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x? Ex. V In the cissoid

(ax--xx) and y:=x: (0–1)+; consequently fyr, or the space AKMPA = fata (a —%)Now (ax-xx)} = the semi-segment AONP; and by art (80) we shall find that fixt i (a—x)}=$ xd (a—a)++ } Sz?" ;

P (a—)Hence fai o (0-2)-4 3: (ar— ~~)*—2+ (ar—æx)*; or APMKA=3APNOA– 4 ANP= 3 AONA – ANP. Consequently the infinitely extended space MKABQ is triple the generating semi-circle ANB.

Ex. VI. In the logarithmic curve yä=my, and fyč, or ABMP=my+C. But when y=1= AB, the space ABMP becomes

Therefore C=—m, and ABMP=m (7-1)= the rectangle OIQM. If we make y=0,

요 we shall have the infinitely extended space BXYA=-m= the rectangle PQIT.

Ex. VII. Let there be proposed the curve BM, whose equation is y=x*, we shall have (art. 84), the space ABMP = Jx* r =

22 x3

+ &c.) 35

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zero.

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x (1 –

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And when AP=PM=1, we have the space

1 ABMP =1

1 1 1 +

-&c.=0.7834 30497589. 22 33 44 55 Ex. VIII. In the curve of sines AMA'M, &c. of

M which the equation is y=sin t, we shall have

р APM=Så sin r=C-cos xIf we make x=0, we

n

M

c

votain C=1, and APM=1.-cos°r. Let r=180° 57, and we shall have AMA'A=2=twice the square of the radius. If we suppose x=27=AA", we shall have the space AMA'B+A'M'A”A'=0, as is evident, since one is positive, and the other negative. In general, if x=2k 7, the space will equal zero, and if x = (2k+1) 7, the space will =2.

If for the origin of 2 we take the

MB point A, the middle of A'A', we shall bave

PA y = cos r.

Consequently the space ABMP=sin x,

the
space

ABA'A=1, and AMBA'A=0, or =2, if we leave out of consideration that of its two parts, one is positive, and one negative.

B 109. If the ordinates proceed from a fixed point C, we may find the quadrature of the curve as fol. lows. Draw two radii CM, Cm, and from the centre

A c, and with the radius CM

2 describe the arc Mr; then the triangle CMM Mr XCM + Mmr.

But 2 when the point m is indefinitely near to the point M, the space Mmr vanishes, and there remains flux

Mr XCM (COMC)

P

N 2 then Mr=, CM=y, and we shall have COMCE Syc+C.

If we denote by Q the angle formed by CM with a fixed line issuing from the point C, or the arc which measures this angle iv a circle whose radius is 1, we shall have Mr = yo, and COMC= *j yy ©+C.

110. Ex. I. Let the curve AM be the conchoid, Pits pole, call PM=y, QM=a, PB=b, and the angle APM=. We shall

B 6 have 0:6::1: PQ= cos

M м

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M

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2

6

+ a. Consequently the space APM=IS cos P

cos? 22 + abs +

tan © £ abl tan (45° + }() + " , COSP without any constant. Therefore – APMFPBQ, or ABQM=abl tan (46° +40)+" ,

a' Q, and AAMM=2abl tan (45° +10). Ex. II. In the cissoid, if we make AM = y, MAB = Q; AQ =

cos p

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AKMOA=IS.Cooo

A А

P

a?
-faa 0 + $ faa

B в cos? p = aa tan aa + aa (1 sin cos 0+50)= aa tano + $aa sin 20-ap. Hence AKMPA = 204 a sin 2 + is a sin 40; and the indefinitely extended space ŘKABQ= al o=3 AONB.

Ex. III. In the spiral of Archimedles, AGFBN=r AGFBA = C, CM = y, CA = a, Mr = ye, flux

В

J)

N

M

a

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360°, the elementary triangles 4 yy ö contain those already computed.

It would be easy to supply this defect by calculating the elementary trapezia comprised between two adjacent spires. The same inconvenience will be found to occur in the common formula fux, if several ordinates correspond to one abscissa. Ex. VI. In the hyperbolic spiral

a:-:::y: Mrsmyx

a

A

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کی

+

P B

MB=y+

5a4

1

1.3 QM=S =X+

+ Vaa—xx) 2 8aa9.4 1.3.5 x?

+&c.; therefore the arc 5a4 ' 2.4.67 ao

A
y 1.3 ys
+

+ &c.

2 Saa 2.5 Example II. In the parabola AM = sy (1 + 499 =pPy (rp+yy). Now we PP р

? have shewn (71) that. ver + aa =

Q & C+ x^(2x +aa) + 4 aal (r + Virr +aa) Therefore AM=C+In (yy + $ 211) +

P pl 3y+v(yy +

A y=0,

M

PP

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