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(52—a2) §2+(52 — (2) n2 = 52—1;

(56)

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thus this equation represents a surface, and consequently the double definite integral, whose element is dn d, expresses the volume contained within the given limits between this surface and the plane of (, ). This surface is represented in fig. 45, where oc = 1, for since a2 and 32 are both less than 1, when (=1,n=0; and consequently the surface intersects the (-axis at c. For values of (a little less than 1, the equation gives impossible values to § and 7; but gives possible values when is greater than 1; thus the surface lies above the point c. When 2+2 = 1, (= ∞; consequently if a right circular cylinder, whose radius = OA = OB = and whose axis is oc, is described as in the fig., the surface (56) is asymptotic to it. Also as (56) is unchanged when is replaced by, another surface equal and similar to CRQ lies below the plane of (§, n), to which the cylinder is also asymptotic. Thus the integral (55) expresses the volume contained within the cylinder and between those two surfaces. Hereby also we have a further interpretation of the definite integral; for if we give to a value greater than 1, and take it to be constant, while έ and ŋ vary, that is, if we cut the surface by a plane parallel to that of (§, n), (56) shews that the curve of section is an ellipse; such is RQ, of which the semi-axes are

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η

(57)

and of which therefore the area =

π (2-1)
(Š2 —— a2) 1 (¿2 — ẞ2)

(58)

If varies, the area of this ellipse will also vary by a narrow

elliptical ring, of which the area = ́а.

2-1

; and

(¿2 — a2) § ( §2 — ẞ2)} consequently we may consider the element of the volume, which is expressed in the right-hand of (55), to be

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Thus by this substitution, and mode of interpretation, due to

M. Catalan *, the double integral in (37) is reduced to the single integral in (60).

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241.] In review of the processes of the two preceding articles let it be observed, that in equations (39) and (45) we have

ds=

{b c (sin a)® (cos B)

ds=

+ c a (sin a) (sinß)* + a®b® (cosa)?} sin a dad3, a2 b2 c2 sin n dn dự

2

;

{ a2 (sin ŋ)2 (cos √)2 + b2 (sinŋ)2 (sin √)2+c2 (cos n)2 } 2 the former of which is irrational, and the latter is rational; so that by means of the following substitutions we have been able to transform an element-function involving irrational quantities into an equivalent in terms of rational quantities only; viz. by substituting a sin ŋ cos

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sin a cos B

=

b sin n sin y

Ρ

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(61)

= a2 (sin ŋ)2 (cos)2 + b2 (sin ŋ)2 (sin¥)2+c2 (cos n)2; and

where pa

1

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Also equating the values of the elements given in (41) and (44),
we have
p3 sin a da d3 = abc sin n dn dự.

η

Hence by the substitutions of (61), the double integral

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(62)

{ b2 c2 (sin a)2 (cos ẞ)2 + c2 a2 (sin a)2 (sin ẞ)2 + a2 b2 (cos a)2 } } '

in which u is a rational function of sin a cos ß, sin a sin ẞ, and cos a, may be transformed into the following, which involves only rational quantities, viz. into

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a2 (sin n)2 (cos y)2 + b2 (sin ŋ)2 (sin y)2 +c2 (cos n)2 ·

the limits of the new variables being easily obtained from those of the former variables by means of equations (61).

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*See Liouville's Journal, Vol. IV, page 323. The same method is extended to integrals of higher orders and more variables; the discussion however of which is beyond the scope of the present work.

242.] As position in space may be determined by a system of polar coordinates, such as that explained in Art. 165, so may of course the equation to a surface be expressed in terms of these polar coordinates; in which case for the determination of the area it is necessary to express the area-element in terms of them, This may be found by transformation of either of the expressions given in (29) and (31). I propose however first to investigate the subject on principles purely geometrical, and subsequently by analytical transformation to exhibit the identity of the result with those thus determined.

Let P, fig. 42, be the point (r, 0, 4) on the surface whose equation is F (r,0,4)=0; let 0 and 4 vary; and let the radii vectores due to the partial variation of 0, to the partial variation of 4, and to the total variation of 0 and 4, respectively, intersect the surface in the points U, V, R, so that PURV is the surface-element. From P draw PS at right angles to OP, and cutting ou in s; then PS = r de, sv = de. Also from P draw PT at right angles to OP, and cutting ov in T; then PT = p dp = r sin 0 do,

TV =

dr

(ar) a

(d) dø. Also let or be produced to z; then from the

geometry, as explained in Art. 165, it is plain that PS, PT, PZ form a system of lines at P such that each one is at right angles to the other two; and consequently they may be considered as constituting a system of rectangular coordinates originating at P; we will employ them as such, and take PS, PT, PZ to be respectively the x —, y—, z—, axes. In reference to this system P is (0, 0, 0); u is (rde, 0, (dr)de); v is (0, r sin 0 dp, (dr)dø).

Hence, if da denotes the surface-element, which is approximately a parallelogram, of which P, U, V, R are the angular points,

2

da = { r2 (sin 0)2 + (sin 0)2 (dry)3

+

r de do.

аф

= ƒ ƒ { r2 (sin 0)2 + (sin 0)2 (dry)2 + (dr)" } *
SS

A =

аф

(63)

r do do; (64) this integral being definite, and the limits being assigned by the circumstances of the problem.

243.] The preceding expression for the surface-element may also be derived by transformation of coordinates from the last of (29) by means of the formulæ given in (70), Art. 165. It will be

convenient to effect the transformation by first replacing x and y in terms of p and p, and subsequently replacing p and z in terms of r and 0.

Since x = p cos, y = p sin p, therefore dx dy = p dp do: and as after substitution the equation to the surface takes the form F (p, p, z) = 0, then, as in (190), Art. 107, Vol. I,

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By a similar process, since z = r cos 0, p = r sin 0,

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(65)

Hence, substituting for (d), squaring and adding,

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If the equation to the surface is given in the explicit form r = f(0, 4), so that r-f(0, 4) = F (r, 0, 4) = 0, then

dr

= 1, ・ (dr), (d)
(d)

and (67) becomes,

A

==

2

==

2

dr

аф

do;

= ƒ ƒ { r2 (sin 0)2 + (sin 0)2 (db) 2 + (dr) * } *r do dp ; (69)

which is the same expression as (64),

If the surface whose area is to be determined is a surface of revolution about the z-axis, then the equation to it is independent

of 4, and (69) becomes,

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If the whole surface is required, the p-integration may be effected for the limits 2π and O, and

A = 2π {(r dr2}1r

2 = [ { (r d0)2 + dr2 } 1 r sin 0.

(71)

This result is manifestly identical with that given in equations

(17) and (21) of the present chapter. general, and includes all the preceding.

Indeed this process is

244.] The following examples are illustrative of the preceding general formulæ.

Ex. 1. To find the area of a spherical triangle.

Let ABC be the triangle whose angles are A, B, C; and o the centre of the sphere. Let oc be the z-axis; and let the central plane containing the side AB be inclined to the plane of (x, y) at an angle = a, and intersect it along the x-axis. Then the equation to this plane is tan 0 cot a cosec p. Let do be the 4-coordinate to A, and consequently 4%+c is the 4-coordinate to B. Letra be the equation to the sphere, and let

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= a2{c-cos-1(sina cos(+c))+cos-1(sin a cos)};

But from the geometry, cos a = sin a cos do,

..

cos (TB) sin a cos (po+c);

the area of the triangle = a3 {A+B+C-T}.

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