SPECTRUM MATHEM MAVICS-I11 100 product P has maximum value at MAXIMA AND MINIMA EXERCISE 3 (a) A and =k-*-y noints of extreme values, if any, of the function f defined by Sx, y) =x -3xy-y - 4. Find extreme values of the function ( Sa.)=x *y'+ 3y» (ii f. y) = 3x + 4xy +y+r - 1- in fx.y) =* - y+y -2y+ 3x -7 (iv) fr,y)-2 - 3xy +y (vi) fx,y)= 3 x - y2 +x Show that (0, 0) is the only critical point of the function f(x, y) =y+ry+ ' and that fhas a y () Sa,y) = x* +y* +6 x+ 12 product is maximum when the numbers are equal. (P.TU. 2016, 2017, 2 Exaryple 13. Find point on the surface =xy+ 1, nearest to origin. Sol. The equation of the surface is :=xy+l minima at that point. ind all the critical points of the function fx, y) = x +xy and examine for maxima, minima 3. (b) or neither. Find the extreme values of the following functions: (*+8y' +6xy Find the local maxima, local minima and saddle point, if any,of the function Let P(x, y. :) be any point on surface (1) and origin be O (0, 0, 0) 4. (a) u =| OP | = ya +y2+:2 (i) 12x+8 xy+y + 8 Let =x+y+? fa) 2xy-5x* -2y2 +4x-4 Let f.y) =x+y+xy+ 1, where f«.)= Find extreme values or the following function () x, y) =y (a-x-y) Show that f(x. y) = -* **- 2)" has a minimum at (2, 2) Find the extreme values of the function sin x sin y sin (r + y) 2x+ 2yx in Sx, y)=y+xy+ a* S For Stationary points -0 find the values of x, y, z for which fr, y, z) is maximum subject to 2x+y = 0 Iff , y, 2) x+ 2y +4z 8. and 2y+x =F0 condition xyz = 8. Adding (2) and (3), we get 3 x + 3 y=0 9. Divide 24 into three parts so that continued product of first, square of second and cube of third x+y=0 may be maximum. Subtracting (4) from (2), we get x = 0. The temperature at any point (r, y, z) in space is 400 xyz. Find the highest temperature on 10. surface of unit spherex +y' +z=1. 11, Show that if thee perimeter of triangle is constant its area is maximum, when triangle is equilateral. (P.T.U. 2011) Subtracting (4) from (3), we get y=0. we have x=0, y = 0 ANSWERS ow A 2.B - -1,.c-2-2 1. No extreme point ox 2 () Neither maxima nor minima at (0, 0); maximum value lat (-1,-1) At (0, 0) A =2, B = 1, C =2 AC B = (2) (2)- (1 =4 - 1 =3>0. Also A =2 >0 (i) min. value (0, 0) is a point of minima From=xy+ 1, when x = 0, y= 0, we have z=1 z=t (i) At (0, 0), neither maxima nor minima; min. value = required points on the surface are (0, 0, 1) (iv) At (0, 0), neither maxima nor minima (v)minimum value of= 3 at (-3,0) () maximum value of = 4 at (-2, 0) SPECTRUNM MATHEMATCS 102 MAXIMA AND MINIMA For this equation to be = o 3. () (0. 0) Is saddle point aqation to be satisfied identically, coeffs. of dr, dy, dz should be separately zero. 7 (0, 0) is saddle point (b) (0, 0) is saddle point; minimum value =- x (4) Extreme value, at (1) At0, 0). no extreme value: at (a. 0) no extreme value; at (0, a) no extreme y (5) and minima if a < 0 and min. value ma. if a> 0 and max. value - .(6) (6) give us the values of x, y, z, A for which f(x, y. :) 15 maximum or Equations (), (4), (5) and (6) P (7)at (0, 0). point of minima if a 2neither maxima nor minima ifa < minimum. ges and disadvantages of Largrange's Method of undetermined Multipliers 33 at and max. value = .4.Ad 7. at (0. 0). no extreme value , maxima at 8 (P.TU. 2010. 2017) -3 Advantages The stationary values/of f(,y, 2) can be deternmined from p (x. y, z) = 0 and min. valuei (1) minima at 8 and 0 even without dternining 9. 4, 8. 12 dx 8. maxima at (4, 2, 1) 2) This method can be extended for a function of any number of variables and subject to any 10. maximum temperature 50 at number of constraints. Disadvantages his Iagrange's method does not put us in a position to find whether for stationary point. function is 3.3. Lagrange's Method of Undetermined Multiplier (P.T.U. 2017,20 imium or minimum. For we find d°F determining this fact, further investigations are needed. For Let f(x. y. :) be a function of x, y, z which is to be examined for maximum or minimum val. the variable be connected by the relation minimum value and, maximum value d^F value d~F <0 and for minimum value d*F >0. This method is very trouble some. Nate. We explain thu method with the help of examples. x.y. )=0 ILLUSTRATIVE EXAMPLES Since f(x. y. 2) is to have a maximum or minimum value 0 - Faole 1. Find the maximum and minimum value of r+f subject to the condition 3 -4 ry+6y= 140. Sol. Let fr, y)=r +y The constraint is d=0 3+4xy+6y = 140 Let F (x, y)= x*+y+1 (3x* +4xy+ 6y- 140) where is Lagrange's multiplier. ...(1) Differentiating (), we get, For extreme points, o =2 x+2 (6x+ 4y)=0 Multiplying (2) by 1, (3) by 2 and adding, we get, 2y+(4x+ 12 y) =0 (1+3 A)x+2Ay=0 or .2) and 2x+(1+62)y = 0 (3) 104 SPECTRUM MATHEMATIC A Since r and y are both non-zero MAXIMA AND MINIMA 105 1+3 A=0 and 2 1+6 (4) (13 A)(1 6)-4 = 0 149+1 =0 -981-56 9t5 For stationary values, 2 -= - 28 28 - Take - From (3), we get from (2), x=-2y y From( from (1). 12*- 8y +6)*- 140 =0 =14 -4 56 ..5) , from (5), y= y=d » y=a When x +56 + 14 70 Take - stationary points are (a, a), (-a, a) from (2), y=2 x stationary points are (-a, - a), (a, - a) Similarly when x' = from (1), 3xr+ 8x+24 x = 140 *-4 -4x 16 Case I: At (4, a), A = 8, B=4, C = 8 AC B'=64- 16=48. Also A =8>0 4x= 16 (a, a) is point of minima +=16+4 20 Hence maximum value of x +f is 70 and minimum value is 20. Also z Exaaple 2. Find the minimum value of r* +y+z subject to the condition that xyz=d a When x a,y=a, Z= za (P.T.U. 2010, 201 a.a Sol. We have to find minimum value of min value = af +a +a=3a Case II: At (a, - a), A =8, B=- 4, C =8 Sx,y. 2)=++2 AC B 48 >0. Also A =8>0 (a, a) is point of minima, andz= - ¢ when xyz - a 0 From (2), 2 min value d +d+d=3d Case IlI: At (-a, a), A =8, B=- 4, C-8 AC B'= 64 16 48> 0. Also A =8>0 (-a, a) is point of minima and z = - a and min value = a* +a+a = 3ad Case IV At (-4,- a), A =8, B = 4,C 8 AC B= 64 - 16= 48>0. Also A>0 (-a,- a) is again point of minima and: = a and min value = a+d+d-3* . Putting this value in (1), we get Sa) =r+y?+ 2a OX TIMATICS-Il SPECTRUM MATHEMA 107 106 Example 3. Use Lagrange: method of undetermined coetticients to show that the rectangular (P.T MA AND MINIMA 8' (d d (da maximum volume that can be inscribed in a sphere is a cube Sol. Let V be the volume a dy dz + a dz dr+ a dx dy) 6) where X.y 2 x, 2y, 2 z wh d'F of the rectangular parallelopiped whose edges are We have to find the maximum value or pifferentiating (2), we get, V= 8 rr when a++:-a' = 0 or 1 0 ultyle where 1s Lagrange's mul. we get, Consider the function F (x, ). 2) -8xyz +2 Putting x = y=- be detemined. d d = 0 a a For extreme points, F, 8y +2.-0 F 8zx+ 0 from (6), we get, F. 8 x .=0 a Multiply ing (3) by x. (4). by y. (5) by z and adding, we get, 24 ry22 0 or 24xyz + 22 (1)=0 i-12xy: from (3). 8yz - 4*y which is negative definite form. a ulie. when rectangular solid is a cube V is max. at | Similarly y and max. value = 8 a a 8a aa a Stationary point is Example 4. Find the shortest distance from the origin to the hyperbola r+8ry+7=225. Sol. Let r be the distance of the origin (0, 0) from any point (, y) A ..2) where x +8 xy+7y = 225 Let F (x. y)=r+y+l (r+8xy +7y-225) d'F =2F. (de + 2E F,, dy d: = 22), (d, az+16 Eady dz a where A is Lagrange's multiplier. SPECTRUM MATHEMATICS.1. 108 109 For extreme points, MAXIMA AND MINIMA Find the shortest distance from the origin to the surface Ix+ my +nz =p. Example 5. Find the shortest Sol. Let (r, Y 2) De any point e r+my+nz =0 =2rÀ(r+ 81) -0 any point on the surfac 1) 2+À (8r+ 141) =0 where r is the distance of (0, Let Fr y. 2) =xr+¥+?+1 (dx+ r Let-r +y+2 2) 2(14)r + 8Ay=0 8ix+2y(1 + 7) =0 0,0) from (x, y, 2) A or nz-p) Since x and y are both non-zero 21+A) 8A For extreme values 0 2(1+7) F, 2x+2l=0 SA (3) F,-2y+2m =0 1+ 4 |=0 42 1+7a (4) F 2z+n=0 (1+)(1+74) - 16l' = 0 71 +82+1 - 162* =0 91-8-1 =0 St 64-368t0=1, - From (3), (4), ), we get, 18 18 Take i = 1 Putting these values of x, y, z in (1), we get, from (3). 2(1+ 1)x+8y =0 2y from (2) 4*-16)+7-225 -5y=225 - S =45, which is not possible. 2P Take = 1+m +n pt pm pn from (3). 2| 1 1 +m +n Now dF =2 F (de)y+ 2 X F,.dy dz = 2 [(drj + (dy + (de}] +0 2 [(dty +(chy} + (d2} from (2) 4 4y7-225 which is positive definite =225 ris minimum =20 =5 P=+y-5+20 25 Now d'F=F,, (dr) +2 F., dt dy + Fy (dy)* =2 (1 +2) (dr)' + 16l dr dy + 2 (1 +72) (d»)} p'm +n+n (+m2+)2 (2 +m +n P+m2+ pn P minimum value ofr = p1 Example 6. If u = dr + by +C# where+=1, show that stationary values of u are given by 0 which is positive definite. .ris minimum and minimum value ofr is 5. F.y, 2)-ar + by+è# + - Sol. required distance = 5. ATICS-Il ONC sPECTRUM MATHEMAT 110 MAXIMA AND MININ For extreme value, we have D MINIMA 111 tor extreme values - =2(x -x) +2=0 oX (5) OF 2 (y -yi) +21 =0 (6) Ff-20 oF 2 - x)+ 0 d (7) From (1). a'r - 4 =by-c? OF -2(y-y)+ (8) 1/3 -c- K AyA Eliminating A1 between (5) and (6), we get 2(r-xi)-4 0-y) =0 *-X = 20-y) 9) Putting all these values in +-+- = 1, we get Eliminating A, between (7) and (8), we get la+b+c) =I K=a+b+c *-x1)+2-) =0 4y -x) = 9x, O-y) ..(10) .2 Dividing equation (10) by (9) we get S 4 Example 7. Find the shortest distance between the line y = 10-2r and the ellipse nd -1 - * Sol. The given line is y = 10-2r The given ellipse is Let A (x, y) be any point on (1) and B (x1, Jyi) be any point on (2) From (9)x-2y =x,-2, .(11) AB r-x)+0-)* From (3) 2r + y = 10 Let f, y, xi. yi) = (r- x1)+-n On solving (11), we get r= 91*20 2+4y +10 5 ...(12) subject to the conditions 2x+y-10 0 When from (12) we get x = 1-0 and ... The distance between AB= 2+(0) 4+i = /5 Now, Lagrange's function F = («-x)'+(-y)+21 (2x+y- 10) +2 6 from from (12) we get x =y= 22 When xi= sPECTRUM MATHEMA MAXIMA AND MININMA Example 9. Find the 9. Find the maximum value of "y when x + y + a. 15 36+9=V45 3s AB But s Sol. Let u =r"" Let Flx. y. )=r""P+ +2(r+y+: - a) shortest distance between the line and the ellipse = 5. akes on where Lagrange's Multiple tmple 8. Find the greatest and the least values that the function y (x, V) = x y tako he For Extreme values, mx"'y'?+l = u+i =0 0) Sol. Here J (T. V) =rV and given ellipse i -1 =0 oy y .(2) E-u+=o (3) Largrange function is given by F (x. y) =xy+ x For maximum and minimum value, we have From (1), dF Putting these values in x *y+z =a, we get (m+n +p) = a 1= n + n+ p) S and =x +Ay=0 -mua am an ap Multiply (3) by r and (4) by y and subtracting. we get, 2| u(m+n+p) m+n+p m+n+ P m+n+P (am x (an)" x (ap)P a Pmmn"pP m+n+ p)"(n+m+ p" (n+ m+ p) m+n+ p ie, ie. x = 4y P :l+0] Putting value of x in (2), we get To maximise u, we have : dF f (de + 22s d de 0 ie., y=1 or y= From (5), we get x*= 4 rt 2. extreme points are given by (2. 1), (2,- 1), -2, 1) and (-2, - 1) Now at these points values of f(x. y) are given by (m+ n* P -(de = + PAP +2drdy + 2udyds + 2d: d a m p at (2, 1). fo. y) =2, at (2,- 1).f(x. y) = -2, at (-2, 1).fa, )) = -2, at (-2,-1),ft, y) =2. maximum value off (x, y) = 2 and Minimum value of f(x. y) =-2. By using values of x, y, z]| (m +n+ p)* t+2drdy-2(dr+ dy) *+y+z =a dr + dy + d: =0 or dz=- (d +dy) s- SPECTRUM MATUEMATICS l14 INIMA 115 + d + 2ddy - 2(dr + d? m+1 + p) Example 11. Find t ++= 1. MAXIMA AND MIN, ind the maximum and minimum distances of the point (3, 4, 12) from the sphere sol. The given sphere is x+ ? Let Pa +=1 mp- lh' -L(d+ d| n ) v. z) be any point on the sphere. Given point is Q (3,4, 12) P x-32 +(y-4y +(z-12 (m+n+ p) distance | PQ f,y, 2) = (POY = (x-3 +(y-4}+ ( -12)* a Let and given condition is (x,y, 2) =r+y+?-1 =0 F(,y. :) is maximised. Example 10. Find the stationary values ofx++# subject to a x +by + c: = l and /r +MS Sol. Let u =r+y+:subjected to conditions ax +bys +e:' =1 .(2) Consider Lagrange's function F.y, ) f%.y. 2) +2 ? x, y, 2) and Ix+ my+ nz =0 (x-3+y-4) +(-12)+2 (x++?- 1) Consider F(a. y, :)=r+y++2, (ax*+ by'+ c::- 1)+2lx+ my+n2) For extreme values, For stationary values 2 r-3) +2Ax=0 2x2 ax)+1,1=0 ox =0 2(y-4) +2Ay=0 oy S ...(4) 2y+2, (2 by) +2z m =0 F = 0 2(:- 12) +2z =0 .5) 2+À,(2 cz) +A2 n=0 .. Multiplying (3) by x, (4) by y, (5) by z and adding, we get 2++-6x-8y-24z+ 22(+y +)-o Multipling equations (3) by x, (4) by y. (5) by z and adding 2+y+)+A,2ax'+2 by?+2 c:)+, lx+ my+n:) =0 2 u+24, (1) =0 >21=- u [ of 2) and () or 2-6x-8y-24 z +21 =0 (6) or 3x+4y+ 12z=1 + from (2), 2x[1+a(-u)] =- A2l * la- From (3), (4) and (5), From (3), 2m An 3 12 ..() ... 2(bu -1) 2(cu -1) Putting all these values in (2), weget Putting these values of x, y, 2 in (6), we have 916 I+2 1+2 1+ 144 =1+ or (1 + = 169 or 1 +2= 13 A,12 m 2au-1) 2(bu-1) 2(cu -1) 2 =0 A = 12 or - 14 Here 2 0(: IfA2=0x=y=z = 0, which does not satisfies equation (1)) when 12, from (7), we get n au-1 bu -1 cu-1 gVEs stationary values of u. X SPECTRUM MATHEMAT l16 117 when = - 14, from (7), we get MAXINMA AND MINIMA (7) Similarly from (4) and (2), we get z = x or u = from (5). (6) and (7), we get Thus. we get two points A and B on the sphere 2 either y=? or u* maximum or minimum distance from the given point we have **ytz = || Now and 4= 6 wheny =2, we get 3x = 1 xy is the stationary point. #F F. (d)*+ FE, (o +Fdf +2F, drdy + 2F , dyde +2F, dt dr. -2(d? +2d* +2(d) +24 z drdy + 24 x dyde +24y dz dr FFy -F--2 3 4 12 minimum distance is at pont A 13 13 Also F,z, F= 4x, F4y -2(a3 2ud6+Adf + 20dkedy +2(6t+26 maximum distance is at point B 13 13 -2 (dt+ d+ d:) >0 Example 12. Use Lagrange's method to find the minimum value ofx' +y + z subject to the +y+= l and xy: + 1 =0. conditio Fy. ) is minimum at and the minimum value = S Sol. Consider Lagrange's function EXERCISE 3 (b) F.y )=+y+2+ia+y+z - 1)+u (yz + 1) Let f: R°> R be defined as f%y, 2)=xyz. Determine x, y, z for maximum of f where For maximum or minimum value xy+2yz+2zx= 108. 2. Find the maximum or minimum value of x* +y+? where ax+by+ c?+2/yz+2gzx + 2hxy= 1. 3. Find the Minimum Value of? + y+2when 2x++4yz =0 .. OF =2y+i+p zx =0 dy ... ()x+y+z = 3a (i) ax+by+ cz =a+b+c 4. Prove that the volume of the greatest rectangular parallelopiped that can be inscribed in the (in) xy+yz + zx =3a 2+l +u ry=0 ... Subtracting (3) from (2), we get 2(ry)+4 z(y-x) =0, -y)(2-u 2) = 0 5. Find the volume of the largest rectangular parallelopiped which can be inscribed in the eitherx-y=0 or 2-uz = 0 2 () ellipsoid 4 i) sphere r +y+?=3 X=y oru .) 6. Which point on the spherex+y+?=1is at the maximum distance from the point (2, 1,3)? Similarly from (3) and (4), we get y= z or = 7. Find the shonest and longest distances from the point (1, 2, -1) to the sphere x+y+* = 24 6 (P.T.U. 2011, 2013)) SPECTRUM MATHE ANWSIN Sing lagrange's method of multipliers, find the pomt on the plane 2r-3. the origin in R 118 3v6:= 49 I19 MANIMA AND MINIM ANSWERS ind the dimensions of Rectangular box open at top or maximum capacit 432 sq cm and r= 6. y = 6, 2 = 3 1 h 0 Find the Mavimum and Minimum distances from the origin to curve 5 +6 values of u are given hN where x + +:= I. PrOve that stationary values of , =0 give e max. or min. value of u. 11. If u 2. Roots ofh b 8 -.V= a+b+c a+b+c a+ 0. ( minima at (a, a, a) and minimum value =3. 3. 12. Find the point on the surface of: =x* +y* + 10 nearest to the planer+2v_. aa +b+ c) b{a +b + c) c(a+ b+c) cn minima a24h? +c +b +c a +bb +c and minimum value a+b+c) ab- 13. Find the maximum value of u = x"y": when the variables x, , z are suhi COnd ax -br+c: 5p *q+r. that the stationary minima at (a, a, a), (Fa, - a4, - a) and minimum value = 3 a 14. Ifu= ar + by* + cs where r +y'+:= l and Ix + mytnz=0, prove that th. (in) = 0. of u satisfies the equation 64 cubic units (i) 8 cubic units 6. a- 0-u C-u +=1, show that a stationary value i givg 15. If u =dr+b'y+c: where 1. 6.3 6 8. (2,-3,6) 9. 12, 12,6 ax=by=cz and this gives a maximum or minimum if abc(a+b +c)> 0. S 10. Minimum distance = 1, x = maximum distance = 4, r =2., y = 2 subject to the condi 16. ind the extreme values of wherer=* 1./r+ my +nz =0. 12 17. Find the maximum and minimum values of r +y+ # subject to the constraints a 16 1-a2 1-b2 1-c?2 m* 0. This equation gives extreme values of r =1 and z =x+y 17. Maximum value = 10, minimum value 18. Find the maximum and minimum values of where d x +bb y + e 2, subiect a constraints +y+#=1,/x+ my+nz= 0. bject tw 19. Find the maximu.il and minimum radii vectors of the section of the surtace 0. This equation gives maximum and minimum values of r. 18. +y+ = l'r+ni y+ made by the plane ax+ by+ cz =0. 19. 0. This equation gives maximum and minimum values of r. 2 20. Find the lengths of the axes of the section of ellipsoid + n T by the plane a al 2 2 C n b m a?-2 b2-2 Ix+my+nz=0. 20. =0. This equation gives maximum and minimum values of r. 21. Find the maximum and minimum radii vectors of the section of the surface ax + by tcr: by the planelx+my+n: = 0. m n -=0. This equation gives maximum and minimum values of r. 21. T OR 1- ar- -br 1-cr Find the extreme value ofr++# subject to the conditions a xr +by+ cz= I and Ix + mytx?* MULTIPLE INTEGRAL Eyaluation of Double Integral ALS 121 4 4.2. 1 of the double integral depends upon the evaluation The ves, Let the region R is bounded by the curves D MULTIPLE INTEGRALS r X1,* 2 and y *y,y: When Xi» X» i J2 Are constants. Here R is the I2 rectangle ABCD. Here t rst integrate from (ABCD. Here the order of integration is immaterial, we the strip P to Q and then slide it from 4.1. Double Integrals can n or we can integrate first along the strip P'Q' and then Y We have studied that the Definite Integral fa) dr is the AB to C yy slide it from AC to BD R Iimit of the sum f (ri) dx +f (t2) dr; +.. (T,) Ox, when n and each of the lengths dxj, dX2 , Öx, tends to zero. Thus Where dx . drz, .. Öx, are the n-subintervals into which Range b-a has been divided and xi, X2. , are values of ar lying in the .2"" nth Sub intervals respectively. A double integral is defined in two dimensions. Let f(x, y) (b)When Xy X are the functions of y and y, ya are constants. Let AD and BC are the curves x =fi (y) and = #). Take a horzontal strip PQ of width dy. be a bounded and single valued function of the variables in the D region R of the - plane. onsider a pa Divide the region R into sub-regions by drawing lines parallel to coordinate axes. Consider Here the Integral first evaluated w.r.t. x and then w.r.t. (,.y) in the ith rectangle whose area is dA, yof the resulting expression from yi to y2. Then consider the sum Thus Sr.) drdy = ||say ld R Sy) ÖA t f(x,. y,) dA, (c)When y. Ja are the functions of x and x,2 are constants. Let AB And CD are the curvesy =flx) and y ft) i=l Let the number of these sub-regions increases indefinitely, such that the largest linear dimension, i diagonal approaches to zero. The limit of the sum (1), if exists, irrespective of the way of subdivisioni YA Take a vertical strip P'Q with width dx. Here we first integrate w.r.t. y along the strip P Q and then the resulting called the Double integral off(x,y) over the region R and is denoted by|fx, y) dA expression w.r.t. x fromxj to x2. Lt X2 2y,) A, -|| s)da Thus|| Sx.y) dt dy or no 4,0 i R R B or can be written as| , y) drdy or ay) dy d« Note: In (6) and () Integration is frst performed w.rt. that variable having variable limits and then w.r.t. the variable with constant limits. R R 120 SPECTRUM MATHEMA 122 MULTIPLE INTEGRA 2 4 ALS 123 ILLUSTRATIVE EXAMPLR Etamph 1. Evaluate ( d d (PTU. 2018) (2) uin +3y)d d PTU From (1) and (2), L.H.S, = R.H.S. 3 Sol. Exampké 3. Evaluate 0Sxs 0Sys 23 Mere f.y)=r*y15 Continuous over A where A = iy):0 sx s1.0 SyS 1} 0Sxs 0sysl S unfu-3)d-j|j«iy* -f'y+y1,d Exapple 4. Evaluate: 00 2- +y) dy de (P.TU. 2013) (n xy dy d. (P.T.U. 2012) -(0+0)9+3 12 0 Exappte 2. Veritya)dy dr= |y) dr dy X Sol. LHS. +d de - dd L - 124 sPECTRU N MATIEMA MULTIPLE INTEGR.A (i) The given integral is RALS 125 3 9 30+14-30 = (0+0-0)= 35 105 0 [(2e-e)-(0-) -e+1-2= é-1 x4-4x+-)a h.What is the value of |xy r dy over the positive quadrant of the circle y =1? Exam (P.T.U. 2018) Sol. The equation of circle is x+ y2= The region A of integration is given by (4 x-4x +x -r) d =. A- ):0sxsl-,0sys1 12 Exapple 5. Sketch the region of integration and evaluate the integral Tv- d -oj S * d dy 0 0 Sol. () The given integral is log8 logy ye y / y= log 0 Here r varies from 0 to log y i.e. x= 0 to x = log y ie. y=e and y varies from 1 to log 8 ie. y= I to y= log 8 log8 logey Exapple 7. Evaluatee**dr dy over the triangle bounded by the lines x= 0., y =0 and x+y= 1. y1 (0, 1) Sol. Here A={,y):0 sxs1,0sys1-x} log8 logy dkdy =| eedr dy 1(1-x 2x+3y -* 0 3 0 L log8 A 00 y =1 x =0 y= =0 log8 log8 log8 Ledy = (dog8 - )e oga-0]e log8 - y-dy -e = (log 8- 1)8-(e°**-e') =8 log 8-8-8+e 8 log 8 + e sPECTRUM MATIEMATC 126 - MULTIPLE INTEGRALS T/2 127 n2 e-D e De - D12e - e- I sin 0.cos 0.2 sin 0 cos 0 do 2 sin0 cos 40 c-)(c- D(2c+ )=(e-1)° (2e + 1) 1.9.7.5.3.1 T_ 945 12.10.8.6.4.2 2 46080 Etample 8. Evaluate 2 2(1- x - ) dr d over the region A bounded by the the triange w -2(01)'da vertices are (0, 0). (1, 0) and (0. 1). Sol. The vertices of the triangle OQR are 0. 0). (1. 0) and (0. 1). Put equation of line QR is 2 sin4 cos o do = 26.4.2 2x 48 96 9.7.5.3.1 945 945 =| or x +y=l =0 y=1 945 A =a. p):0 sxs1.0 <ysl-x (xy2(1-x-y'dr dy =- 96 T X- 46080 945 480 where A is the region. y =0 Ae 9. Evaluate xy(X + y) dr dy over the region A bounded by the curve y =r and the line (1-x 1 1 3y 0-xr-1' ad - [220-r-dl y= Sol. Consider the equations x. S Put y(1-r)1, d=(1-x) dt y=* (1) and y =r From (1) and (2), x* ==x x-x =0 x(*-1) =0 When y=0, 1=0 2) When y=l -x, I= x 0, 1| 0-x -y' drd = [l[20-»2 20-0-0'a-)a |d from (2), y=0, 1 (1) and (2) intersect in points (0, 0), (1, 1). A {(x. y):0 sxs l, x sy <x} (1.1 - 20-?120-1?a ld - (0-)2 d (- a 9 Jxyx+ )dk dy - }|}*y(x+ y)d|d 2a 0 9 Oo. 0) Let Ix 2 (1-)2 dr Put x sins 4, dr 2 sin 4 cos 0 de When r=0, 0 = 0 When x=1,6=5 56 SPECTRUM MATHEM TEIATSIII 128 MULTIPLE INTEGRALs .aEvaluate|(y-2x jdrdly where R is the Region inside the Square z| yl=1 129 anded by Eampfe 10. Evaluate dd where A is the region in the tirst quadrant boba R Exapple 12. () Evalu Evaluatey dr dy, where R = {(x, y): |x|+ ly| s ) b ) R Sol. A 1.y):0Srs a, 0s ys-va* - a Region R is x as shown in fig. It is a square with Sol. (9 vertices (1, 0, (0, 1),1,0) and (0, -1). Limits of integration for the region in first quadrant OAB, are a D-1 0Sx Sl-y and 0 s ysi A 0 2 0 0 Example Evaluate||y dr dy where A is the region bounded by the parabolas j*=4 x and A 0 Sok Consider the equations -4x and r=4 yy o 0 -2 -2y2+yhd- From (1) and (2), we get, 0 4x or x - 64 x =0 EXERCISE 4 (a) x(r-64)=0 x=0,4 1. Evaluate Aay):0S xs4sys2Nx +2) dy d (in 5) 0 0 0 Now fx. y)=y is continuous over A (v) xy (1+*+y) dyd dy dr ,0) (vi) 42x oa-x-y*) 01 0 A 2 (4 2. Verify that T3 3T 161024=1652 160 3. Show 130 R CSE SPECTRUM MATUEMATICS-lll FOR G 131 4. Evaluate MULTIPLENTEGRALS cos (x + y) dx dy over the region bounded by x = 0, y= 0, x + y = 1. 13. Evaluate cos (+ 2x y dd verify that JJ *y ) dk dy J +y") dy dx where A is the triangle formed by the lines ii) x1,y=0 and y = x. ate x y dx dy where A is the domain bounded by the x-axis, ordinate x =2 a and arc of 0 dy dr 15. Eval (vi) + the parabola x* = 4 ay. over the area bounded by x = 0, y-0, x+y =1 and 5y-3. 16. valuatedd a- (v) x sin y drd ate x dx dy where A IS the region bounded by parabolas y = 4 ax and x = 4 ay. 17. Evaluate x dx dy where. 5. Show that cos x+ sin y| dr d» exists. Cualuate ya dy where A is the region bounded by the parabolas y 2x =0 and -2y= 0. Evaluate|3y dx dy, where A 1s the region common to the circes x*+ y =x, *+ =y. 6. Evaluate the double integral(4 x -y')d d if the region D is bounded by the straigh 19. CS lines x = 0, x = 1.j =0 and jy= ANSWERRS 123 7 Evaluate ( ) á d where A is the rectangle, bounded by the lines x - 0, x = I and 2 2 4 y0. = 2. ()log a log b (vi) 3. Evaluate | dd over the region +y s 1 67 8 4. 12 iv) 2 9. Evaluate 120 dr dy JJa (nloe 1v 2 (vii) a 6 )-1 where A {a,y):x 20. r+y= 1}. 9 1. 9. 4R )15 8. 10. Evaluate|ydr dy where A is the region in the first quadrant enclosed by x = 0, y = 0 and 6. 2 24 r+=1. 11. 13. sin + cos I - I 0 10. 96 11. Evaluate y dr dy where A is the region x +y sE1 6 18. 5 19. 16. sin 17. 12. Evaluate 4x - y dr dy where A is the triangle bounded by the lines y - 0.y = x, x=1. 96 132 rORC SPECTRUM MATHEMATICS-II 43. Double Integral in Polar Co-ordinates MULTIPLE INTEGRALs 4 r12 D a cos 20)-a d0 =a Il+cos 262-1] do=a 1(2cos 02 - 1]do The integral fr.e) dr de is double integral -T/4 -/4 r4(0) B in Polar co-ordinates bounded by the lines ,. 0 0 and the curves (cos0-1 d0 =2 (2 cos0-1 do s) =2[s(x) rif s(-z) = f() ri.r. We first integrate w.r.t. r 0 and then wr.L e between the limits =6,e =6 Note: If e, = a. 0 = b and ri = c, integration can be evaluated separately w.r.t. 0 andr. then -2af2 sin 0157-2a ie .0 drd ® - (r)de fd EXERCISE 4 (b) ILLUSTRATIVE EXAMPLES 1. Evaluate asin& dr de T a(l+cos 6) T of Examp 1. Evaluate ||drde over the Area included between the circlesr=2 sin & and r = 4 sin A a/2 a cos 4 sine Y Sol rarde d Evaluate J Ne-2ar ld. -2 Sin 6 -72 2sin8 r4 sin Evaluate dr db, over the area bounded between the circles =2 cos and r= 4 cos (256 sine-16sin 0) de Evaluate sin Gdr de over the area of the cardioid = a (+ cos ) above the initial line. -r2 4. 2 sin ; 240sin e.de= 240 .sin 0.de 5. Show that || sin drd6 = where R is the region bounded by the Semicircle r = 2 a cos -r/2 3 -1/2 R T2 above the initial line. 2x60 sin de 120 x - 457 0 ANSWERS Examgle 2. Evaluate | over one loop of the lemniscate d cos26. 25 ya+2 Sol. Given leminsacte is r=a cos 26. Now for the one loop of the Lemniscate r varies from 0 to acos 26 and 6 from to 4 4a 4. 3. 3 T/4 aycos 20 2=d cos 2 44. Change the Order of Integration (a? +2y2.2r dr de While evaluating double integrals if the limits of integration are variables, then the change of order of imegration changes the limits of integration. In changing the order of integration sometimes it is required to split up the region and express the given double integral as a sum of the number of double integrals with changed limits. Sometime if is advisable to draw rough sketch of the region of integration. T4 ycos 20 de A 4 4 The following examples will illustrate the idea more clear. 134 SPECTRUM MATIEMATICS- GRALS 135 MULTIPLE INTEGA oiven integral on cnange ol order of integration becomes ILLUSTRATIVE EXAMPLES d r. y)d Fxaple 1. 0Evaluate d d by reversing the order of integration. 0 3 Change the order of integration in the integ. I 1I= -a 0 () Change the order of integration in a h.1d (PTU Sol. The given integral is I = ty)dr d Here r varies from 0 to ya -y2 along horizontal strip -a Sol. () Given integral is r3 0 3y Here the region of integration is PQ and y varies from-a to a. On n changing the order of integration, we first integrate Cha R (r. 1):0 sysl:3ysrs3 valong a vertical strip PQ which extends from w.r.t. ie. it is bounded by the curves .-a- to y=va* -x and x varies from x = 0 to y=- 3y.T=3 :y=0,y =1 While changing the order of integration the horizontal strip is changed into vertical strip. y0 A ya-r ya? _2 I-dfx.)d or 1= sa. ydydr Region R can be written as R={(a, ):0srS3:0<ys* 0 -ya-2 ya-r given integral on change of order of integration becomes x/3 Esample 3. Change the order of integration and hence evaluate y dr d (P.TU. 2011, 2017) 0 2-r Sol. Given integral is I = |y dr dy (in Giyeh integral is df(x.y)d y2-x 0 X Here y varies from x to 2 - X along vertical strip PQ and x varies from 0 to 1. Here the region of inteyralion is B (0, 2) A-x. y): r Sys Vx ;0 <x< 1} A(1. 1) In order to integrate the given integral, we split the r=0 given region OAB into two parts OAC and ABC ie., it is bounded by y =x, y = x ;x= 0, x = 1 A(1, 1) Now for region OAC, limits of integration for x While changing the order of integration vertical strip is changed to horizontal strip (0, 0) are from x =0 to x = y and those of y are from y = 0 B toy=I and for the region ABC, limits of integration of x are from x0 to x = 2 - y and those of y from y = | toy=2. So. region R can be written as R .)y sxSy:0 sy s l 136 SPECTRUM MATIEMATICS-IL ATICS-I\ OR 137 MULTIPLE INTEGRALs sol. () he given integral is | LHere x varies from y to 4 along horizontal strip PQ and y varies from 0 to 4 ue change the order of integration we change the Wnen PO to vertical strip P2 which gives the limits dhx I=4 horizo on of y as y =0 toy*x and X varies from x = 0 to fintegra 4 r P Example 4. Evaluatee*" dyd by change of order of integration. P.T.U. 2010, 2014 Ctan1-tan od - dr = x(4-0)=z 0 Sol. The given integral is - dyd YA a The region of integration (The given integral is l= | +y>) dydr xa R {(r. ) :0 S x So,0sysx} is as shown in fig. A (a. 1) CS In the given integral, the integration is first w.r.t. y and then w.r.t x. To change the order of integration, take an elementary strip PQ starting from P parallel to x - axis. Thus, the limits for x are ; x =y to x =o and y =0 to y = Here y varies from toand x varies from 0 to a. The point of intersection of the curves y=^, y= and a a y=0 the line x = a is (a, 1). The limits of integration of y varies -*y dx \dy x along the vertical strip PQ and those of x as x = 0 tox =a along horizontal. In order to change the order of integration, the vertical strip PQ is changed to horizontal strip P'Q', which gives the limits of x as x = ay to x = ay and y varies from y = 0 0 0 Put x =t 2xdr = dt , we get toy=. xa ay 0 xa ay /a a Example 5. Evaluate (i) 3aa a 84 20 28 20 xa SPECTRUM MATUEMATICS. 138 LTIPLE INTEGRALS 2ar 139 tample 6. Change the order ot integration and evaluate the integral; 2a a 4 2ay 4a /4a 0 /4a 0 0 OL 4a 2a va(ta2 64a _162 162 Sol. () The given integral is 3/2 3.4a aa- 4 Here x variesfrom a-a- to ata* -y Example 7. Evaluateedrdy by changing the order of integration. 0 4 and y varies from 0 to a. AsT varies from a- a -I to a +ya -y Sol. The given integral is I = | 0 4y I= 4y y1 =a ta- (r- a)'=a -y - 2ax = 0 O(0.0 (a. 0) (2a,0 Here x varies from 4 to 4 along horizontal strip PQ x=4 which is the circle with centre at (a, 0) and radius =a This integration is Performed along the horizontal strip PQ. Now change the strip horizontal to veri P, theny varies from 0 to 2ar- and x varies from 0 to 2a. and x varies from 0 to1. After the changing the order of integration the ntal strip will become vertical strip P'Q' which rtical sh 2a 2ar-r? X 2a a- extend fromy=0 toy = x/4. aV2ar- y=0 y2ar-de= --g Now y varies from y =0 to y = x/4 with vertical strip 0 0 P'Q' and r varies fromr=0 tox =4. La 4x/4 --(a-a.sin S Sin a 22 2 0 4y a 2ar Putr= t 2r dt= dt so that (in) The given integral is 2 4ay when x = 0,1 = 0 and when x = 4, t= 16 04a 0 :Aax -y=4 6 A (4a, 4a) Here y varies from -to 2ax and x varies 4a 18 from 0 to 4 a. r= 4a Here integration is carried along the vertical strip PQ. When we change the order of integration the vertical strip must be changed to horizontal strip Example 8. Evaluate after changing the order of integration O y drdy P'Q where x varies from x = to = 2ay 4a laax)a-y 0 and y varies from 0 to 4a. The point of intersection of the cures is A (4a, 4a). dydr (P.T.U. 2011) 0