CC\).

Let \(BDDB\) be the side of an equilateral prism of \(n\) sides described about the cylinder, and put \(AC = r\). Then the base of the prism is a regular polygon of \(n\) sides whose area

\[ = nr^2 \tan \frac{\pi}{n} \quad (\text{art. 25}). \]

Therefore the volume of the prism

\[ = nr^2 h \tan \frac{\pi}{n} = nr^2 h \cdot \frac{\tan \frac{\pi}{n}}{\pi}. \]

But when \(n\) is increased indefinitely, the prism evidently coincides with the cylinder,

\[ \tan \frac{\pi}{n} = 1; \]

and

\[ \frac{\pi}{n} = 1; \]

\[ \therefore \text{the cylinder} = \pi r^2 h. \]

Or if \(m^2\) be put for the area of the base,

the cylinder \(= m^2 h\).

Hence the volume of a cylinder is equal to the area of its base multiplied by its height.

Example.—Find the volume of a cylinder whose base is 4 feet in diameter and its height 8 feet.

Here \(r = 2-25, h = 8\).

\[ \log_2 = 3-521825 \]

\[ \log(2-25)^2 = 7-043650 \]

\[ \log \pi = -4971499 \]

\[ \log 8 = -9030900 \]

Volume \(= 127-235\). log volume \(= 2-1046049\)

69. To find the Surface of a Cylinder.—The lateral surface of the prism in last article \(= n \times \text{parallelogram } BD\),

\[ = 2nr \cdot Aa = 2nr \tan \frac{\pi}{n} = 2nr \cdot \frac{\tan \frac{\pi}{n}}{\pi}. \]

But by indefinitely increasing \(n\), the surface of the prism will coincide with that of the cylinder, and ultimately

\[ \tan \frac{\pi}{n} = 1; \]

\[ \therefore \text{convex surface of cylinder} = 2\pi rh. \]

Or, putting \(p\) for the circumference of the base,

the surface \(= ph\).

Hence the convex surface of a cylinder is equal to the circumference of its base multiplied by its height.

If to the convex surface we add the areas of the ends of the cylinder, each \(= \pi r^2\) (art. 55), we obtain

the whole surface \(= 2\pi r(r + h)\).

Example.—Find the surface of a cylinder whose base is 4 feet in diameter, and its height is 8 feet.

\(2r = 4-5, h = 8\).

\[ \log_4 = -6532125 \quad \text{By art. 35}, \]

\[ \log 8 = -9030900 \quad \text{surface of ends} = 31-808, \]

\[ \log \pi = -4971499 \quad \text{convex surface} = 113-098. \]

Log convex surface \(= 2-0534524\) whole surface \(= 144-906\).

The Sphere.

70. To find the Surface of a Sphere or of a Spherical Segment or Zone.—Let the sphere be generated by the revolution of the semicircle \(ABD\) round \(AD\); and put \(AC = r, AF = x, BF = y\).

Then

\[ y^2 = 2rx - x^2, \quad \frac{dy}{dx} = \frac{r - x}{y}; \]

and the surface generated by the revolution of the arc \(AB\)

\[ = 2\int_0^x y \left(1 + \left(\frac{dy}{dx}\right)^2\right)^{\frac{1}{2}} dx = 2\int_0^x rdx = 2\pi rx. \]

Hence the convex surface of a segment whose height is \(h\)

\[ = 2\pi rh. \] Also, if \( AG = a \), \( AF = b \), \( FG = h \), the surface of the zone generated by \( BH \)

\[ = 2 \int_{b}^{a} r dx = 2 \pi (a - b) = 2 \pi rh : \]

and, if \( AD = d \), the surface of the whole sphere

\[ = 2 \int_{0}^{2r} r dx = 4 \pi r^2 = \pi d^2 . \]

71. Otherwise: let \( BE \) be a tangent to the circle. Then if the figure revolves about \( AD \), \( BH \) will describe a spheroidal, and \( BE \) a conical surface; and if \( EG \) be taken indefinitely near to \( BF \), the two surfaces will ultimately be equal.

Let \( AC = r \), and \( ACB = \theta \).

Then \( BF = r \sin \theta \), and \( BE = FG \cos \theta \).

Also the conical surface

\[ = \pi \cdot BE(BF + EG) \text{ (art. 67)} \]

\[ = 2 \pi \cdot BF \cdot BE \text{ ultimately,} \]

\[ = 2 \pi r \sin \theta \cdot FG \cos \theta \]

\[ = 2 \pi r \cdot FG . \]

Now, as this value is independent of \( \theta \), and is ultimately the surface of the elementary zone generated by \( BH \); it is evident that the surface of any segment or zone is equal to \( 2 \pi r \) multiplied by the sum of the heights of the elementary zones of which it is composed. Hence if \( AG = h \), the surface of the segment generated by \( ABH = 2 \pi rh \); or if \( FG = h \), the surface of the zone generated by \( BH = 2 \pi rh \).

Also, the surface of the sphere is obviously

\[ = 2 \pi r \times 2r = 4 \pi r^2 . \]

72. (i.) Hence the surface of a sphere is equal to four times the area of its great circle, or of a section by a plane passing through its centre.

Example.—To find the surface of a sphere whose diameter is 7.

The surface of the circle whose diameter is 7 has been found (Ex. (ii.), art. 65) to be 38-4845.

Therefore the surface of the sphere \( = 4 \times 38-4845 = 153-938 \).

(ii.) The convex surface of a segment, or zone of a sphere, is equal to the circumference of a great circle multiplied by the height of the segment or zone.

Example.—Find the convex surface of a segment or of a zone whose height is 3, the diameter of the sphere being 5. The circumference of the circle whose diameter is 5 has been found (Ex. (i.), art. 45) to be 15-70796.

Therefore the surface of the segment or of the zone

\[ = 3 \times 15-70796 = 47-12388 . \]

73. To find the Surface of a Lune, a Spherical Triangle, and a Spherical Polygon.—It will be shown in spherical trigonometry—

(i.) That the area of a lune included between two great circles of a sphere whose inclination is \( \theta \),

\[ = 2 \theta r^2 ; \quad \text{where } r \text{ is the radius of the sphere}; \]

(ii.) That the area of a spherical triangle whose angles are \( A \), \( B \), and \( C \),

\[ = r^2(A + B + C - \pi) ; \]

(iii.) That the area of a spherical polygon of \( n \) sides, the sum of whose angles is \( P \),

\[ = r^2 \left[ P - (n - 2)\pi \right] . \]

74. To find the Volume of a Sphere.—If we retain the figure and notation of art. 70, the volume of the segment generated by the revolution of the figure \( ABF \),

\[ = \int_{0}^{2r} y^2 dx = \int_{0}^{2r} (2rx - x^2) dx = \frac{\pi r^2}{3} (3r - x) . \]

Also the volume of the whole sphere

\[ = \int_{0}^{2r} y^2 dx = \frac{4}{3} \pi r^3 . \]

75. Or, if we suppose a regular polyhedron of \( n \) faces described about the sphere, by indefinitely increasing \( n \), the surface and volume of the polyhedron will be ultimately equal respectively to the surface and volume of the sphere.

Now, if \( a^2 \) be put for one of the faces of the polyhedron, its volume is equal to \( n \) pyramids, whose bases are each \( a^2 \), and their common height the radius of the sphere.

Therefore the volume of the polyhedron

\[ = n \times \frac{1}{3} a^2 r = \frac{1}{3} \pi n r^2 , \]

\[ = \frac{1}{3} \pi \times \text{surface of polyhedron}, \]

\[ = \frac{1}{3} \pi \times \text{surface of sphere, ultimately}, \]

\[ = \frac{1}{3} \pi \times 4 \pi r^2 . \]

Whence the volume of the sphere \( = \frac{4}{3} \pi r^3 \).

From this it follows that if \( d \) be the diameter of the sphere,

\[ \text{the volume} = \frac{4}{3} \pi r^3 . \]

\[ \therefore \text{Since } \frac{4}{3} \pi = 4 \cdot 1887902048 , \text{ and } \frac{1}{3} \pi = 5235987756 ; \]

to find the volume of a sphere, multiply the cube of the radius by \( 4 \cdot 1887902 \), or the cube of the diameter by \( 5235988 \).

Example.—Find the volume of a sphere whose radius is 12.

The volume \( = (12)^3 \times \frac{4}{3} \pi = 1728 \times 4 \cdot 1887902 \)

\[ = 7238-2293 . \]

76. It obviously follows from arts. 70 and 74 that in a sphere

\[ \text{radius} = \frac{1}{2 \sqrt{\pi}} \cdot \sqrt{\text{surface}} = \frac{3}{4 \pi} \cdot \sqrt{\text{volume}}; \]

\[ \text{diameter} = \frac{1}{\sqrt{\pi}} \cdot \sqrt{\text{surface}} = \frac{6}{\pi} \cdot \sqrt{\text{volume}}; \]

\[ \text{surface} = \frac{3}{\pi} \cdot \sqrt{\text{(volume)}}^3; \quad \text{volume} = \frac{1}{6 \sqrt{\pi}} \cdot \sqrt{\text{surface}}^3 . \]

77. To find the Volume of the Segment of a Sphere.—The volume of a segment whose height is \( h \)

\[ = \int_{0}^{h} y^2 dx = \int_{0}^{h} (2rx - x^2) dx = \frac{\pi h^3}{3} (3r - h) . \]

78. The same result may be obtained by supposing the parallelogram \( ABEF \), the quadrilateral \( ABGF \), and the segment \( ABCD \), to revolve about \( AB \) so as to generate a cylinder, a frustum of a cone, and a segment of a sphere.

If \( AB \) be divided into equal parts, \( Ab, bc, \ldots \), each \( = c \), and the rectangles \( be, bi, ch, cf, \ldots \), constructed, these rectangles will generate cylinders; and by indefinitely increasing the number of parts into which \( AB \) is divided, the sum of the cylinders generated by \( be, cf, \ldots \), will ultimately be equal to the segment of the sphere generated by \( ABCD \), and the sum of the cylinders generated by \( Ah, bi, \ldots \), will ultimately be equal to the frustum of the cone generated by \( ABGF \); while the cylinders generated by \( Ah, bm, \ldots \), constitute the cylinder generated by \( ABEF \).

Now \( bh^2 = Op^2 = Ob^2 + Oq^2 = lp^2 + bq^2 \);

\[ \therefore \pi \cdot bh^2 = \pi \cdot bp^2 + \pi \cdot bq^2 ; \]

or the cylinder generated by \( bm \) is equal to the sum of those generated by \( be \) and \( bi \); and because this is true of every corresponding set of cylinders, we have,

Sum of cylinders generated by \( Ah, bm, \ldots \) = sum of cylinders generated by \( be, cf, \ldots \) + sum of cylinders generated by \( Ah, bi, \ldots \).

Therefore since these sums are ultimately equal respectively to the cylinder generated by \( ABEF \), the sphere generated by \( ABCD \), and the conic frustum generated by \( ABGF \), we have

Cylinder = segment of sphere + frustum of cone. Now, if AF = r and AB = h, we have

\[ Bg = r - h; \]

therefore (art. 65) the frustum of the cone

\[ = \frac{\pi}{3} (3r^2 - 3rh + h^2). \]

Also, the cylinder (art. 68) \(= \pi r^2 h\); therefore the segment of the sphere

\[ = \pi r^2 h - \frac{\pi}{3} (3r^2 - 3rh + h^2) \]

\[ = \frac{\pi}{3} (3r - h). \]

If we put \(h = 2r\), we obtain the volume of the whole sphere \(= \frac{4}{3} \pi r^3\), as in art. 74.

Example.—Find the volume of the segment of a sphere; the radius being 12, and the height of the segment 6.

The volume \(= \frac{\pi}{3} \times 36 (36 - 6) = 314159 \times 360 \)

\[ = 11309733552. \]

79. It is evident that we may prove, in like manner, that the cylinder generated by AOHF is equal to the sum of the hemisphere and cone generated by ADHO and AOF. Whence it easily follows (arts. 63, 68), that if we have a cone and sphere inscribed in a cylinder,

the cylinder \(= \text{sphere} + \text{cone} = \text{sphere} + \frac{1}{3} \text{cylinder}; \)

\[ \therefore \text{the sphere} = \frac{1}{3} \text{cylinder}, \]

and

\[ \text{cone : sphere : cylinder} : : 1 : 2 : 3; \]

a relation also obviously true from their ascertained volumes, which are respectively \(= \frac{4}{3} \pi r^3\), \(= \frac{4}{3} \pi r^3\), and \(2 = \pi r^3\).

80. Since, from the equation to the circle, \(y^2 = 2rx - x^2\); putting \(a\) for the radius BC of the base of the segment generated by ABCD, we have

\[ a^2 = 2rh - h^2; \]

\[ \therefore r = \frac{a^2 + h^2}{2h}. \]

Whence, substituting, we obtain the segment of a sphere, whose height is \(h\), and the radius of its base \(a\),

\[ = \frac{\pi}{6} (3a^2 + h^2). \]

Example.—The height of a segment of a sphere is 5, and the radius of its base is 7.

Volume \(= \frac{\pi}{6} \times 5 (3 \times 49 + 25) \)

\[ = 523599 \times 860 = 45029. \]

81. To find the Volume of the Frustum of a Sphere—(i.) When one of the terminating planes passes through the centre of the sphere.

Let BO = \(l\), then AB or \(h = r - l\); and the frustum generated by the revolution of BCHO = hemisphere generated by AOH — spherical segment generated by ABCD,

\[ = \frac{2}{3} \pi r^3 - \frac{\pi}{3} (r - l)^3 (2r + l) = \frac{\pi}{3} (3r^2 - l^2). \]

Example.—Find the volume of the frustum of a sphere; the radius of the sphere being 12, the height of the frustum 6, and one of its terminating planes passing through the centre.

The volume \(= \frac{\pi}{3} \times 6 (3 \times 144 - 36) \)

\[ = 314159 \times 792 \]

\[ = 24888143816. \]

(ii.) When neither of the terminating planes passes through the centre.

If \(h\) be the height of the frustum, and \(a_1, a_2\) the radii of its ends,

the volume \(= \frac{\pi}{6} \left[ 3 (a_1^2 + a_2^2) + h^2 \right]. \)

SECTION IV.—THE ELLIPSE.

82. To find the Length of an Arc of an Ellipse.—If AC = \(a\), BC = \(b\), CM = \(x\), and MP = \(y\), the equation to the ellipse

\[ a^2 y^2 + b^2 x^2 = a^2 b^2; \]

whence

\[ \frac{dy}{dx} = - \frac{b^2 x}{a^2 y}. \]

The length of an arc BP

\[ = \int_{0}^{x} \left( 1 + \frac{dy}{dx} \right)^{\frac{1}{2}} dx = \int_{0}^{x} \left( \frac{a^2 - e^2 x^2}{a^2 - x^2} \right)^{\frac{1}{2}} dx, \]

where \(e = \frac{a^2 - b^2}{a^2}. \)

Putting \(x = az\),

the arc BP \(= \int_{0}^{z} \left( \frac{1 - e^2 z^2}{(1 - z^2)} \right)^{\frac{1}{2}} dz \)

\[ = \int_{0}^{z} \left[ 1 - \frac{1}{2} e^2 z^2 - \frac{1}{2 \cdot 4} e^4 z^4 - \frac{1}{2 \cdot 4 \cdot 6} e^6 z^6 + \ldots \right] dz. \]

To obtain the length of a quadrant of the ellipse, we must integrate from \(x = 0\) to \(x = a\), or from \(z = 0\) to \(z = 1\); when we obtain (see Fluxions)

the quadrant APB \(= \frac{\pi}{2} \left[ 1 - \frac{e^2}{2} - \frac{1}{2 \cdot 4} e^4 - \frac{1}{2 \cdot 4 \cdot 6} e^6 + \ldots \right] \)

a rapidly converging series when \(e\) is small.

83. We shall obtain more and more accurate values of the circumference of the ellipse according to the number of terms of the above series we employ. The following is the first approximation:

Since \((1 - \frac{1}{2} e^2)^{\frac{1}{2}} = 1 - \frac{e^2}{2} + \ldots, \)

we have the elliptic quadrant \(= \frac{\pi}{2} \left( 1 - \frac{e^2}{2} \right)^{\frac{1}{2}} \)

\[ = \frac{\pi}{2} a \left( 1 - \frac{a^2 - b^2}{2a^2} \right)^{\frac{1}{2}} = \frac{\pi}{2} \left( \frac{a^2 + b^2}{2} \right)^{\frac{1}{2}}; \]

and the circumference of the ellipse

\[ = \pi \left( \frac{(2a)^2 + (2b)^2}{2} \right)^{\frac{1}{2}} \]

nearly.

Hence the circumference of an ellipse is obtained approximately by multiplying the square root of half the sum of the squares of its axes by 3.14159.

Example.—Find the circumference of an ellipse whose axes are 3 and 4.

The circumference \(= 31416 \left( \frac{9 + 16}{2} \right)^{\frac{1}{2}} \)

\[ = 31416 \times 5 \sqrt{2} \]

\[ = 111 \text{ nearly}. \]

84. To find the Area of an Ellipse.—The area of a quadrant ACB (fig. art. 82)

\[ = \int_{0}^{a} y dx = \frac{b}{a} \int_{0}^{a} (a^2 - x^2)^{\frac{1}{2}} dx = \frac{\pi ab}{4}. \]

85. Or if the ellipse ABA' and the circle ADA' be divided by lines cd, c'd', &c., indefinitely near to each other, and parallel to CD, the polygons ACDA', ACBB', will be ultimately equal to the circular and elliptic quadrants, in which they are inscribed; and if AC = a, BC = b,

\[ \frac{bc}{ed} : \frac{be}{cd} : \frac{cd}{b} : a. \] (Anal. Geom., art. 92, p. 553.)

Therefore the trapezoid be'c' is to dec'd in the same ratio (art. 14); and hence also the polygon ACBb' is to the polygon ACDa' in the same ratio of a to b.

Therefore, since the polygons are ultimately equal to the elliptic and circular quadrants, and the latter \(= \frac{1}{4}\pi a^2\), we have

\[ a : b :: \frac{1}{4}\pi a^2 : \text{elliptic quadrant ABC}. \]

Whence the elliptic quadrant \(= \frac{1}{4}\pi ab\);

and the ellipse \(= \pi ab\).

As this is the same as \( \frac{1}{4}\pi \times 2a \times 2b \), the area of an ellipse is equal to the product of the major and minor axes multiplied by 78539.

Example.—Find the area of an ellipse whose axes are 3 and 4.

Area \(= 3 \times 4 \times 78539 = 80110613.\)

Spheroids.

86. To find the Surface of a Prolate Spheroid.—The prolate spheroid is generated by the revolution of an ellipse (fig. art. 82) about its major axis AA', and the equation to the generating curve is therefore

\[ a^2 y^2 + b^2 x^2 = a^2 b^2. \]

Whence \( \frac{dy}{dx} = - \frac{b^2 x}{a^2 y} \); and if \( e^2 = \frac{a^2 - b^2}{a^2} \),

\[ y \left(1 + \frac{dy}{dx}\right)^{\frac{1}{2}} = \frac{b}{a} \left(a^2 - e^2 x^2\right)^{\frac{1}{2}}. \]

Therefore, putting CM = h, the convex surface of the frustum of the spheroid generated by the figure BCMP

\[ = 2\int_0^h y \left(1 + \frac{dy}{dx}\right)^{\frac{1}{2}} dx = \frac{2\pi a^2}{b} \int_0^h \left(a^2 - e^2 x^2\right)^{\frac{1}{2}} dx \]

\[ = \frac{\pi ab}{e} \left\{ \sin^{-1} \frac{eh}{a} + \frac{eh}{a} \left(1 - \frac{e^2 h^2}{a^2}\right)^{\frac{1}{2}} \right\}. \]

87. To obtain the surface of the whole spheroid, we must integrate from \(x = -a\) to \(x = a\), when we obtain

surface of spheroid \(= \frac{2\pi ab}{e} \left\{ \sin^{-1} e + e \left(1 - e^2\right)^{\frac{1}{2}} \right\} \)

\[ = \frac{2\pi a^2}{e} \left[1 - e^2 + \left(1 - e^2\right)^{\frac{1}{2}} \cdot \sin^{-1} e\right]. \]

Example.—Find the surface of a prolate spheroid whose axes are 5 and 3.

Here \(a = \frac{5}{2}, b = \frac{3}{2}; \therefore e = \frac{4}{5}. \)

Hence the surface \(= 2\pi \left(\frac{5}{2}\right)^2 \left\{ \frac{9}{25} + \frac{3}{5} \cdot \frac{5}{4} \cdot \sin^{-1} \frac{4}{5} \right\} \)

\[ = 12\cdot5 \times \left\{ \frac{36}{25} + \frac{75}{4} \cdot \sin^{-1} \frac{4}{5} \right\}. \]

Now log 0·8 + 10 = 9·90309 = L sin 53° 7' 49"

\[ = L \sin 53° 130 \text{ nearly}; \]

therefore (art. 51)

\[ \sin^{-1} 0·8 = \frac{\pi}{180} \times 53° 130 = 69547, \]

and the surface of the spheroid

\[ = 3\cdot14159 \times 12\cdot5 \times \left(36 + 69547\right) = 41\cdot448. \]

88. To find the Surface of an Oblate Spheroid.—The oblate spheroid is generated by the revolution of an ellipse A'B'C' about its minor axis BB'. Hence if CM = x, and MP = y, the equation to the generating curve will be

\[ a^2 x^2 + b^2 y^2 = a^2 b^2; \]

\[ \frac{dy}{dx} = - \frac{a^2 x}{b^2 y} \text{ and } y \left(1 + \frac{dy}{dx}\right)^{\frac{1}{2}} = \frac{a^2 e}{b^2} \left(c^2 + x^2\right)^{\frac{1}{2}}; \]

where \( e^2 = \frac{a^2 - b^2}{a^2} \), and \( c = \frac{b^2}{ae} = \frac{a}{e} \left(1 - e^2\right). \)

If CM = h, the convex surface generated by the revolution of the figure ACMP

\[ = 2\int_0^h y \left(1 + \frac{dy}{dx}\right)^{\frac{1}{2}} dx = \frac{2\pi a^2}{b} \int_0^h \left(c^2 + x^2\right)^{\frac{1}{2}} dx \]

\[ = \frac{\pi a^2}{e} \left\{ h \left(e^2 + h^2\right)^{\frac{1}{2}} + e^2 \log \frac{h + \left(e^2 + h^2\right)^{\frac{1}{2}}}{e} \right\}. \]

89. By integrating from \(x = -b\) to \(x = b\), we obtain the surface of the spheroid

\[ = 2\pi a^2 \left\{ 1 + \frac{1 - e^2}{2e} \log \left(1 + e\right) \left(1 - e\right) \right\}. \]

Example.—Find the surface of an oblate spheroid whose axes are 5 and 3.

Here \(a = \frac{5}{2}, b = \frac{3}{2}; \therefore e = \frac{4}{5}. \)

Surface \(= 2\pi \left(\frac{5}{2}\right)^2 \left\{ 1 + \frac{5}{2} \cdot \frac{9}{25} \cdot \log \frac{1 + \frac{4}{5}}{1 - \frac{4}{5}} \right\} \)

\[ = 12\cdot5 \times \left(1 + \frac{9}{40} \log 9\right). \]

But \( \frac{9}{40} \log_e 9 = \frac{9}{40} \times 2\cdot1972246 = 4943755; \)

\[ \therefore \text{surface} = 12\cdot5 \times 3\cdot14159 \times 1\cdot4943755 = 58\cdot685. \]

90. To find the Volume of a Spheroid.—(i.) For the prolate spheroid the equation to generating curve (art. 86) is \(a^2 y^2 + b^2 x^2 = a^2 b^2. \)

Whence the volume \(= \int_{-a}^{+a} y^2 dx \)

\[ = \frac{\pi a^2}{b} \int_{-a}^{+a} \left(a^2 - x^2\right) dx = \frac{4}{3}\pi a^2 b. \]

91. Or if we conceive the figure of article 84 to revolve about AA', so as to generate a spheroid and a sphere, we may conceive these solids to be ultimately equal to two series of elementary cylinders described by the rectangles, whose bases are BC, be, &c., and CD, cd, &c., and whose common altitude is Ce = ec = &c.

Then, because BC : CD :: (be)^2 : (cd)^2 :: b^2 : a^2;

\[ \therefore \text{volume of spheroid} = \frac{4}{3}\pi a^2 b. \]

Whence the corresponding pairs of elementary cylinders which constitute the sphere and spheroid are in the constant ratio of \(a^2\) to \(b^2\); and since the sphere \(= \frac{4}{3}\pi a^2\);

\[ \therefore a^2 : b^2 :: \frac{4}{3}\pi a^2 : \text{spheroid}. \]

Hence the spheroid \(= \frac{4}{3}\pi ab^2. \)

92. (ii.) For the oblate spheroid the equation to the generating curve is \(a^2 x^2 + b^2 y^2 = a^2 b^2\) (art. 88).

Whence the volume \(= \int_{-b}^{+b} y^2 dx \)

\[ = \frac{\pi a^2}{b} \int_{-b}^{+b} \left(a^2 - x^2\right) dx = \frac{4}{3}\pi a^2 b. \]

* To obtain the logarithm of any number to the base e, multiply its common logarithm to the base 10 by 2·3025850929.

Thus log_e 9 = 2·3025850929 × 2·3025850929 = 2·1972246. The same result may also be obtained by the method of article 91.

93. On comparing the expressions for the volumes of oblate and prolate spheroids, it appears that either volume is equal to two-thirds the area of the circle generated by the revolving axis of the ellipse, multiplied by the length of the axis about which it revolves; or the spheroid is two-thirds of the circumscribing cylinder. Also, for both species of spheroid, the volume is equal to the continued product of the fixed axis, multiplied by the square of the revolving axis, and by \( \frac{1}{2} \pi \), or \( \frac{52359878}{6} \).

Examples.—Find the volumes of an oblate and prolate spheroid whose axes are 20 and 12.

The volume of the oblate spheroid

\[ = \frac{4}{3} \pi \times 100 \times 6 = 800 \pi = 2513274. \]

The volume of the prolate spheroid

\[ = \frac{4}{3} \pi \times 36 \times 10 = 480 \pi = 1057962. \]

94. To find the Volume of a Segment of a Spheroid.

(i.) The prolate spheroid.—The segment is generated by the revolution of AMP about AM (fig. art. 82).

If A be the origin, and AA' the axis of x, the equation to the ellipse is

\[ y^2 = \frac{b^2}{a^2} (2ax - x^2); \]

and if AM = h, the volume of the segment

\[ = \frac{\pi}{3} \int_0^h (2ax - x^2) dx = \frac{\pi}{3} \cdot \frac{b^2}{a^2} (3a - h). \]

Example.—Find the volume of the segment of a prolate spheroid, where the axes of the generating ellipse are 20 and 12, and the height of the segment 8.

Here \( a = 10, b = 6, h = 8. \)

The volume

\[ = \frac{\pi}{3} \cdot \frac{36 \times 64}{100} (30 - 8) = 314159 \times \frac{12 \times 64 \times 22}{100} = 5308034. \]

95. (ii.) The oblate spheroid.—The segment is generated by the revolution of BMP (fig. art. 83) about BM; where, if B be the origin and BB' the axis of x, the equation to the ellipse is

\[ y^2 = \frac{a^2}{b^2} (2bx - x^2); \]

and if BM = h, the volume of the segment

\[ = \frac{\pi}{3} \int_0^h (2bx - x^2) dx = \frac{\pi}{3} \cdot \frac{a^2}{b^2} (3b - h). \]

Example.—Find the volume of a segment of an oblate spheroid, the axes being 20 and 12, and the height of the segment 2.

\[ a = 10, b = 6, h = 2; \]

and the volume

\[ = \frac{\pi}{3} \cdot \frac{100 \times 4}{36} (18 - 2) = 1861658. \]

96. To find the Volume of the Frustum of a Spheroid when one of the Terminating Planes passes through the Centre.

(i.) The prolate spheroid.—The frustum is generated by the revolution of the figure BCMP (fig. art. 82) about CM; and if CM = h, the volume of the frustum

\[ = \frac{\pi}{3} \int_0^h (a^2 - x^2) dx = \frac{\pi}{3} \cdot \frac{b^2}{a^2} (3a^2 - h^2). \]

If \( b, b' \) be the radii BC, PM of the terminating planes of the frustum, since from the equation to the ellipse

\[ b'h^2 + a'b'^2 = a^2b^2; \quad \therefore \quad h^2 = \frac{a^2(b^2 - b'^2)}{b^2}. \]

Whence, by substitution, the volume of the frustum

\[ = \frac{\pi}{3} \cdot \frac{b(2b^2 + b'^2)}{a^2}. \]

Example.—Find the volume of the frustum of a prolate spheroid whose axes are 20 and 12, the height of the frustum being 2, and one of its ends passing through the centre of the spheroid.

\[ a = 10, b = 6, h = 2. \]

The volume

\[ = \frac{\pi}{3} \cdot \frac{36 \times 2}{100} (300 - 4) = 2231787. \]

97. (ii.) The oblate spheroid.—Adopting the figure and notation of article 88, the frustum is generated by the revolution of ACMP about CM. The equation is

\[ a^2x^2 + b^2y^2 = a^2b^2; \]

and if CM = h, the volume

\[ = \frac{\pi}{3} \int_0^h (b^2 - x^2) dx = \frac{\pi}{3} \cdot \frac{a^2}{b^2} (3b^2 - h^2). \]

Also, if \( a, a' \) be the radii of the terminating planes, since from the equation to the ellipse

\[ a^2h^2 + a'^2b'^2 = a^2b^2, \]

if we substitute for \( h^2 \) we obtain the volume

\[ = \frac{\pi}{3} \cdot \frac{b(2a^2 + a'^2)}{a^2}. \]

Example.—Find the volume of the frustum of an oblate spheroid whose axes are 20 and 12, the height of the frustum being 4, and one of its ends passing through the centre of the spheroid.

\[ a = 10, b = 6, c = 4. \]

The volume

\[ = \frac{\pi}{3} \cdot \frac{100 \times 4}{36} (108 - 16) = 1070469. \]

It appears also from the last two articles that the volume of a frustum, of either a prolate or oblate spheroid, one of whose ends passes through the centre of the generating ellipse, is obtained by adding the area of the smaller end to twice that of the greater, and multiplying the sum by one-third of the altitude of the frustum.

SECTION V.—THE HYPERBOLA.

98. To find the Area of an Hyperbola.

(i.) The area of the segment AMP.

Let C be the centre, and CM the axis of x. The equation is

\[ a^2y^2 - b^2x^2 = -a^2b^2; \]

and if CM = h, PM = k, the area AMP

\[ = \int_a^b ydx = \frac{b}{a} \int_a^b (x^2 - a^2)^{1/2} \]

\[ = \frac{b}{a} \left[ \frac{h}{2} (h^2 - a^2)^{1/2} - \frac{a^2}{2} \log \frac{h + (h^2 - a^2)^{1/2}}{a} \right] \]

\[ = \frac{1}{2} hk - \frac{ab}{2} \log \left( \frac{h}{a} + \frac{k}{b} \right); \]

where \( b = \frac{ak}{(h^2 - a^2)^{1/2}}. \) Example.—Find the area of a hyperbolic segment whose base is 48 and altitude 20; the transverse axis of the curve being 60.

\[ a = 30, \quad k = 24, \quad x = 30 + 20 = 50; \] also \( b = \frac{30 \times 24}{(2500 - 900)^2} = \frac{720}{40} = 18. \]

The area AMP

\[ = \frac{1}{2} \times 50 \times 24 - \frac{30 \times 18}{2} \log_2 \left( \frac{50}{30} + \frac{24}{18} \right) \] \[ = 600 - 270 \log_2 \left( \frac{50}{30} + \frac{24}{18} \right) = 600 - 270 \times 0.986123 \text{ (note, art. 89)} \] \[ = 3033756. \]

Hence the whole segment, which is double AMP,

\[ = 6067513. \]

99. (ii.) The area of the sector ACP.

The area ACP = CMP - AMP = \( \frac{1}{2} h k - AMP \)

\[ = \frac{ab}{2} \log \left( \frac{h}{a} + \frac{k}{b} \right). \]

100. To find the Area bounded by a Rectangular Hyperbola and its Asymptotes.

The equation to the hyperbola referred to its asymptotes is

\[ xy = \frac{a^2}{2}. \]

Let CM = \( x_1 \), CN = \( x_2 \), PM = \( y_1 \), QN = \( y_2 \).

The area PMNQ = \( \int_{x_1}^{x_2} y dx \)

\[ = \frac{a^2}{2} \int_{x_1}^{x_2} \frac{dx}{x} = \frac{a^2}{2} \log \frac{x_2}{x_1}. \]

From the equation:

\[ \frac{x_2}{x_1} = \frac{a^2}{2}, \quad \text{or triangle CMP = triangle CNQ.} \]

Whence sector PCQ = area PMNQ.

101. If the ordinates PM = \( h_1 \), QN = \( h_2 \), and MN = \( h \), be given, putting \( x_1, x_2 \) for CM, CN, as before,

\[ x_2 h_2 = \frac{a^2}{2} = x_1 h_1 \quad \text{and} \quad x_1 - x_2 = h. \]

Whence \( \frac{a^2}{2} = \frac{h_1 h_2}{k_1 - k_2} \),

and area PMNQ = \( \frac{h_1 h_2}{k_1 - k_2} \log \frac{k_2}{k_1}. \)

Example.—Given the ordinates PM = 20, QN = 45, and MN = 50, to find the area.

Here \( h_1 = 20, \quad h_2 = 45, \quad \text{and} \quad h = 50. \)

The area \( = \frac{50 \times 20 \times 45}{45 - 20} \log \frac{45}{20} \) \[ = 1800 \log \frac{9}{4} = 145967. \]

102. To find the Volume of an Hyperboloid.—The solid is generated by the revolution of the hyperbolic segment AMP about AM (fig. art. 98).

If, therefore, we put AM = \( h \),

the volume \( = \int_{a}^{a+h} y^2 dx = \frac{b^2}{a} \int_{a}^{a+h} (x^2 - a^2) dx \) \[ = \frac{b^2 h^3}{3a^2} (3a + h). \]

103. If PM = \( h \), from the equation to the hyperbola (art. 98),

\[ a^2 - b^2 (a + h)^2 = -a^2 h^2. \]

Whence \( \frac{b^2}{a^2} = \frac{h^2}{2ah + h^2}. \)

Therefore substituting, we obtain the volume of the hyperboloid,

\[ = \frac{\pi b^2}{3} \cdot \frac{3a + h}{2a + h}. \]

Example.—Find the volume of an hyperboloid, the radius of whose base is 24, its height 20, and the transverse axis of the generating curve 60.

Here \( a = 30, \quad k = 24, \quad h = 20. \)

The volume \( = \frac{\pi \times 20 \times (24)^2}{3} \cdot \frac{90 + 20}{60 + 20} \) \[ = 5280\pi = 163876092. \]

104. The expression for the volume may be put into the form

\[ \frac{\pi b^2}{2} \cdot \frac{2a + \frac{3}{2} h}{2a + h}, \]

whence the following rule:—As the sum of the transverse axis of the generating hyperbola, and the height of the solid, is to the sum of the transverse axis and \( \frac{3}{2} \) of the height, so is half the cylinder of the same base and altitude to the volume of the hyperboloid.

105. To find the Volume of the Solid generated by the Revolution of a Rectangular Hyperbola about its Asymptote.

The solid is generated by the revolution of PMNQ (fig. art. 100) about MN.

If CM = \( x_1 \), CN = \( x_2 \) and MN = \( h \),

the volume \( = \pi \int_{x_1}^{x_2} y^2 dx = \frac{\pi a^2}{4} \int_{x_1}^{x_2} \frac{dx}{x} \) \[ = \frac{\pi a^2}{4} \left( \frac{1}{x_2} - \frac{1}{x_1} \right) = \frac{\pi a^2}{4} \cdot \frac{a^2}{x_1 x_2}. \]

Now if \( r_1, r_2 \) be the radii PM, QN of the ends of the solid,

\[ r_1 x_1 = r_2 x_2 = \frac{a^2}{2}; \]

therefore \( x_1 x_2 = \frac{a^4}{4r_1 r_2}. \)

Whence the volume \( = \pi r_1 r_2 h; \)

or the volume is equal to the continued product of the radii of its ends, its height, and \( 3:14159. \)

Example.—Find the volume of the hyperbolic solid generated by PMNQ, where PM = 20, QN = 45, and MN = 50.

The volume \( = 20 \times 45 \times 50 \times \pi \) \[ = 141371669. \]

SECTION VI.—THE PARABOLA.

106. To find the Length of an Arc of a Parabola.—The equation to the parabola being \( y^2 = 4mx \),

the length of the arc AB, where AD = \( h \),

\[ = \int_{0}^{h} \left( 1 + \frac{dy}{dx} \right)^{\frac{1}{2}} dx \] \[ = \int_{0}^{h} \left( 1 + \frac{m}{x} \right)^{\frac{1}{2}} dx \] \[ = (mh + h^2)^{\frac{1}{2}} + \frac{m}{2} \log \left( \frac{m + 2h + (mh + h^2)^{\frac{1}{2}}}{m} \right). \]

If BD = \( b, \quad b^2 = 4mh; \quad \therefore m = \frac{b^2}{4h}. \)

Whence the arc BAC

\[ = (b^2 + 4h^2)^{\frac{1}{2}} + \frac{b^2}{4h} \log \left( \frac{b^2 + 8(h^2 + (b^2 + 4h^2)^{\frac{1}{2}})}{b^2} \right). \] Example.—If AD = 4, and BC = 12; then b = 6, and h = 4.

The arc BAC = \( \frac{9}{4} \log_{\frac{9}{4}} \)

\[ = 10 + \frac{9}{4} \times 1.9236493 = 14.3282. \]

107. To find the Area of a Parabola.—In the figure of last article the area ABD

\[ = \int_0^x ydx = 2m^4 \int_0^x x^4 dx \]

\[ = \frac{4}{3} m^4 x^{\frac{5}{3}} = \frac{4}{3} xy. \]

Or if AD = h, and BD = b,

the area ABD = \( \frac{3}{2} bh. \)

Whence the area ABC = \( \frac{3}{2} AD \cdot BC. \)

108. Or if PT be a tangent to the parabola AP, AB a diameter, and P' a point in the curve indefinitely near to P; then P' will ultimately be on the line PT, and the parallelograms PC, P'B will be equal.

But because AB = AT (Conic Sections, vol. vii., p. 235, prop. ix.), PD = \( \frac{1}{2} PC, \) therefore the parallelogram P'B is double the parallelogram PD; and, if we take a succession of points, such as PP', indefinitely near to each other, the areas APB, APD may be conceived to be made up of parallelograms, every parallelogram in the area APB being double the corresponding parallelogram in APD. Hence the area APB is double APD, or APB is \( \frac{3}{2} \) the parallelogram ABPD.

From this it follows that PA = \( \frac{3}{2} PD \); or the area of a parabola is two-thirds of the circumscribed parallelogram.

Example.—The base of a parabolic segment is 10 and its height 4.

The area = \( \frac{3}{2} \times 10 \times 4 = 26. \)

109. To find the Volume of a Paraboloid.—The parabola ABD (fig. art. 106) revolving about AD = h, generates the paraboloid whose volume

\[ = \int_0^h y^2 dx = 4 \pi m \int_0^h x dx = 2 \pi m h^2. \]

Or if BD = b, since \( 4mh = b^2, \)

the volume = \( \frac{3}{2} \pi b^2 h. \)

The cylinder whose base is the circle described by the revolution of BD, and whose altitude is AD, = \( \pi b^2 h. \)

Whence a paraboloid is equal to one-half its circumscribing cylinder.

110. The same result may also be obtained thus:—If AGC and BHD be two equal parabolas, whose vertices are A, B, and if the figure revolve round AB, ABCD will describe a cylinder, and the parabolas will describe paraboloids. If we also suppose that the solids are cut by a number of planes indefinitely near to each other and parallel to the base of the cylinder, we may conceive the solids to be made up of elementary cylinders constituted between the contiguous planes.

Now, if \( y^2 = px \) be the equation to the parabolas,

\[ EG^2 = p \cdot AE, \text{ and } EH^2 = p \cdot EB; \]

Also if we put \( h \) for the height of the cylinders whose bases are in the plane described by EF, their volumes will be \( \pi h \cdot EG^2, \pi h \cdot EH^2, \pi h \cdot EF^2; \) and since \( \pi h \cdot EG^2 + \pi h \cdot EH^2 = \pi h \cdot EF^2, \) and a similar relation exists for every corresponding set of cylinders, it follows that the elementary cylinders which constitute the paraboloids are ultimately equal to those which constitute the whole cylinder.

Hence the two paraboloids are together equal to the whole cylinder; and since the paraboloids are equal to each other, each is equal to half the cylinder.

Example.—The diameter of the base of a paraboloid is 10, and its height 4.

Here \( h = 4, \) and \( b = 5. \)

The volume = \( \frac{3}{2} \times 25 \times 4 = 157.0795. \)

111. To find the Volume of the Frustum of a Paraboloid.—The frustum is generated by the revolution of BDMP (fig. art. 106) about DM, where if AD = \( h_1, \) AM = \( h_2, \) DM = \( h_3, \) BD = \( b_1, \) MP = \( b_2, \)

the volume = \( \pi \int_{h_1}^{h_2} y^2 dx = 4 \pi m \int_{h_1}^{h_2} x dx \)

\[ = 2 \pi m(h_1^2 - h_2^2) = 2 \pi m(h_1 + h_2)h_3 \]

\[ = \frac{3}{2} \pi(4mh_1 + 4mh_2)b_3 = \frac{3}{2} \pi(b_1^2 + b_2^2)b_3. \]

Or otherwise, the volume of the frustum is equal to the difference of the paraboloids generated by ABD, AMP

\[ = \frac{3}{2} \pi(b_1^2h_1 - b_2^2h_2) = 2 \pi m(h_1^2 - h_2^2) \]

\[ = \frac{3}{2} \pi(b_1^2 + b_2^2)b_3. \]

Now since \( \pi b_1^2, \pi b_2^2 \) are the areas of the circles generated by the revolution of BD and MP, it follows that the volume of the frustum of a paraboloid is equal to half the sum of the areas of its ends multiplied by its height.

Example.—The radii of the ends of the frustum of a paraboloid are 3 and 6, and its height is 3.

The volume = \( \frac{3}{2} \times (36 + 9) \times 3 \)

\[ = 212.058. \]

112. To find the Volume of a Parabolic Spindle.—The parabolic spindle is generated by the revolution of the parabola AEB about AB, a line at right angles to CE, the axis of the curve.

If CM = \( x, \) and MP = \( y, \) the equation to the curve is

\[ y = a - \frac{x^2}{p}, \text{ where } EC = a. \]

Wherefore, if AC = \( b, \) we have \( b^2 = op, \) and the volume generated by the revolution of ACE

\[ = \int_0^b y^2 dx = \pi \int_0^b \left( a^2 - \frac{2ax^2}{p} + \frac{x^4}{p^2} \right) dx \]

\[ = \frac{8}{15} \pi a^2 b. \]

Hence the volume generated by the revolution of AEB is equal to \( \frac{8}{15} \) of a circumscribed cylinder.

Example.—The length of a parabolic spindle is 12, and its middle diameter is 8.

Here \( a = 4, \) \( 2b = 12. \)

The volume = \( \frac{8}{15} \times 16 \times 12 = 321.699. \)

113. To find the Volume of the Frustum of a Parabolic Spindle.—The frustum is generated by the revolution of CEPM about CM. If PM = \(a\), and CM = \(h\), from the equation to the parabola

\[a' = a - \frac{h^2}{p}\]

and the volume

\[= \int_{-h}^{h} \left( a' + \frac{2ax^2}{p} + x^3 \right) dx\]

\[= \frac{15}{15} (8a^3 + 4aa' + 3a'^2).\]

**Example.**—Find the volume of the frustum of a parabolic spindle, the radius of the end passing through the centre of the entire spindle being 16, the radius of the other end 12, and the length 20.

The volume

\[= \frac{20\pi}{15} (8 \times 256 + 4 \times 16 \times 12 + 3 \times 144)\]

\[= 13605.19.\]

**SECTION VII.—ON THE DETERMINATION OF THE VOLUMES AND SURFACES OF SOLIDS OF REVOLUTION, GENERATED BY PLANE FIGURES WHOSE AREAS AND CENTRES OF GRAVITY ARE KNOWN.**

**114.** Let a solid be generated by any plane figure CD, revolving round an axis AB, in the same plane, but which is supposed not to cut CD. The area CD, and the distance AG of the centre of gravity of CD from the axis AB, being known, it is required to find the volume and surface of the solid.

(i.) To find the volume of the solid. — Let the plane of the figure CD, in its initial position, be the plane of \((x, y)\); let AB be the axis of \(x\), and let \(\theta\) be the angle \(GAG\) through which CD revolves. Then an elementary area \(dxdy\) of the figure CD, in revolving through an angle \(d\theta\), will generate an elementary solid whose volume is \(ydxdy\). Therefore the whole solid

\[= \int_0^\theta \int_0^{\infty} ydxdy = \theta \int_0^{\infty} ydxdy.\]

The limits of \(x\) and \(y\) will depend upon the nature of the revolving curve. But if \(g\) be the distance AG of the centre of gravity from AB, we shall have, from the nature of the centre of gravity,

\[g = \int_0^\theta \int_0^{\infty} ydxdy;\]

the limits of \(x\) and \(y\) being the same as before. Therefore the whole solid

\[= \theta g \int_0^{\infty} dxdy = \text{arc } Gg \times \text{area } CD.\]

A more elementary proof may be obtained as follows:—

If \(A_i\) the area CD, be made up of the elements \(a_1, a_2, \ldots, a_n\), whose respective distances from AB are \(y_1, y_2, \ldots, y_n\); then the solid generated by the element \(a_i\) at E

\[= a_i \times \text{arc } Ee = a_i \times BE \times \text{angle } EB = a_i y_i \theta;\]

and the whole solid

\[= \theta (a_1 y_1 + a_2 y_2 + \ldots + a_n y_n) = \theta g A,\]

\[= \text{arc } Gg \times \text{area } CD.\]

Hence if any plane figure revolve about an axis which lies in the same plane with it, but does not cut it, the volume

\[= \text{perimeter of } CD \times \text{circle whose radius is } Ag.\]

**115.** (ii.) To find the surface of the solid.—The surface generated by an element \(ds\) of the perimeter of the figure CD, revolving through an angle \(d\theta\), is \(yd\theta\); therefore the whole surface

\[= \int_0^\theta \int_0^{\infty} yd\theta ds = \theta \int_0^{\infty} yd\theta ds.\]

The limits depend upon the nature of the figure CD; but if \(g = AG\), the distance of the centre of gravity of the perimeter of CD, from AB,

\[g = \int_0^{\infty} yd\theta ds.\]

Therefore the surface of the solid generated by CD

\[= \theta g \int_0^{\infty} d\theta ds = \text{arc } Gg \times \text{perimeter of } CD.\]

The same result may be obtained, if in the second demonstration of the last proposition we substitute the elements of the perimeter of CD for those of the surface CD.

Hence if any plane figure revolve about an axis in the same plane with it, but which does not cut it, the surface of the solid which is generated is equal to a rectangle whose base is the perimeter of the revolving figure, and whose altitude is the length of the path described by the centre of gravity of the perimeter.

If the figure CD make a complete revolution, the surface of the solid

\[= \text{perimeter of } CD \times \text{circle whose radius is } Ag.\]

**116.** To find the Volume and Surface of a Circular Ring. —Let \(a\) be the distance of the centre of the generating circle from the axis round which it revolves to generate the ring, and \(r\) the radius of the generating circle.

Then the path described by the centre of gravity, either of the area or perimeter of the generating circle \(= 2\pi a\).

Also area of circle \(= \pi r^2\),

Perimeter of circle \(= 2\pi r\).

Hence volume of ring \(= 2\pi ar^3\).

Surface of ring \(= 4\pi ar\).

**Example.**—Find the volume and surface of a ring 4 inches thick, and whose internal diameter is 6 inches.

Here the radius of the generating circle is 2 inches, and the radius of the circle described by the centre of gravity is 5 inches.

Volume of ring \(= 2\pi \times 5 \times 2^2\)

\[= 40 \times 9.8696044 = 394.784 \text{ cubic inches}.\]

Surface of ring \(= 4\pi \times 2 \times 5\)

\[= 40\pi = 394.784 \text{ square inches}.\]

**SECTION VIII.—ON THE APPROXIMATE DETERMINATION OF THE AREAS OF CURVES, AND THE VOLUMES OF SOLIDS, BY MEANS OF EQUIDISTANT ORDINATES, OR EQUIDISTANT SECTIONS.**

**117.** Let AEB be any curve, and let the ordinates Ce, Dd, &c., be drawn perpendicular to AB, dividing it into equal parts. Having given the lengths of the ordinates and their common distance, it is required to determine, either accurately or approximately, the areas of the whole curve AEB, or of the portion CEHHE; and also the volumes of the solids generated by the revolution of these figures about the line AB.

118. To find the Area of the Curve.—(i.) As the ordinates Cc, Dd, &c., are supposed to be near to each other, if straight lines be drawn between AC, CD, &c., these lines will nearly coincide with the curve, and its area will be nearly equal to the sum of the two triangles ACE, BHA, and the trapezoids CDeE, DEEd, &c.

Hence, putting

\[ Ce = a_1, \quad Dd = a_2, \quad Hh = a_n, \quad \text{and} \quad Ac = cd = &c. = h, \]

we may assume (arts. 10, 14) that the surface AEB is nearly equal to

\[ \frac{1}{2}a_1h + \frac{1}{2}(a_1 + a_2)h + \frac{1}{2}(a_2 + a_3)h + \ldots + \frac{1}{2}a_nh; \]

or area \(= h(a_1 + a_2 + a_3 + \ldots + a_n)\) nearly.

Similarly it may be shown that the area of the figure CEHHE

\[ = \frac{1}{2}h\{a_1 + a_2 + 2(a_2 + a_3 + \ldots + a_{n-1})\} \text{ nearly}. \]

Therefore the area of any figure contained by a straight line and a curve is equal to the sum of the equidistant ordinates multiplied by their common distance; or if the figure be contained by a straight line, two perpendiculars at its ends, and a curve, its area is equal to the two perpendiculars added to twice the sum of the intermediate ordinates, and multiplied by half their common distance.

Example 1.—In a figure AEB, the ordinates are 5, 7, 9, 13, 8, 6, and 4, and their common distance 3.

The area \(= 3(5 + 7 + 9 + 13 + 8 + 6 + 4) = 156\) nearly.

Example 2.—In a figure such as CEHHE, where the ordinates taken in order are \(a_1, a_2, \ldots, a_n\); the common distance of the ordinates \(= 1\), and their values are as follows:

\[ \begin{align*} a_1 &= 109087121 \\ a_2 &= 113137085 \\ a_3 &= 116189500 \\ a_4 &= 118321596 \\ a_5 &= 119582607 \\ \end{align*} \]

Sum \(= 576317909\)

area \(= 688279433\)

119. (ii.) A very close approximation to the areas of any curvilinear figures, such as AEB or CEH (fig. art. 117), may generally be obtained by the following method:

Let \(eh\) (art. 117) be divided into any even number of equal parts; and let ABDC (art. 119) represent a portion of the figure CEH (art. 117), such as CeeE terminated by two odd ordinates Ce, Ee; then a curve, whose equation is

\[ y = m + nx + px^2; \]

may be supposed to pass through the points A, Ee, C (art. 119); and since these points are near each other, the curve will coincide either accurately or very nearly with the curve AEC.

Let F be the origin, and put \(AB = a_1, EF = a_2, CD = a_n,\) and \(BF = FD = h.\)

Then when \(x = 0, y = a_1;\)

\[ x = h, \quad y = a_2; \]

\[ x = -h, \quad y = a_n; \]

and substituting in the equation, we obtain

\[ a_2 = m \] \[ a_1 = a_2 - nh + ph^2; \] \[ a_n = a_2 + nh + ph^2. \]

Whence \(n = \frac{a_2 - a_1}{2h}; \quad p = \frac{a_2 - 2a_1 + a_n}{2h}. \]

We have then the area AECDB

\[ = \int_{-h}^{h} ydx = \int_{-h}^{h} (m + nx + px^2) dx \]

\[ = 2 \left( mh + \frac{nh^3}{3} \right) = \frac{1}{2}h(a_1 + 4a_2 + a_n). \]

This result is usually obtained thus:

The area AECDB is the sum or difference of the trapezoid ABDC and the parabolic segment AEC.

Now, \(ABDC = \frac{1}{2}(AB + CD) BD = (a_1 + a_n)h\) (art. 14);

and \(AEC = \frac{1}{2}AGHC\) (art. 108).

\[ = \frac{1}{2}BD \cdot EI = \frac{1}{2}BD (EF \times IF) \]

\[ = \frac{1}{2} BD \left( EF \times \frac{AB + CD}{2} \right) = \frac{1}{2}h \left( a_1 + \frac{a_1 + a_n}{2} \right) \]

Therefore the area AECDB

\[ = h(a_1 + a_n) = \frac{1}{2}h \left( a_1 + \frac{a_1 + a_n}{2} \right) \]

\[ = \frac{1}{2}h(a_1 + 4a_2 + a_n). \]

120. If now we put (fig. art. 117) the ordinates \(Ce = a_1, Dd = a_2, Ee = a_3, \ldots, Hh = a_n,\)

we have \(CeeE = \frac{1}{2}h(a_1 + 4a_2 + a_n);\)

\(EegG = \frac{1}{2}h(a_1 + 4a_2 + a_n), \ldots \)

Therefore the figure CeH

\[ = \frac{1}{2}\{a_1 + a_n + 2(a_2 + a_3 + \ldots + a_{n-1}) + 4(a_1 + a_2 + \ldots + a_{n-1})\}. \]

121. The area ARGD may be derived from CeH by putting \(a_1 = a_n = 0.\) Substituting \(a_1, \ldots, a_n\) for the remaining ordinates \(a_2, \ldots, a_{n-1},\) we then obtain the figure ABGD

\[ = \frac{1}{2}\{a_1 + a_n + 2(a_2 + a_3 + \ldots + a_{n-1}) + 2(a_1 + a_2 + \ldots + a_{n-1})\}. \]

122. From the preceding demonstrations it follows, that the area of any curvilinear figure, such as CeH (art. 117) whose base is divided into an even number of equal parts by equidistant ordinates, is obtained by the following rule:—Add together the two extreme ordinates, twice the sum of the intermediate odd ordinates, and four times the sum of the even ones. Multiply the result by the common distance of the ordinates, and one-third of the sum is the area, either accurately or approximately. A similar rule may be derived from the formula of article 121 for areas such as ABGD.

According to the relation subsisting between the ordinates \(a_1, a_2, a_n,\) the equation of art. 119 will be that of a straight line or a parabola; hence the result obtained by the above rule will be strictly accurate in the case of all figures, such as CeH (fig. art. 117), having those portions of their boundaries CDE, EFG, which are terminated by the odd ordinates, either straight lines or parabolic arcs. The curve CEH may therefore be either wholly convex or wholly concave, towards AB; or partly convex and partly concave, provided in the latter case the points of contrary flexure occur only at the odd ordinates; for otherwise the intermediate arcs could not be, even approximately, parabolic. When points of contrary flexure occur, ordinates may be drawn at those points, and the intermediate areas, being found separately, may be added to obtain the whole surface. In other cases where there are numerous points of contrary flexure, the formula of art. 118 will probably give as good an approximation to the area of the curve as that obtained by the more complex formula of art. 121.

**Example.—To find the area of a quadrant of a circle. Draw DE bisecting AC at right angles, then BD is an arc of 30°, and the quadrant ABC = 3 sector DBC = 3(BCDE - CED);

\[ DE = DC^2 - EC^2 \text{ if we put } DE = a_1 FJ = a_2 \text{, we have } a_1^2 = 144 - 36 = 108, \] \[ a_2^2 = 144 - 25 = 119, \text{ etc.;} \]

\[ \therefore a_1 = 10\sqrt{3923048}, \] \[ a_2 = 10\sqrt{9087121}, \] \[ a_3 = 11\sqrt{3137085}, \] \[ a_4 = 11\sqrt{6189500}, \] \[ a_5 = 11\sqrt{8321596}, \] \[ a_6 = 11\sqrt{9582607}, \] \[ a_7 = 12\sqrt{0000000}. \]

The area is then calculated as follows:

\[ a_1 = 10\sqrt{9087121}, \] \[ a_2 = 10\sqrt{9087121}, \] \[ a_3 = 11\sqrt{3137085}, \] \[ a_4 = 11\sqrt{6189500}, \] \[ a_5 = 11\sqrt{8321596}, \] \[ a_6 = 11\sqrt{9582607}, \] \[ a_7 = 12\sqrt{0000000}. \]

The quadrant whose radius is unity, obtained by dividing by 144, is 785396, which is correct to five places of decimals. The area of BCDE, calculated by the rule of art. 118, is 688279433 from which the quadrant whose radius is unity, is found to be 789396, a result correct only to two places.

**123. To find the Volume of a Solid of Revolution.**

The solid is supposed to be generated by the revolution of the figure CEH (art. 117) about the line AB.

Let ABCD (fig. 119) be a portion of the figure (art. 117) such as CEe, terminated by two odd ordinates; and assuming EF, FD as axes, let

\[ y^2 = m + nx + px^2 \]

be the equation of a curve of the second degree passing through the points A, E, C. Then if we put

\[ AB = a_1, EF = a_2, CD = a_3, \text{ and } BF = FD = h, \]

it will be found, as in art. 119, that

\[ m = a_1^2 - a_2^2, n = \frac{a_2^2 - a_3^2}{2h}, p = \frac{a_3^2 - 2a_2^2 + a_1^2}{2h^2}. \]

Hence, the volume generated by ABDCE

\[ = \int_{-h}^{h} y^2 dx = \int_{-h}^{h} (m + nx + px^2) dx = \frac{\pi h}{3} (a_1^2 + 4a_2^2 + a_3^2). \]

**124. A less rigorous demonstration of the same result may also be obtained by assuming that the points A, E, and C, are taken so near each other, that the solid generated by ABDCE differs little from the frustum of a cone; for, by art. 66, its volume will evidently be

\[ \frac{1}{3} BD(a_1^2 + 4a_2^2 + a_3^2) = \frac{1}{3} h(a_1^2 + 4a_2^2 + a_3^2). \]

**125. If now we put in the figure of art. 117 as formerly,**

\[ Ce = a_1, Dd = a_2, Hh = a_3, \text{ and } cd = de = ec = h, \]

the volumes generated by CEec, EGge, &c., will be respectively

\[ \frac{1}{3} h(a_1^2 + 4a_2^2 + a_3^2), \frac{1}{3} h(a_1^2 + 4a_2^2 + a_3^2), \ldots \]

\[ \frac{1}{3} h(a_1^2 + 4a_2^2 + a_3^2). \]

Whence the whole solid of revolution generated by CeH will be

\[ \frac{1}{3} h[a_1^2 + a_2^2 + 2(a_1^2 + a_2^2 + \ldots + a_n^2)] + 4(a_1^2 + a_2^2 + \ldots + a_n^2)]. \]

**Example.—To find the volume of the frustum of a sphere generated by the revolution of BDEC, in the figure to art. 122, we have the values of \(a_1^2, a_2^2\) as follows:**

| \(a_1^2\) | \(a_2^2\) | |----------|----------| | 119 | 108 | | 135 | 144 | | 143 | 1588 | | 397 | 536 | | 4 | 2376 |

\[ \frac{1}{3} \times 2376 = 24881418816; \]

which agrees accurately with the value found by the formula of article 81.

**126. As it may sometimes be more convenient to measure equidistant circumferences of the solid of revolution than equidistant radii; let \(c_1, c_2, \ldots, c_n\) be the circumferences of the circles described by the revolving points C, D, &c. (art. 117).**

Then, since \(a_1^2 = \frac{c_1^2}{4\pi}, a_2^2 = \frac{c_2^2}{4\pi}, \ldots, a_n^2 = \frac{c_n^2}{4\pi}\) (art. 56),

the volume of the solid of revolution will be

\[ \frac{h}{12\pi} \left[ c_1^2 + c_2^2 + 2(c_1^2 + c_2^2 + \ldots + c_n^2) + 4(c_1^2 + c_2^2 + \ldots + c_n^2) \right]. \]

**127. Also, since \(wa_1^2, wa_2^2, \ldots, wa_n^2\) are the areas of the circles described by the revolving ordinates \(a_1, a_2, \ldots, a_n\), if \(s_1, s_2, \ldots, s_n\) be put for these areas, the volume of the solid of revolution will be**

\[ \frac{h}{3} \left[ s_1^2 + s_2^2 + 2(s_1^2 + s_2^2 + \ldots + s_n^2) + 4(s_1^2 + s_2^2 + \ldots + s_n^2) \right]. \]

**128. In the figure AEB (art. 117) \(a_1 = o_1\) and \(a_2 = o_2\); therefore the volume of the solid generated by its revolution**

\[ = \frac{h}{3} \left[ a_1^2 + a_2^2 + \ldots + a_n^2 + 2(a_1^2 + a_2^2 + \ldots + a_n^2) \right]; \]

and the formulae of articles 125, 126 may be similarly modified.

The equation of art. 123 may either be that of a straight line, a circle, an ellipse, or a hyperbola, according to the relations which may subsist between the ordinates \(a_1, a_2, a_3\). Hence the formulae of arts. 124-128 will be strictly accurate for solids generated by the revolution of areas bounded by any of these lines, subject to the limitation stated in art. 122.

**129. If the curve which generates the solid of revolution be everywhere convex to the axis of the solid, the volume may be obtained by a somewhat simpler formula than that of art. 124.** Let the base of the figure be divided into any number of equal parts by the ordinates \(a_1, a_2, \ldots, a_n\), whose common distance is \(h\).

Then if a parabola be described passing through the points AE (fig. art. 119), and whose axis is the line BF, it will coincide either nearly or exactly with the curve AE; and by art. 111 the paraboloid generated by AEBF,

\[ = \frac{1}{3} \pi (a_1^2 + a_2^2) h. \]

Similarly if another parabola be described passing through the points E and C, the volume generated by ECDF

\[ = \frac{1}{3} \pi (a_2^2 + a_3^2) h. \]

Therefore proceeding as in art. 125, the volume of the whole solid of revolution

\[ = \frac{1}{3} h \left( a_1^2 + a_2^2 + 2(a_2^2 + a_3^2 + \ldots + a_n^2 - 1) \right). \]

**Example.**—The volume of the segment of a sphere which is calculated in the example to art. 124, may be found as follows:

\[ \begin{align*} a_1^2 &= 119 \\ a_2^2 &= 128 \\ a_3^2 &= 135 \\ a_4^2 &= 140 \\ a_5^2 &= 143 \\ 665 & \\ 2 & \\ 1330 & \end{align*} \]

\[ \therefore \text{volume} = \frac{\pi}{2} \times 1582 = 2485 \text{ nearly}. \]

**130. To find the Volume of any Solid.**—Let \(s_1, s_2, \ldots, s_n\) be the areas of equidistant sections, taken so near each other that every two consecutive sections may be regarded as sensibly similar. Then, since it has been proved in art. 39 that the volume of any prismoidal solid, of which \(s_1, s_2, s_3\) are the areas of the ends and middle section, and \(h\) half the height, is \(\frac{3}{2}(s_1 + 4s_2 + s_3)\); if we proceed as in art. 125, we shall find that the volume of the solid is expressed by the formula of art. 127, which therefore applies not only to solids of revolution, but approximately to all solids whatever.

---

**ON GAUGING.**

Gauging denotes the measurement of casks or of substances liable to excise duties.

**To Compute the Contents of a Cask from its Length and Diameters at the Middle and End.**—Let the axis of the cask MN be trisected in L and Q; then the usual form of a cask is such that we may assume AB, DE to be straight lines, and BCD the arc of a parabola whose axis is CO. The portions of the cask AI, DF will then be frusta of cones, and DI will be the frustum of a parabolic spindle.

Put \(CG = a, AH = b, BI = c,\) and \(ML = LQ = QN = \frac{1}{3} l.\)

Then the volume generated by AL or DN

\[ = \frac{\pi l}{180} (5b^2 + 5bc + 5c^2) \quad (\text{art. 65}); \]

and the volume generated by BQ

\[ = \frac{\pi l}{180} (8a^2 + 4ac + 3c^2) \quad (\text{art. 113}). \]

Now if AB be produced to T, because TB is a tangent to the parabola BC, TK = 2CK, and by similar triangles TK : BL = AM : BK : ML, or \(2(a - c) : c - b :: 1 : 2;\) from which \(c = \frac{1}{2}(4a + b).\) Therefore substituting for \(c,\) and adding, we obtain the whole content of the cask

\[ = \frac{l}{360} (39a^2 + 25ab + 25b^2) \quad \text{Mensuration.} \]

If \(a, b,\) and \(l\) are expressed in inches, since there are 277-3 cubic inches in a gallon, and

\[ \frac{\pi}{360 \times 277.3} = 0.0003147, \]

the contents of the cask in imperial gallons

\[ = l (39a^2 + 26ab + 25b^2) \times 0.0003147 \quad \text{(1)} \]

Again, because \(39a^2 + 26ab + 25b^2\)

\[ = (39a^2 + 26ab + 26b^2) \left(1 - \frac{b^2}{39a^2 + 26ab + 26b^2}\right); \]

and since \(b\) is generally about \(\frac{2}{3}a,\)

\[ 1 - \frac{b^2}{39a^2 + 26ab + 26b^2} = 1 - \frac{1}{129} = \frac{128}{129}; \]

therefore the content of the cask in gallons

\[ = l (39a^2 + 26ab + 26b^2) \times \frac{128}{129} \times 0.0003147 \]

\[ = l \left(\frac{3}{2}a^2 + ab + b^2\right) \times 0.000812 \text{ nearly} \quad \text{(2)} \]

**Example.**—Let the bung and head diameters of a cask be 32 and 24 inches respectively, and its length 40 inches; required its content in gallons.

By formula (1) the content

\[ = (39 \times 32^2 + 25 \times 24^2 + 32 \times 24 \times 26) \times 40 \times 0.0003147 = 93.29 \text{ gallons}. \]

By formula (2) the content

\[ = \left(\frac{3}{2} \times 32^2 + 24^2 + 32 \times 24\right) \times 40 \times 0.000812 = 93.54 \text{ gallons}. \]

The contents of casks of any forms to which the preceding formulæ may not apply, may be determined with great accuracy by the following method.

If \(a\) and \(b\) be the bung and head diameters, \(c\) the diameter equidistant from the bung and head, and \(l\) the length of the cask, all in inches, it is obvious, by article 124, that the capacity of the cask in gallons

\[ = \frac{4\pi l}{3 \times 4 \times 277.3} (2a^2 + 2b^2 + 8c^2) \]

\[ = l \left(a^2 + b^2 + (2c)^2\right) \times 0.0047205. \]

**Example.**—Let the bung and head diameters be 32 and 24, the middle diameter 30-2, and the length of the cask 40. The capacity

\[ = 40 \times \left\{(32)^2 + (24)^2 + (60.4)^2\right\} \times 0.00472 = 99.1 \text{ gallons}. \]

**To find the Ullage of a Cask.**—The quantity of liquor contained in a cask partially filled, and the capacity of the portion which is empty, are termed respectively the wet and dry ullage.

(i.) The ullage of a standing cask is found by the method of article 124 as follows:

Add the square of the diameter at the surface, the square of the diameter at the nearest end, and the square of double the diameter half way between; multiply the sum by the length between the surface and the nearest end, and by 0.00472. The product will be the wet or dry ullage, according as the lesser portion of the cask is filled or empty.

(ii.) The ullage of a lying cask is found approximately on the assumption that it is proportional to the segment of the bung circle cut off by the surface of the liquor. The rule adopted in practice is,

\[ \text{ullage} = \frac{4}{3} \text{ content of cask} \times \text{segmental area}. \] ### Table I.—Areas of Polygons

| Number of Sides | Polygon Side = l | Area | Log Area | |-----------------|------------------|------|---------| | 3 | Equilateral Triangle | 0.4330127 | 9.6355005 | | 4 | Square | 1.0000000 | 0.0000000 | | 5 | Pentagon | 1.7204774 | 0.2335490 | | 6 | Hexagon | 2.5980762 | 0.4146519 | | 7 | Heptagon | 3.6339125 | 0.5603745 | | 8 | Octagon | 4.8284271 | 0.6380568 | | 9 | Nonagon | 6.1818247 | 0.7911166 | | 10 | Decagon | 7.6942268 | 0.88851640 | | 11 | Hendecagon | 9.3655407 | 0.9715375 | | 12 | Dodecagon | 11.1951024 | 1.0490688 |

### Table II.—Surfaces and Volumes of Regular Polyhedrons

| Polyhedron | Number of Faces | Surface | Log Surface | Volume | Log Volume | |-----------------|-----------------|---------|-------------|--------|-----------| | Tetrahedron | 4 equilateral triangles | 1.7320508 | 0.2320508 | 0.1125511 | 0.0715226 | | Hexahedron | 6 squares | 2.4494853 | 0.3849485 | 0.2449485 | 0.1649485 | | Octahedron | 8 equilateral triangles | 2.9642798 | 0.4764279 | 0.3964279 | 0.2496428 | | Dodecahedron | 12 pentagons | 3.8944330 | 0.5894433 | 0.5894433 | 0.3794433 | | Icosahedron | 20 equilateral triangles | 4.3301270 | 0.6330127 | 0.7330127 | 0.4730127 |

### Table III.—Functions of π

| Number | Logarithm | Number | Logarithm | |--------|-----------|--------|-----------| | π | 3.1415927 | 0.4971499 | 9.6969044 | 0.9942997 | | 2π | 6.2831853 | 0.7981790 | 1 | 0.0000000 | | 4π | 12.5663706 | 1.0992099 | 6π² | 0.0168869 | 8.2275490 | | 1/π | 0.3183099 | 9.5028501 | 2√π | 0.2830948 | 9.4503951 | | 4/π | 1.2732395 | 0.1049101 | √π | 1.2407010 | 0.0936671 | | 1/4π | 0.0795775 | 8.9007901 | √3/4π | 0.6203505 | 9.7926371 | | 180° | 57.2957785 | 1.7581226 | log π | 1.1447299 | 0.6587030 |

### Table IV.—Lengths of Circular Arcs (Radius = 1.)

| Degrees | Arc | Degrees | Arc | Degrees | Arc | Degrees | Arc | |---------|-----|---------|-----|---------|-----|---------|-----| | 1 | 0.0174533 | 41 | 0.6460581 | 121 | 2.1138454 | 1 | 0.002009 | 1 | | 2 | 0.0349066 | 82 | 0.6921041 | 162 | 2.1203017 | 2 | 0.004018 | 2 | | 3 | 0.0523599 | 123 | 0.7381491 | 203 | 2.1267579 | 3 | 0.006027 | 3 | | 4 | 0.0698132 | 164 | 0.7841941 | 244 | 2.1332142 | 4 | 0.008036 | 4 | | 5 | 0.0872665 | 205 | 0.8302391 | 285 | 2.1396695 | 5 | 0.010045 | 5 | | 6 | 0.1047198 | 246 | 0.8762841 | 326 | 2.1461258 | 6 | 0.012054 | 6 | | 7 | 0.1221731 | 287 | 0.9223291 | 367 | 2.1525821 | 7 | 0.014063 | 7 | | 8 | 0.1396264 | 328 | 0.9683741 | 408 | 2.1590384 | 8 | 0.016072 | 8 | | 9 | 0.1570797 | 369 | 1.0144191 | 449 | 2.1654947 | 9 | 0.018081 | 9 | | 10 | 0.1745330 | 410 | 1.0604641 | 490 | 2.1719510 | 10 | 0.020090 | 10 | | 11 | 0.1919863 | 451 | 1.1065091 | 531 | 2.1784073 | 11 | 0.022099 | 11 | | 12 | 0.2094396 | 492 | 1.1525541 | 572 | 2.1848636 | 12 | 0.024108 | 12 |

### Table V.—Areas of Segments of a Circle (Radius = 1.)

| Height | Area | Height | Area | Height | Area | Height | Area | |--------|------|--------|------|--------|------|--------|------| | 0.01 | 0.018883 | -26 | 2.39997 | -51 | 531.563 | 76 | 1.698445 | | 0.02 | 0.035317 | -27 | 2.53358 | -52 | 549.053 | 77 | 1.114885 | | 0.03 | 0.069754 | -28 | 2.67333 | -53 | 566.552 | 78 | 1.134372 | | 0.04 | 0.140994 | -29 | 2.81315 | -54 | 584.052 | 79 | 1.153904 | | 0.05 | 0.209232 | -30 | 2.95299 | -55 | 601.552 | 80 | 1.173479 | | 0.06 | 0.277470 | -31 | 3.09283 | -56 | 619.052 | 81 | 1.193055 | | 0.07 | 0.345708 | -32 | 3.23268 | -57 | 636.552 | 82 | 1.212630 | | 0.08 | 0.413946 | -33 | 3.37253 | -58 | 654.052 | 83 | 1.232206 | | 0.09 | 0.482184 | -34 | 3.51238 | -59 | 671.552 | 84 | 1.251782 | | 0.10 | 0.550422 | -35 | 3.65223 | -60 | 689.052 | 85 | 1.271358 | | 0.11 | 0.618660 | -36 | 3.79208 | -61 | 706.552 | 86 | 1.290934 | | 0.12 | 0.686898 | -37 | 3.93193 | -62 | 724.052 | 87 | 1.310510 | | 0.13 | 0.755136 | -38 | 4.07178 | -63 | 741.552 | 88 | 1.330086 | | 0.14 | 0.823374 | -39 | 4.21163 | -64 | 759.052 | 89 | 1.349662 | | 0.15 | 0.891612 | -40 | 4.35148 | -65 | 776.552 | 90 | 1.369238 | | 0.16 | 0.959850 | -41 | 4.49133 | -66 | 794.052 | 91 | 1.388814 | | 0.17 | 1.028088 | -42 | 4.63118 | -67 | 811.552 | 92 | 1.408390 | | 0.18 | 1.096326 | -43 | 4.77103 | -68 | 829.052 | 93 | 1.427966 | | 0.19 | 1.164564 | -44 | 4.91088 | -69 | 846.552 | 94 | 1.447542 | | 0.20 | 1.232802 | -45 | 5.05073 | -70 | 864.052 | 95 | 1.467118 | | 0.21 | 1.301040 | -46 | 5.19058 | -71 | 881.552 | 96 | 1.486694 | | 0.22 | 1.369278 | -47 | 5.33043 | -72 | 899.052 | 97 | 1.506270 | | 0.23 | 1.437516 | -48 | 5.47028 | -73 | 916.552 | 98 | 1.525846 | | 0.24 | 1.505754 | -49 | 5.61013 | -74 | 934.052 | 99 | 1.545422 | | 0.25 | 1.573992 | -50 | 5.74998 | -75 | 951.552 | 100 | 1.565098 |

(w.s.)