HIGHER ENGINEERING MATHEMANICS Example 6-22. Trace the curve r? = a? cos 20. yy (Poona, 1990) (i) Symmetry : The curve is symmetrical about the pole. (ii) Limits : (a) The curve lies wholly within the circle r =a. (b) No portion of the curve lies between the lines 6 = 1/4 and O=3n/4. . Fig. 610. +20 (ii) Points : (a) tan 9 =r a cot 20 = tan = 0/2 when 0 = 0. ie. 9= 9420 + 9=0, when O=— 1/4; Thus the tangent at O is @=~n/4 and the tangent at A is to the initial line () The variations ofr and 0 are given below : As 8 varies from ___rvaries from Portion traced D ton/4 a tod ABO 3n/4 ton Otoa ocp As 6 increases from x to 2n, we get the reflection of the arc ABOCD in the initial line. Hence the shape of the curve is as shown in Fig. 6-10. This curve is known as Lemniscate of Bernoulli. Problems 6:7 Trace the following cury 1. y? (a +x) =x" (a -x) (V.T.U., 2000) 2. y? (a? + 2 (a? - x?) (Kuvempu, 1998) 8. y= (x24 D7? - 1) 4. ay? =x? (4-2) (Rewa, 1990) 5. a? yx") (Poona, 1990) 6.x =a cos? 0,y=6 sin” 0 4. x =a (8-sin 6),y =a (1 cos 0) (0<0¢2n) (a cos t+ log tan (72), y =a sint 9.7 =a cos 28 (lysore 1999) 95 38 (Kuvempu, 1998) 11. rsa] ~cos 6) +3.cos 0 (Marathwada, 1998 $} 6 10. AREAS OF CARTESIAN CURVES ot Sinisa aaa b bis fa yds. (1) Area bounded by the curve » =f (x), the x-axis and the ordinates x Let AB be the curve y =/(x) between the ordinates LA (x =a) and MB (x =). Let P (x, 7), P” (x + &, y + By) be two neighbouring points on the curve and NP, N'P’ be U respective ordinates. area PNN'P’ = 6A. Complete the rectangles PN’ and PN’. ¥. ‘Then the area PNN'P’ lies between the areas of the les PN’ and PN. ie, &A lies between ydx and (y + By) 8x ®4 jies between y and y + By. Now tuking limits as PP ie., be O(and 2. By > 0), ae dA. oO dx 7area ALMB= |. yds. 0 3 Thus Fig.612.* (2) Interchanging x and y in the above formula, we see that the area bounded by the curve ® eft), thy axiz end the abscissae y = 0,7 -Lis {, xdy. Obs. 1. The area bounded by a curve, the x-axis and wo ordinates is called the area under the curve. ‘Te process of finding the area of plane curves is often called quadrature. Obs. 2, Sign of an area, An area whase boundary is described in the anti-clockwise direction is considered positive and an area whose boundary is described in the clockwise direction is taken as negative. b In Fig. 6-13, the area ALMB | = I, y dx |which is described in the anticlockwise direction and lies above the =axis, will give a positive result. y, Fig, 613. Fig.614. 3 Fig. 6-15. 6 In Fig. 6-14, the area ALMB [ i y dx |which is described in the clockwise direction and lies below the Sloe -axis, will give a negative result! land the b In Fig. 6-15, the area ALMB [- I y ae not consist of the sum of the area ALN'| 6 1 ea NIB (- J y «| but their difference. Thus to find the total area in such cases the numerical value of the area of cach portion must be faluated separately and their results added afterwards. Example 6-23, Find the area of the loop of the curve ay’ = Let us trace the curve roughly to get the limits of integration. (i) The curve is symmetrical about x-axis. (ii) It passes through the nrigin. The tangents at the origin are ay" anode. (iii) The curve has no asymptotes. (a- x). (Osmania, 2000S) ory =2., +. OriginHIGHER ENGINEERING MATHEMANICS (iv) The curve meets the x-axis at (0, 0) and (a, 0). It meets the xis at (0, 0) only. From the equation of the curve, we have = Pa Ma -x) For x >a,y is imaginary. Thus no portion of the curve lies to the right of the line x =a. Also as Fig. 616. Thus the'curve is as , Area of the loop = 2 (area of upper half of the loop) @ a. 0 = 2 =2f) yax-2 f° Nee )$h fa-(a-x)] la - x) de a 32/7 572|" v2 3/2) ly = (ax) 2. | (an xy" Bs, la (a x)? ~ (a ~ x)°7] dv = 2Na |S] - Je | ea ° 4 (9 03/2) 4-9 3a 0-0) +27 (0 Example 6-24, Find the area included between the curve y? (2a ~ x) = x° and its asymptote. (Ranchi, 1998) ‘The curve is as shown in Fig. 6-2. Area between the curve and the asymptote 2a 20 2 _ eof yar feo" Put x.=2a sin? 0m. 0 jo" (2a so that dx = 4a sin 0 cos 0 d0 ie 22) eV ans 2a sin? =o + 4a sin 8 cos 0 d9 | 2a cos? 0 = 1602-3212 _ gyq2 = 16a? af sint 0d0 = 16a? 24.5 = 3na?, Example 6-25. Find the area enclosed by the curve x? =? (2a -y). Let us first find the limits of integration. (i) The cui ve is symmetrical about y-axis. (ii) It passes through the origin and the tangents at the origin are or x=0,x=0. .. There is a cusp at the origin. iii) The curve has no asymptote. (iv) The curve meets the x-axis at the origin only and meets the y-axis at 0, 2a). From the equation of the curve, we have x=%vy a-y) For y 2a,x is imaginary. Thus the curve entirely lies between Fig. 617. = 0 (x-axis) and y = 2a, which is shown in Fig. 6-17. a aim Put y =2 a *. Area of the curve = 2 J xdy =2f, y Vy (20-9) dy ca ae Bias eisa(0- sin U),y #4 (4 = cos U) june tis Ua (VALU, 20007 To describe its first arch, 0 varies from 0 to 2nie.x varies m0 to 2an (sce Fig. 6-18). ee 2ra +, Required area =f, wove . O@=07 B=2n) where y =a (1 ~cos v), dx =a (1 - cos 0) d0. i ike Fig. 6-18. Jo 29 7 (1-08 0) +a (1 ~ cos 0) do 1" x 2a? J. (1 ~ cos 0)? do = 8a sin' S40 n/2 = 16a? J, sin! 9d, putting 0/2 = 9 so that d0 = 2do. 23-16 anita Example 6-27. Find the area of the segment cut off from the parabola x? = 8y by the line -2y+8=0. 2 Given parabola is x? = 8y a) \d the straight line is x - 2y +8 =0 alii) Substituting the value of y from (ii) in (i), we get x? = 4 (x +8) or x?7- dx -32=0 (x-8)(@@+4)=0 2. x=8,-4. , Thus (i) and (ii) intersect at P and Q where x = 8 and x =~ 4. (Fig. 6-19) Required area POQ (i.e. dotted area) = arca bounded by st. line (ii) and x-axis from :-4 tox =8 — area bounded by parabola (i) and x-axis from x =- 4 tox = 8. fp FP [ . = J_, yds, from (ii) - J y dx, from (i) 3 8 Em 3 2 Miao. 1 1 i = 3 ((B2 + 64) — (- 24)) ~ 5 (512 + 64) = 36. -4 Pelee tas? S don, Example 6-28; Find the area common to the parabola y? = ax and the circle x* +)" = dox. (Raipur, 1998) Given parabola is ysar . see ae wi 4 the circle is Paya dar, mHIGHER ENGINEERING MATHEMALICS, Y Both thi dand Gi) for curves are symmetrical about x-axis. Solving we have Jax or x (x ~ Sa) = 0 or = 0, 3a, ‘Thus the two curves intersect at the points where x =0 O17 and x= 3a. (Fig. 6-20), Also (i) meets the x-axis at A (a, 0). Area common to (i) and (i) i . the shaded area = 2 (Area ORP + Area PRA} (by symmetry) 3a 4a 2 fe y dx, from (i) + Si, y dx, from (ii) ba to =2 [J Vax) dx + I \(dax - x*) a, 372|54 ser nake 2 =a [ES ; +2),0 Vida? (&- 2a)71 de aa (8a)? + 2[2 (© ~ 2a) Vida? ~ (& ~ 2a)?) = ANS a? + 2110 ava) + 20? tn/2-~ x/6I1 = 4N3 a? ~808 $f son) Problems 6.8 2 2 1, (i) Pind the area of the ellipse 35 +5 =1, a8 (ii) Find the area bounded by the parabola y* ‘ind the area bounded hy the curve y = x(x (-~ 5) and the x-axis. (Ranchi, 1990) 3. Find the area included between the curve ay? 4, Pind the area of the loop of the curve : (i) Bay? = x (x -a)* (Osmania, 1995) (ii) y? (a +x) = fi Pde B Hence the required area.OAB = 3 if 7 d0. Example 6-29, Find the area of the cardioid r =a (1 -cos 0), The curve is as shown in Fig. 6-22. Its upper half is traced from g= x. =0toQ=n. “Area of the curve=2-2[",2a0=a2 [ca = cos 0) do a . 2/0 ‘0 . x 5 =a? J (2 sin? 0/2)? do = 4a? fo sin’ 0/2-d0 #2 = Ba? is sin! 99, putting 0/2 =o and d0 = 2do. 2 = 8a? Se. Sra’ poteqaa ome Example 6-30, Find the area of a loop of the curve r =a si The curve is as shown in Fig. 6-9. It consists of three loops.GHC ENGINEERING MAttIAAnCS pucting re O, 0 = Our avo which are the limits for the first loop cv of loop = (= cos GO) do Qbe The Hinits oF inerration For woop ofr a ain Over a oO when r= 0 0 are the two consccutive values of Example 631, Prove that the area of a loup of the curve x" y3 = Jaxy is da? /2. Changing to polar form (by putting x =r cos 0, y Putting r sin 0 cos 0 = 0, 2 0 =0,x/2, which are the limits of integration for its loop (See Fig. 6-5). +. Area of the loop ier : ufc? Aaa sco 220 hacrr 0 (cos? 0 + sin? 0)? 9a? fie tan? 0 sec? 0 4g 2/0 (+ tan’ 6)? (Dividing num. and denom. by « an? -F 1 a putting 1 + tan? 0=¢ and 3 tan’ 0 sec? 0 6 = « £ Example 6-32. Find the area common to the circles r=a\2 and + ‘The equations of the circles are r=a V2 wi) und r= 2a cos 6 () represents a circle with centre at (0, 0) and radius a 2. (ii) represents a circle symmetrical about OX, with centre at (a, 0) and radius a. The circles are shown in Fig. 6-23. At their point of intersection P, eliminating r froin (i) and ue), a2 = 2a cos Bi.e., cos 0 = 1V2 or O=n/4. <. Required area = 2x area OAPQ (by symmetry) = 2 (area OAP + wrea OPQ) fam ype, z Es Pap, for@+3 fr 7 do, for (i?) Ms w/2 (uw V2)? do-+ le (2a cos 0)? do sin 20|"7 _ = 2a" (1/4 - 0) + 2a? 3 O+ WA§. Show that the area incivued beiween the solrum x” aay and its asyniplote is equal to t ef loop. (Kurutkshetr in 4. Prove that the area of the lonp of the curve x9 + y? = Saxy is three times the area of the loop of t curve =a” cos 20. (Dethi, 3: $, Find the area outside the circle r = 2a cos @ and inside the cardinid r = a( + cos 0). 6, Show that the area included between the cardioids r = a (1+ cos 0) and r=a (1 ~¢0s 0) is a” (3x ~ 61/2. (Mangalore, 1999) 611. LENGTHS OF CURVES (1) The length of the are of the curve y = (x) between the points where x =a andx=bis 2 ‘| ‘dy EN [+ (22) Jaw Let AB be the curve y=/(x) between the points and B where x.=.a.and x= b (Fi Let P (z, y) be any point on the curve and arc AP that it is a function of x. (1) of p. 173] 6 4 2 i‘ Fig. 6 24. d) ds =b fe [+(2) Jerr -aeet = (value of s for x = 6) - (value of s forx =a) =areAB-0 2 6 ‘ i Hence, the arc AB =f vi 2) Je (2) The length of the arc of the curve x = f(y) between the points where y =a and y = 6, is i =6, is | 4 I [1 Je [Use (2) of p. 173) (3) The length of the arc of the curve x=f(t),y =O!) between the points where tzaand t=b, is a aeeee &) (z sc 13) p. 17; i (e) (2) Joe [se (3) p. 173) (4) The length of the arc of the curve r is between the points where 0= cand 0=B. és j, fe-(@) + acy [Wse (1) of p. 1741HIGHUT LGN CNG Mansrttante wth of the are of the curve Do f(rysetwesn the ports where re aand eo ble in vi: ife at) |ue [Use 120 of po V4) I Keanple 633, Pind the length of the are of the parabola x day meunured from the verter tvity of the latus-rectum, (Mangalore, 199% be the vertex and Lan oxtremity of the Iatumrectum so that at A, 299 and ot 1 G25), ¥, t slo so that“ dy eae Ney Vp. (i } Je un li aay de Fg. 625. ‘ta? sinh’? 7 inh?! 1) sa [v2 + log (1 + ¥2)) [ee sinh |x olog be eta» 0?) G34, Find the perimeter of the loop of the curve y = a “(Ranchi, 1998) raat i Gay 6a") 4 VW iax) Toa) a (3x4 ax V4 de ce 17% i eee ay M. oy pple 685. Find the length of one arch of the eyelid xea(t~ainl, yall ~ cox ty (Kuvempu, (998; Hamirpur, ide 8 Ava point moves from ong end O to thy othar and of (ts first arch, the parameter Cineres Seyi Ot 2, fuee Fig, GB)sto}, sin t72dt na = da [(- cos n) = (~ cos 0)] = Ba Example 6:36, Find the entire length of the cardioid r =a (1 + cox 0). (Mysore, 19. MALE, 1997 Wi Also show that the upper half is bisected by 0= 7/3. (Mangalore, 1999) ‘The cardioid is symmetrical about the initial line and for its upper half, 0 increases from 0 on (Pig. 6:27) P Also . 2 =. Length of the curve = 2f 24 (22) Jao ° a 3 Rego erigleae Tea Nila (1 + cos 8)1? + (“a sin 04) d0= 20 | VI2 (1 + cos 6)] dO a cos 0/2.d0 = 4a | 0/2 4 Bain n/2 = sin 0) = 8a -. Length of upper half of the curve is 4a. Also length of the arc AP from.0 to r/3. Pa 1/3 =of, V2 (1 + cos @)) dO = raf, cos 0/2-d0 = 4a |sin 0/21?” = 2a = half the length of upper half of the cardioid Problems 6.10 x" from the vertex to the ordinate L. Fird the length of the are of the semi-cubical parabola x= 5a. 2. Find the length of the curve y = log [(e' ~ 1)/(e" + H)] from x= tox = 2. (Bhopal, 1997) 4. Find the length of the are of the parabola y® = dax (i) from the vertex to one end of the latussrectum (ii) cus off-by the line 3y = Bx. Wagpur, 188s 4 Find the perimeter of the loop of the following curves Be exhia-x (S. Gujarat, 1990) (ii) 9y" (x= 2) 0-577, 4, G, Show that the whole length of the curve x? (a? = x") = 8a? y? is na V2. 7. Find the length of an arch of the cycloid x =a (0-4 sin Oy #0 ff = cos On 9 Find the length of the curve y? = (2x ~ 1)° cut off by the line x 6, Find the whole length of the curve x # a cost, yzastn"t ¥. Find the length of the loop of the curvo xa (yet 673, (Manithiwada, 189s s+