50 SPECTRUM MATHEMAL ro. 151 _ 3e 8 22 4 MIL TIPLE NTEGR x+ yt si 8 = 0 uSine (a'-r*) Evample 11. Evaluate rd dr d9 x is an odd function of rand rdr = 0| asine (a -p<) 2 asin a sine Sol. Let I- | rde dr d-J 1l dr do rta-)dr dal Similarly yddy d-0 :d dy de =o r d6 0 0 0 y+s r+y+sl a sine sin sin' eldo d9 EXERCISE 4 (d) valuate the following Integrals xyz d: dhy dr dr 64 0 0 Example 12. Show that (ar +by+ c:) dr dy dz = 0. log 2 x + log y logye (iv) ety dz dy dr uJ log: dz dy dr 0 0 0 Sol. Since x++is i1 2 a cos 6Na-r s1, +ys 1,x+y+#s1 r de dr de (vi) Jo'in o dpdode 0 0 0 0 -ISxS 1,- v1-r sysyi-x -1- -y s:syi-x* - 2. Evaluate +y+z) dr dy dz over the region defined b 20, y 2 0,: 20, x +y + : S I Now (ax + by + cz) dr dy dz a EvaluateJxy:dr dvde over the lipsoid ANSWERS 13log3 (i)(+8e+3e*) +y+s1 s 1. (i) 9 6 8 a 0+b.0 +c 0=0 (vi) 25007 ()log 2 (v) 3. 24 52 Cs-W FOR SPECTRUM MATHEMATICS. 4.6. Jacobian of Two Functions 15 48. Change of Variatble the volume V of Variables in Triple Integral in xyz space is mapped into the region V' in u v w space by the transformations IT u and v are two functions of two variables r andy possesing partial derivatives af MULIPLE INTEGRALS the first on w), z fi(u,v, w) u ox is called the jacobian of u, v w.r.t. x. y. Ox,y,z) a(u, v, w) oy oy is know as Jocobian of the transformation. w Jacobian of Three Functions If u, r and w are three functions of three variables x, y and z poSsessing partial derivatives a where 49. Cha to indrical Coordinates Let OM=r, XOM =0. Then x=r cos 6, y= r sin 0, z = MP order then the f Ou u ou Let P(r, y, ) be a be any point in the region V. From P, draw PM L xy-plane. Joint OM. Ow (r cos 6, r sin 8, 2) are called cylindrical coordinates of P. ZA is called the jacobian of u, v, w W.r.t. x, y, z. cos -r sin 6 0 r cos sin Pr cos , r sin 6, ) 4.7. Change of Variables in Double Integral Let the variables x and y in xy- plane are transformed to u, v n iw - plane by using the transforma p (u, ) and y = y (u, v) 0 0 Y The Sx. ) d dy changed to || 1ou, v), v(u, v)|J| du do =r cos 0 + r sin 0 = r (coss 0 + sin* 0) =r (1) =r R Here V= V R' Where R is the region in xy- plane and R° its transformed region in u v-plane. Moreover the term | J | is given by :JJfx,y.2)dt dy t = ]]]fr cos , r sin 6, z)r dr d9 dz 4.10. Change to Spherical Coordinates Let P x,y, z) be any point in the region V. From P, draw PM Lry-plane. Join OM. Let OP =r, | XOM=8, |ZOP=¢ dx dx = Ox)= du d known or Jacobian of transformation from (, y) to (u, v) co-ordinates du, v) du Note: Change to Polar- Co-ordinates Here x= r cos 6, y =r sin 6 x=rsin Ù cos 6,y =r sin p sin 6, z = r cos o Or sin o cos r coso cos -r sin osin 6| = sin o sin rcoso sin 6 -psin dx J= O,)by a (r,6) dx cos dysin6 rsin ocos 0 sin -r sin r cos =r cos Or sa.)dr dy- ||fr cos0, r sin 0) r dr de Here V V R R' Jr. 2) d* d de - [fresin øcos0,r sin gsin Or cos $ sin g dr d d 154 SPECTRUM MATIEMATICS-1U 155 a sphere. Note 1. 1The polar spherical coordinates are useful when the region of integration is a part of a MULTIPLE NTEGRALS T 2cos 4- cos 0 -r sin 0 r dr de ZA 2 cos6 0 Y 372 cos0 3 4-4 cos0)2 -(4)2 ld0- ;sin'-1) do 0 Note 2. Under these transformations V {a, r. :):* +yf +: s a} is mapped onto V ={(r.6. p):0 Srsa, 0 s6s27,0 sp s7}| 8 sin 0 d0+ d ILLUSTRATIVE EXAMPLES b2 dr dy over the positive quadrant of the ellipse Example 3. Evaluate Example 1. Evaluate sin z (x+)dt dy over the circle x + s Sol. Here the region is A= {a ) :+ys l} Changing to polar co-ordinates by x = r cos 6, y =r sin 6, it becomes sa Put -=v ie., X=au,y=bv A-r. 0):0 Srsl:0 s0s27} where x +y=r dr a du, dy = b dv 2T 1 sin 7(r +y') d dy Ifsin er yr dr db- sin ar2.27rdr =I in the positive quadrant of xy-plane transforms into the circle uf +v' = ellipse )0 0 in the positive quadrant of the u v-plane. --cos TrI=- [cosn-cos 0] =--1-1]-2. JO)ou (u,v) Example 2. Evaluate 4-x* -y dr dy over the region bounded by the semi-circle x +y-2r b and the coordinate axes lying in the first quadrant. Sol. () Consider the circle x+ = 2x or x* +-2x-0. Its centre is (1. 0) and radius =1 b ab du do Put r=r cos , y = r sin Pr,6 given circle becomes r cos +r sin 0 = 2r cos 0 i.e., r= 2 cos 6 where A is the region u 2 0,v2 0, u + v s1 region of integration A is shown in the figure and for this O Put u =r cos 6, v =r sin 6 region 6 varies from 0 toand r varies from 0 to 2 cos 6 156 SPECTRUM MATUEMA cu cos6 157 rsin = r (cos + sin )= r MULTIPLE INTEGRALs sin -( dy d rcose Sol. )Let JJ., YA Here r andy. Putrr cos 6,y =r sin 0 and y both varies from 0 to oo Also A r, 0):0srs 1,0s0s + rand tan 0 ie. 0 tan(ylx) 72 1-cos 6-r2 sin2 o varies 0 to oo and 8 from 0 to T/2 2 dex rd [Put ? =t 2r dr = di 1+r cos 0+r2 sin ?e abr dr do dy d - r drd | 0 0 0 T2 Let i) Let 0 0 Putx = r cos 6, y=r Sin & so that dr dy = r dr d9 Put t 2rdr Here r varies from 0 tol and 8 varies from 0 to 4 Ir/2 T/2 -sin 0-)-sin1-sin o- de 0 0 0 Fample 5. Evaluate | (x+ drdy where R is the parallelogram in the xy- plane with vertices (1.0), R 0,1, 2, 2), (0, 1) using the transformation u = x+y and v =x-2y. Sol. The region R and R before and after the transformation are shown below: d dy ab5- a-2 b2 Y R(2,2) P(1,1) Q4.1) Example 4. () Change into polar Co-ordinates and evaluate R (P.T.U. 2011 s(3.1) O u 1 0 Q(0, 1)k (i) Evaluate +y") dr dy by changing to polar co-ordinates. (P.T.U. 2010 >X O P(1,0) S' (1,-2) U-2 R(4,-2) SPECTRUM MATHEMAT MATICS-M ON 158 159 ALS NOW the given transformation is given by u = x +)y and v = x - 2y. om (1) dr d= de dy » de dy =rdr do .3 sin Gcos @drd 0 = rsin 6) (r cos 6) r drd 6) MULTIPLE INTEGRALS Solving it for x and y, we get x (2tt):y=u-) J-C)-u 2/3 3-/3 0 0 -1/3 a ya?-? ax dy de which is in Cartesian form. c(u,U) OU a y=0 7va dr R -21 R 2-x*)dt=0 Sr d=0, when f(x) is odd 7[], = 7x3 =21 a ample nle 7. By changing into polar co-ordinates, evaluate Example 6. By changing into Cartesian coordinate evaluate sin cos 0dr de 0 Sol. given 0 ses7 and 0 S r sa which is upper half of the circle. log (+y* +)dt dy Letr = r cos and y =r sin 6 Then =x+ and tan = yx cal. Given integral is putx = r cOs 6,y= rsin 6 drdy = r dr de ra, .r+y =a Here x varies from -yl- to -2 cydx dy orx l-y*or*+y=1 or 1 r=1| Also dr d=|J| dr dy A ..(1) 0 0 and y varies from -l to I, :. r sin 6= - 1 and r sin = 1 or sin 6 =-1 and sin & = 1 [: r=l] Nowr r+y varies from and and 0= tan - I log,r+ +1) dt dy = -7/2 0 log,(r+1)r drde Putr+1 =1> 2rdr = dt; when r = 0, 1=1; when r=1,t=2 and 2ns -k]- 5121,2-1)- (2 1og, 2-1 SPECTRUM MATHEMATI FoR 160 61 s3) Example 8. Evaluate [[[: (r h d de i; where V- {n 2):**y's \.25 :s1 MULTIPLE INTEGRALS Let = dr, Put r= sin t, dr = cos t di so we Cnange to cylina O VI- Sol. Since the region of integration is a part of a right circular cylinder, so we chano coordinates, r = r cos 8, = r sin ,: = z and dr dy d: =r dr dd az when r= 1,t=0; when r = 1, V(r. O. :):0 srsi,0s0s27,2 s:s 3} 0 Now 2cos 0+r sin 0).r d: de dr Nuo-0 V 0 Also sin do = [-cos o =- cos+cos 0=1| 0 2 0 yd=T'abc 4 Example 10. Prove that dx dy dz Example 9. Evaluate JJJ-?-2- = over the positive octant of the sphere x* +y+?s cS where V (a Sol. Since the region of integration is that part of the sphere of radius I and centred at origin which the positive octant, region of integration is which Sol. Put = V. - W C x au, y = bu, z = Cw so that dr = a du, dy = bdv, dz =c dw dr dy dz = abc diu dv dw V a, y. 3):r2 0,y 0,: 2 0, r+y+#sl} region of integration V transforms into V" = {(u, v, w): uf +v +ws1} Changing to spherical co-ordinates by substituting x=r sin o cos 6, y=r sin o sin 6, z =r cos o and dt dy de =* sin d dr dB dip -dr dy dIJ-u2-v-w* abe dudv d V' V-.4.0):0sr, 050ss,0 s0s% -rsin dr d8 do 0 0 00 sin o dr dd dp 2 Now, Changing to spherical coordinates by substituting u= r sin o cos6, v = r sin o sin 0, w= r cos o 0 0 0 162 163 Ics-W FOR. CS sPECTRUM MATHEMATICS-I MULTIPLE INTEGRALSs Let I 1-rr dr, Putr= sin , dr = cost, de 0 T2 co d42 T2 V T AA 1= -sin r. sin .cos/ dt=sin -cos t di rty 1 x =0 - u =v u 27-0 27 Oy0 A Also, ov=0 Also sin o do=-cos ol = -cos 7+ cos 0 =l+1 =2 Now r-y = u, X*y=v 0 (u + v), y= (v-u) abc2 2=T"abe 16 4 When xF0, u t U=0 > u=-j 4 T When y 0, V-U 0 u=v over the region x" +y' +s1,m> Whenx+y =1, v=1' regjon of integration E transforms into E = {(u, v):0 S vsI, -v SuSv} Example 11. Show that ||+y+" dr dy d = 2 m +3 Sol. Since the region of integration is a sphere bounded by the sphere x* +yf +z= 1, so we chas spherical coordinates by substituting x = r sin o cos 6, y =r sin o sin 6, z =r cos o so that change to Ox dr dy d: =r sinp dr d do Also (u, v) V={(r.9.0):0 SrS 1,0 so S7, 0 s6s27} 1 2z ff "adye =[[ f*y"*sin dde dB dr 00 0 r+ +#=r sin o cos' 0 + r' sin ó sin' 6 +r cos p= 2m+3 vtsin v{sin 1 -sin (-1)} d (sin 1+sin 1) dv 2m+ lo 0 4 1 x 2nx2 = 0 (27-0) (-cos 7 t cos 0) = sin 1 [vsin 1 (1-0)= sin I 2 m +3 2m+3 2 sin I do = sin 1 do Example 12. Using the transformation u =x -y, v=x+y, evaluate coxd dy over the .Evaluate J1 V1422 dy over tne positive quadrant of the circler+=1. EXERCISE 4 (e) region bounded by the lines x = 0, y = 0, 1 =x +y. (P.T.U. 2017) Sol. Let Evaluate -x-y drdy overthesemicirclex +y*= ar in the positive quadrant. and E- {(x, y):r=0, y=0, x+y = 1} 4 SPECTRUM MATHEMATICS-J l FOR CS 165 3. Evaluatef(. y) dA as a suitable iterated integral where ANSWERS MULTIPLE INNTEGRALS () fa. v)= e*and A = {a. v);x+*=1} () fr. 1)=- ndA):r+y's1} 2 (i) 0 Ciinna 7(e-1) (u) ft.y)=(+* and A- {a,y):*+ysd} 6. -2 4. Evaluate a)d dy where A is the region bounded by the four hyperbolas 8. 13. 0 r-2) 4 and xy 2, 4. here a> Evaluate ||a-2-y? ddy over the circle x +y s axin the positive quadrant where Area by Double Integration (0 In Cartesian Co-ordinates Area enclosed by the curves y =Ji () and y =/s(r) and the ordinates x = X1,r=X2 is given by 6. Evaluate d dy over the positive quadrant of the circle r+y=1. YA yJ2r) 36-4x9y Evaluate 36+4x+9 1. d dy over the region bounded by the ellipse 4 x +9y =36 am the coordinate axes lying in the first quadrant. xX1 X2 s. Evaluate :+y+2)aN where V= ,y.x+ysa',0s :sh 4T y/r) 9. Show that ||| dk dy dz=where V= {ay):*++?s 1} 15 Area- dy de dr dy dz Show that ) +2 + an (4-7) over the region x* +y+zsd. and Area enclosed by the curves x =fi(y) and x =f0) and the ordinatesy = y, y=y) is given by: 21T 1. Using the transformation x +y= u, y = m, show that d-x-) dsdy integatia 105 being taken over the area of the triangle bounded by the lines x =0, y = 0, x +y= 1. yy2 12. Evaluate e* dy dr by using the transformation x + y = u, y = 4v. /0) dy dx 13. Evaluate sin |dr dy, where E is the region bounded by the coordinate axes and x+y= X+y E in the first quadrant. 167 losed by the parabolas 4 av and 4av, a>0. rea enCo of the two parabolas The cqalions of the twe dar are amye 2. tnd the. (an In P'olar Coordinates ad ay The area ot the rogm boundot by the curves r (ani the lines P-is given by thom (24a -6-4 a 0 4 -4 a) -0 from (3), r -0, 4a parabolas () and (2) Intersect in (0, 0). (4 a, 4 a). 0, 4a ILLUSTRATIVE EXAMPLES The region of integration is(, y):0srs4aSys 2Na hampk 1. Using double integration, find the area enclosed by the ellipse + rcquired area Sol. The equation of ell1pse is We know that ellise 3 =I is symmetrical about both the axes Aiso in first quadrant. - -ya* -a ,0 sr Sa required area 32a l6a l6a 0 3 3 Example 3. () Find the area bounded by the parabola y x and the line y r+2 ) Find the area of the region bounded by the curves a 0 2- and y= * Sol. The equation ot the parabola is a-k -4 Na-, sin The equation of line is r+2 From (1) and (2), r+2-r 2 0 (r 2)(r+ 1) =0 1,2 2) 2 a -0 o.sino -Tuh y1,4 region of integration A is given by A {0. ):-I Srs2,r' sys2} 168 SPECTRUM MATHEMATICS.m area fa d [t- Jr2-s)a 2 RALS 169 MULTIPLE INTEGRA &. () Using double intergration find t the area bounded by the curves x =2y-y andx = y. Framph hle integration, find the area enclosed by the curves y = x,y =x. (P.T.U. 2011) authle integration find the area of the region bounded by the lines x = - 2. x = 2 and the 2 (n Using double ) Using double nuation y = x is a parabola with vertex at (0, 0) also a parabola with vertex (1, 1) and can be 19307 circle r+y=9. sol. (9 The equation y= ri andx =2y- writen (-1)=-a hen will intersect when =2y- 1, 1) 6 6 6 () The equation of the parabola is =2-* The equation of the line is y=x From (1) and (2), x =2 -x*r+x-2=0 (r+2) a-1) = 0 =1, -2 region of integration A is given by A = {(a.): -2 srsl, x sy s2-x} O - 1) *=1, -2 A(1, 1) o 2-2y=0 A 2-x 2y0-1)=0 or y=0, 1 - x) dx Put y 0, 1 iny =x, we get x = 0, 1 Point of intersection are (0, 0) and (1, 1) ie Example 4. Find the area of the region bounded by x = 0,y =0, x +yf= 1,y= -I dcd-o- fey-y*-)a- Ja-23)d The required area 0 Sol. Since r+ysi 0 0 sI-y as x20 Area 2 fo-y)4-2 A-.y):0sys;,0srsVI- 0 where A is region of integration. The curves are y =X and y = x and will intersect if y=y +=1 x=0 y y(1y) =0 or y=0,1 >*=0, 1 required area = Point of intersection are (0, 0) and (1, 1) 0 y0 0 2 A(1.1) -a-)d 1 0-0 O 522 170 SrECTRUM MATHEMATIO 171 (1) The equation on the cirele + 9 and equation of lines arer-2 andx 2 Area of required region =4 times area of region OPQR e 7. hydouble integration, the area lying inside the circle r = a sin and outside the cardioid MULTIPLE NTEGRALS 9- (P.T.U. 2014) Example dr (from figure) a(l-cos e). (1) Sol. quation of circle is r=a sin 0 2) Equation of cardioid is r= a(l-cos 6) rom (1) and (2) a sin 6= a(l-cos G) sin 6 = 1 - cos sin 6 + cos 6= I Squaring both sides sin + cos 6 +2 sin 0 cos = 1| 4 sin 45 18 sin ' 2/3 1+sin 29= 1 sin 26 = 0 Example 6. Find the volume bounded by the cylinderx *ys4 and the planes y + z=4 and =o 20-0 to z (P.TU. 2011, 201 from Sol. Here. we have to integrate : =4-y over the circle x" +j=4 in the xy-plane. Here r varies 0-0. 4-while y varies from 2 to 2. varies from 0 to also F varies from a (1 - cos ) to a sin 2 ics The given volume V is given by V={ (a. y. :) : T 2 sy s2.0Sxs V4-».0 s:s 4 -y} rdr d Required area = 0 a(l-cose) Ja(l-cos@0) Required volume - a2 sin2 6-a(1- cos 6)]d6 4-) dd -2(4-k*d T2 sin2e-1+2cos6 - cos e)) de - (2 cos-2 cos* 0) de 2fa-ya-y a- 2[4/a-2d-2»ya-y2 -2 -84-ydy 20) yy4-y is an odd function'f)a=0|Sol. The equation of the lemniscate is = cos 26 Example 8. Find the area enclosed by the lemniscate r =a cos 20. Required area is shaded in the figure. 19fya- -1 1o- sin 1] = 32 x = l67. T4 acos 20 Required area 4 dr do 0 r 172 SPECTRUM MATIEMAT MULTIPLEINTEGRALSs olume by Doub ycos 20 cos 2e 173 4 de 2a |cos 20 do uble Integra 412 In Cartesian Co-ordinates 0 tes Surfacez =fx. y). Let the orthoganal projection of its portion S be the Area S. Divide S into piangles of Area Ôxðy by drawing lines parallel to x From each of these Rectangle make a prism of 2 nsin o (0-0)-d. Consi elementary rectangles ofArea y - axis. Fro on Note :Using double integration, find the area of one loop of the lemniscate r=a cos 29 andy length parallel to O2. z/4 acos 20 PTU. S and the Surface z =f (x, y) Sol. Required area = 2 Jrd of this prism between Volume of this pris isz drdy. EXERCISE 4 (f) 1. Find the area of the region in the first quadrant which is bounded by the parabola 2 Volume of the Solid cylinder on S as base, bounded by the Surface with generata parallel to the z - axis the line x =2 a. Lt PI. y)= 0 4 axa ) Find the area lying between the parabola y = 4x-and the line y = x. () Find by double integration, the smaller of the areas bounded by the circle 2+ 2 o0 the line r+y= 3. 9 a cvlindrical co-ordinates, volume = ||zr dr do which is also volume in cylinder co-ordinates. (i) Find the area bounded by the parabola y = x and the line y = 2x +3. 3. Find the smaller of the areas bounded by the ellipse 4x* +9y= 36 and the straight line 2r+3 4. () Find the area bounded by the parabola y = 4-x and the line y -4-4 x. 1LLUSTRATIVE EXAMPLES 3y=6 (i) Find the area enclosed by the curves y= and 4y =x* (P.TU.201 Example 1. Find the volume of the ellipsoid 5. Find the area outside the circler= a and inside the cardioid r = a (1 +cos 4) 6. Using double integration, find the area of the region bounded by the curves y = 0, x* +y =l (P.T.U. 2012) Sol. The given ellipsoid is y 2 (1) 7. Find the area enclosed by the cardioid r = a (1 + cos 6). 8. Find the area bounded by the parabola r = 2 sin 6 and r=4 sin 6. Since ellipsoid is symmetrical about the axis. volume of ellipsoid is 8 times the volume in the first octant. Now in the xy-plane, z = 0 so, the portion of ellipsoid ANSWERS 1.2 becomes 1, z = 0, which is an ellipse. Here x 9 2. 0 3 (in7-2 in 3-) vary from 0 to a and y varies from 0 to b, 4. () 8 (in log3- 2 5. 8+7) 6. 3na 4 7. In the positive octant z is c,| 8. 37 4 SPECTRUNM MAIIEMATO SU 175 Hence the volume of the ellipsoid = 8 | MULTIPLE DNTEGRALS Eample 3. Find the given cylinder are r+y= d Find the volume common to the cylinders r+y = d and?+? =d. a d ..() dd, where 2) and (0, 6) varies from 0 to a and y varies from 0 to va- r andz = va2-2 In first oclant r wcected portion of the surfaces represent the required volume. The intersected va J Required volume = 8 a, 0) 0 Jo ya-x -od 0 32bcl ae Example 2. Find. by double integration, the volume of the sphere r +y +z=9. Sol. The given sphere is r+y+#=9 Since sphere is symmetrical about its axes volume of sphere = 8 times the volume is first octant. A. Find the volume of the Tetrahedron bounded by the Co-ordinate planes and the plane Example (P.T.U. 2014) In r-plane, z = 0, so sphere becomes circle x +y =9, z = 0. Given tetrahedron is 0) Here x varies from 0 to 3 and y varies from 0 to 9-x The required volume is given by In XOY plane (1) becomes =I for which 9 92 |x=3 -8 dy de =8 9-- dy de r varies from 0 to a and y from 0 to b -and 0 0 Put r =r cos6, y = r sin 6 ; dr dy = r dr de9 where 0r s 3,0 s 6 s B a n/2 a Volume= -2r) dr. do The required volume dy dr 0 - -2719 a b(l-x/a) dydr Jo 176 SPECTRI IMATENMATIOS MUTIPLE INTEGRALS In y-plane, z = 0, 177 from (2) r+ = I -I In first quadrant, 6 varies from 1 of integration becomes R = {7,0): 1 srs2,0 s0sI2} region be| (1- v a)' 0(0-1) = abe v-8J-V? rde do4 6 6 do 3/2 JO L m2 Example 5. Find the volume common to the sphere x* + y +z =a' and the cylinder + Sol Given sphere is a + 437 2 = ay. 0 Given cylinder is + =ay The required volume is the part of the sphere x' ++* = a' lying within the cylinder. Due to symmetry of the sphere, half of the volume lies in the first quadrant and half lies in the second quadrant of the plane. Moreover half above and half below the ay - plane. EXERCISE 4 (g) 2. Find the volume of t and the ry - plane. Eind the volume bounded by the xy-plane the cylinder +- l and the planex *y+2 =3. the cylinderr+y-2ax =0 intercepted between the paraboloid r+=2a: Required volume = 4 d d, where : = ya--y ( Find the volume of the solid bounded by the surface f t. y)=4- the planes r- 3. 3. y=2 and the three coordinate planes. ( Find the volume of the solid in the first octant bounded by the paraboloid : = 9- r-3y Find the volume under the plane : =x +y and above the area cut from the first quadrant by the Put r =r cos y = r sin & so that dr d'=r dr dt and 0 sr sa sin 0. 0 ses: X ellipse 4 x +9y* = 36. find the volume of the culinder +y -2 ax=0 intercepted between the paraboloid asin& 72asin60 Volume=4 Va-rdrde=4 +2az and the xy-plane. 0 a sin & 6. Find the volume bounded by the xy-plane, the paraboliod 2 x=x*+y and the cylinder +y* =4. cos' e-ahdo 7. Find the volume bounded by the plane z = 0, surface = x+ y +2 and the cylinder +yi =4. ANSWERS a (Bn-4) 4 3 3. (2 (in 95 Example 6. Find the volume bounded by the cylinder x +y* = 4 and the hyperboloid r - y-z=1. Sol. The given cylinder is r+ =4 and given hyperboloid is x+y-2= I Since the intersected region bounded by the curves (1) and (2) is symmetrical about axes. 4. 10 6. 47 7. 16 7 4.13. Surface Area by Double Integration Let a function =f , y) be continuous and having continuous first order partial derivations over a closed region A of the x y-plane, then the area of the surface z fx. y) of which the projection on r y plane is the area A is given as Required volume is V=8||: dr dy, where R is the region in first quadrant R 8 - )dk dy R Putr=r cos 6, y=r sin ; dr dy =r dr do from (1), we get, = 4 =2 Surface area S = 178 SPECTRUM MATHEMATI 179 ILLUSTRATIVE EXAMPLES IPLE INTEGRALS MLZ 4 9 lying inside the s. Surface area S ample 1. Find the arca of the portion of the surface of the sphere a+ +?-9 lyinp he cyine SolL. The projection of the one-fourth of the desired surface area in the ry-plane is the semicircle x + = 3y in the first quadrant as shown in the fig. For the surface of the sphere +y+=9, where. A = ((x. y); x* +y* s4, 4 4- dx 4y-y dy S we have 4y-y 0 Thus 9-x y2 2(0-0)+2 {sin- sin(- x 2 Thus, the required surface area S is EXERCISE 4 (h) 3 dr dy, where R is the semicircle x +y = 3y in the first quadrant ( Find the area of the portion of the surface of the sphere x+ y + =d that lies in the first octant 9-x* -y cn Show that surface area of the sphere r++ =d is 4r a ics In terms of the polar coordinates x=r cos y=r sin 6. the region is ( Find the area of the surface of that portion of the sphere r +y+=d that lies inside the cylinder x +y = ax. Gn Find the area of the portion of the surface of the sphere r +y+ 9 that lies inside the cylinder x+y*= 3x. 3. (0 Find the area of the cylinderx+ =16 lying inside the cylinder x+y= 16. Find the area of the portion of the surface z =y-* which is cut by the cylinder - y*-d (in) Find the area of that portion of the sphere x+y+= 2a that is cut out by the upper nappe of the cone x* +y' = z R= .005rS3sin G.0S 0s 23sine I3Sin T2 s 4 rar 3 d6= - 12 flo- de 36 (1- cos6) d9 36 0-sin 6 18 (7 2) sq. unit. (iv) Find the area of the portion of the cylinder x + y = 4 y lying inside the sphere Example 2. Find the area of the portion of the cone x + = 3z lying above thexy-plane and insidete cylinder+y-4y. +y+-16 Sol. The projection of the required area on the x y-plane is the region A enclosed by the circle r* +y=4 For the curve 32=x +y, we have, ANSWERS 6 2x, 6:2y or , E 2. 2a (7 -2) ) 18 (7-2) 1 94+y_9: +32124 3.) 128 (iv) 32(T -2) 9 924 181 ence tdnl. where d cVnENis tc, e, c,... 1ts range is t c).This sequence is a 5 SFOUENCES 4. constant sequence. The sequence The sequence (dni, where an=(-1" 5 SEQUENCES ..-)n+1 5.1. Sequence th and nth terms dm and an Tor mznare treated as distinct even if am an ie.. the terms Definition. IfN is a set of natural numbers and X any set, then a function f: N X is calle. Note 1. The n 1. Tne teretnt positions are treated as distinct terms even if they have the same value. sequence or a subset of C, t If X - Ror a subset of R. then fis called a real sequence and if X =C or a subset sequence {ani. the order ot elements cannot be changed. occuring at differ called a complex sequence. then med with real sequences only. Hene In the remaining part of this chapter, we will be concerned with real sequences sequence. we will mean real sequence. The number of terms in a sequence is always infinite where as the range of a sequence may have a te 2. finite number of elements. netimes a sequence has the zero-th term. In this case, its domain is NU {0} so that the Note 3. Sometimes a aulence is i{do , d, d2 an.....} Thus a sequence is a function whose domain is the set N of natural numbers and range i any set x Let a: NR be a sequence. The image of n E N. instead of being denoted by ) gchen this mapping. anis ca are the real numbers associated to 1, 2, 3, ... by this mapping, or an in 2 0 tilarly, we can start a sequence from any positive integer m and in this case, the sequence is written as denoted by an. Thus a1. a2 a3 the general term or the nth tem of the sequence. If the nth term a, of a sequence is given, we can find the first, second, third, terms of the sequeno an In z Note 4. Equal sequences. by putting n = 1, 2. 3. an> Thus a sequence whose nth termm is an is written asianwhere n EN or {an} or u) or Two sequences {an} and ibni are said to be equal if an = b, for every n. Sometimes. it is denoted by writing all its terms within the brackets Tf ranges of two sequences are equal, even then the sequences may not be equal. Le. a a , a3, an .. a=-1, bn=-1)"* Let Another definition ranges of an} and tbn} are equal A set of numbers a , a2 , a3 an Such that to every positive integer n, there correspon number an of the set, is calleda sequence. But a b as a - 1, b = 1 two sequences are not equal. a, az. a;.. are called the elements of the sequence. 5.2. Bounded and Unbounded Sequences Range ofa sequence ( A sequence {an is said to be bounded above if there exists a real number k such that The range of a sequence {an} is the set of values {an , nE N} consisting of distinct terms, withou repetition and irespective of their position. ank nEN. kis called an upper bound of the sequence {an In other words, the set of all distinct terms of sequence {an} is called its range. Examples of sequepces (i) A sequence {an} is said to be bounded below if there exists a real numberh such that hs an V nEN. his called a lower bound of the sequence {an 1 The sequence {an}, where a,= n is {1, 2, 3, ..., n,.. (ii) A sequence {a,} is said to be bounded if it is bounded above as well as bounded below i.e.. if 2. The sequence {ani, where an is there exist two real numbers h and k such that hS a,s k nEN. 3. The sequence {an). where a (-1)" is {-1, 1,-1, 1,- 1, 1...}. Its range is 1, 1 (v) A sequence {a,} is said to be unbounded if it is not bounded. 180 IS2 SPTCTRMMAmEMA 183 ASequcnce u} is said to be unbounded if given A> 0, however large, 3 m EN S such that SEOUENCES Note 1. A sequenee is bounded above. bounded below or bounded according as its range bounded below or bounded. Eramples is bounded a Consider the sequence Note. Ifa sequence a, is bounded. then its range is bounded. o however small, e choose a natural number i such that m Givene0, however smal Note 3. () tf each d,>0 and sequence tal is unbounded. then it is unbounded above n2m, (7) If each dn0 and sequence {a is unbounded, then it is unbounded below. , Evamples. () Consider the sequence tai defined by o Given F>0, however small, we choose a natural number m such that nEN n m From(1) and (2). 0< sI Y n¬N Vnz m, we have an is bounded. H , L.. (i) Consider the sequence {an} defined by an n Vn2 m sequence is {1. 2, 3. ... n,...} Here a 2l 7nEN, but 3 no real number k such that a, k Lt =1 the sequence {a,} is not bounded above. n on+l (ii) Consider the sequence {an} defined by an-n. sequence Converges to I. n+ It is bounded above and 0 is an upper bound. But it is not bounded below () The sequence {an} where an=(-2)" is neither bounded above nor bounded below. Note 1. A sequence {an} is said to be convergent if Lt dn is finite 5.3. Convergent Sequence Note 2. A sequence (a,) is said to be convergent if ay converges to some limit, otherwise (an} is said tobe A sequence an IS said to converge to a limit 7, if given e>0, however small, there exists a positie OE6 integer m (depending upon e) such that |a - | 0, however large, there exists a positive integer m (depending upon k) such that l is called a limit of the sequence {ani and we write it as Lt anl or Lt an = ak n2 mn an as n > o or Simply an. We write it as dn Lt an o or n 184 SPECTRUM MATIEMA 185 )A Sequence {a,} is said to diverge to -o if given k> 0, however large, thero nteger m (depending upon k) such that an<-k V n2 m. StOUENCES Let m= Max. (m, , m) here exists a po Nofe. A sequence ia,} is said to be divergent if Li an is not finite for n2m ..(1) and a -lr| k+2 m 11-"E for every e >0 and Vn2 theonly non -negative realnumber whichis lessthanal positive numbers is zero an-k V n2m wherem k+2, mE N 1-I=0 sequence {an diverges to - (i) Considerthe sequence {an }. where a=n+ 3 n. = our supposition is wrong. ifa sequence is convergent, then it converges to a unique limit. Let k be any positive real number, however large. Now if +3n>k 55. Prove that a convergent sequence is bounded. Is its converse true ? i.e, if >k Proo Let the sequence4n } converge to l. iven e>0, however smal, there exists a natural number m such that Le., if n> Vk Vn2m ifm is a positive integer such that m> Vk ,then an>k Vnzm an e sequence an diverges to + -E K an <[t8 n2m Note 1. A divergent sequence is always unbounded. Let h = Min. ((-6, a1 , a2 Am-1) and k= Max. (+E, aj , az,.. Am-1) Oscillatory Sequence (P.T.U. 201 h an Sk VnEN A sequence which is neither convergent nor divergent is said to be oscillatory. sequence {an} is bounded. Note. () A bounded sequence {an} which is not convergent is said to oscillate finitely. The converse is not true i.e., a bounded sequence may not be convergent. (i) An unbounded sequence {an} which diverges neither to + o nor to infinitely Consider the sequence {an} where a,n = (-1" sequence is{- 1, 1, - 1, 1, - 1, 1,...), which is bounded as I and 1 are its Lub. and glb. 0o is said to oscillæ Examples : () The sequence{(-1"} oscillates finitely. respectively. (i) The sequence {(-1)' } oscillates infinitely. Now we will prove that {an} is not convergent. 5.4. Uniqueness of Limit of a Convergent Sequence If possible, suppose that (an} converges to l. If a sequence is convergent, then it converges to a unique limit. givene >0, however smal, there exists a natural number m such that Proof: If possible, suppose that the sequence {an} converges to two different limits land '. an-| *e Vnz mn Then givent > 0, however small, there exist natural numbers m and m2 such that (1) | am-1| KKe an-l | for n2 m and dn-| for n2 m2 and 4m+1-I| 0, however small, there exists a natural number m such that Prool.( given. Note 1. The converse of () is not true. Vn2m | an-1|e n .n is odd m -E an 0, there exists m EN such that -2n, n is odd Consider the sequence {an where an u- am ..(1) n is even Since the sequence an IS monotonically increasing 2n The sequence is bounded above as an 0VnEN and unbounded below, but it does not diveree to amS an for every n 2 m 2) From (1) and (2), u-e < an n2m Note 3. The three behaviours of a sequence namely convergence, divergence to + o and divergence to- are mutually exclusive i.e., only one of them is true at a time. Also an ute7neN 5.6. Monotonic Sequences in particular an u+& Vn2m (i) A sequence {an is said to be monotonically increasing if an an + 1 nEN. -e < an 0. however small, there exists a natural number m such that (V) A sequence (a,) is said to be strictly monotonically decreasing if an an+ 1 nEN. an-| 0. there exists m E N such that (1) (an) is monotonic decreasing Since an S Am or n2m aml+E..(1) Since the sequence {an} is monotonically decreasing (2) an Sam for evern2 m From(1) and (2), m an-k n2 From (1) and (2), dnl-e VnEN in particular an>-E n m creasing sequence is either convergent or diverges to - o Note monotonic decrea A -E an 0, however large, there exists a positive integer m such that 58. Algebra of Limits .) and {bn be two convergent sequences and Lt a, a, Lt b,n = b, then ank V n2m there are infinitely many terms of {an which are greater than k iCS sequence {an} is not bounded above. 9Lt,=|a| (i) Lt-a,)=-a n n Converse. Assume that monotonic sequence {an} is unbounded above given k> 0, however large, there exists a term am Such that (i Lt (ka,)=ka, where k is any constant 1 (iv)Lt a, +b,) =a+b k Lt ,)=a-b (vi) Lt (a,b,)=ab Im (V) n Since {an} is monotonic increasing (wi) L b,0, bz0 (viii) an 2 am for n 2 m b, 0, ba o. From (1) and (2), nob b Note. If Lt |an eXists, then Lt an may not exist. n ank n 2m n0 Example. Let an =(-1)" Lt an does not exist as the sequence {ant is not convergent sequence {an} diverges to +o () Assume that sequence {an} diverges to - o n o given k>0, however large, there exists a positive integer m such that But a -1FT nEN an-kVn2m there are infinitely many terms of {an} which are less than - k. Lt an=1> Lt an exists an is not bounded below. Note. If Lt (4n t bn) exists, then Lt ay and Lt bn may not exist. n0 n

Missbrauch melden

PDF-Details

Kontaktdaten

m3-upto199.pdf