& ROH INII cottece oF encineerinc & TECHNOLOGY Z Near Anjugramam Junction, Kanyakumari Main Road, Palkulam, Variyoor P.O - 629401 Kanyakumari Dist, Tamilnadu, E-mail: admin@rcet.org.in, Website : www. rcet.org.in DEPARTMENT OF MATHEMATICS ENGINEERING MATHEMATICS -I SUBJECT CODE: MA8151 (Regulation 2017) UNIT -IV MULTIPLE INTEGRALSUNIT -IV MULTIPLE INTEGRALS 4.1Introduction The mathematical modeling of any engineering problem which leads to the formation of differential equation of more than one variable has its solution by the integration in terms of those variables the need of the solution in an integral where many variables are involved motivated the study of integral calculus of several variables, In this chapter all the basic concepts related to the methods to approach such integrals are discussed. 4.2Double integration in Cartesian co — ordinates Let f(x,y) be a single valued function and continuous in a region R bounded by a closed curve C. Let the region R be subdivided in any manner into n sub regions Ry, Ra, Rg, +++, Ry OF areas Ay, Az,A3, ***,An-Let (x), yj) be any point in the sub regionR. Then consider the sum formed by multiplying the area of each sub — region by the value of the function f(x,y) at any point of the sub — region and adding up the products which we denote UF Gey)Ai The limit of this sum ( if it exists) asm — 2 in such a way that each A; — 0 is defined as, the double integral of f(x,y) over the region R, Thus lim 23 F@oy)Ar= Se Fy) dA ‘The above integral can be given as Sp fOuydyde or fy fl%.ydxdy Evaluation of Double Integrals To evaluate J" f°" f(x, y)dx dy we first integrate f(x, y) with respect to x partially, that is treating y as a constant temporarily, between xp and x,. The resulting function got after the inner integration and substitution of limits will be function of y, Then we integrate this fumetion of with respect to y between the limits yp and y,as used. Region of Integration Fels) Case (i) Consider the integral f° 2°) f(@x.y)dy dx Given that y varies from y = fi@) to y= fala) xvaries from x = a tox = b. We get the region R by y = fil),y= f(x), x =a, x = b. The points A, B, C, D are obtained by solving the intersecting curves, Here the region divi ed into vertical strips (dy dx). Case (ii) Consider the integral ff") f(x, y)dx dy Here varies from x = f(y) to x= fo) and y varies from y = c toy = d «the region is bounded by x = f,(y), = fh(y), y=c, y =d. The points P, Q, R, S are obtained by solving the inte ecting curves. Here the region divided into horizontal strips (dx dy) y Problems based on Double Integration in Cartesian co-ordinates Example: 4.1 Evaluate 2 J? x(x + y)dydx Solution: Se Se xl + ydydx = J? [2? + xy)dydx “hey =fp [@x? + 2x) - @? +5] dx =f, [2x2 + 2x2? -D] axfife 2a] +E +32), -G+2)-@+o =2 2 Example: 4,2 Evaluate fo {? xy(x— y)dydx Solution: Soi ye —ydyde = [8 JP Gey —2y2dydx (ce-2h (= we ‘)- @-0) @-b) Example: 4.3 ‘ a pb dedy Evaluate ff fp Solution: RBS = flea], ay aff} (logh — log2)dy E5tog(Z)dy [+ log = toga — togb] oath yay ~ tog losylt = log {loga—log2] [leg] [to] Example: 4.4 Evaluate {2 {2 (x? + y®)dxdy Solution: LEG +ydxdy =f E+y’ 2x). ey £1499) Era) =f, [9+ 3y? 2-29] ay aExample: 4.5 Evaluate fp {2 e*¥ dydx Solution: SB eer? dydx = ff {2 e* erdydx -|f- e* dx|[J2 e dy] “Le*I§le"IB = fe? — elle? - e°) =e? — 1]le? - 1] Note: If the limits are variable, then check the given problem is in the correct form Rule: (i) The limits for the inner integral are funetions of , then the first integral is with respect to y (ii) The limits for the inner integral are functions of , then the first integral is with respect to x Example: 4.6 Evaluate > f)* dxdy Solution: The given integral is in incorrect form Thus the correct form is So KO ayae =(SyN* ax = [2 [Va eax 20-2 Example: 4.7 Evaluate fof." y@? + y*)dxdy Solution: The given integral is in incorrect form Thus the correct form is Se oO yO? + ydydx = fo" (aPy + y*)dydeExample: 4.8 Evaluate {> {°* xy(x + y)dxdy Solution: The given integral is in incorrect form Thus the correct form is Se xv + yayd = f° (ty + xy*)dydx Example: 4.9 a p VT _ax Evaluate > J.) we Solution: The given integral is in incorrect form Thus the correct form is iat Ge NTR __ayax So ae8 feerton™* (a) te “Le laesston oa [+ ean¥() =f ie Fede =F eg ax (tan) = 0) = Sflogle + VTFSAI]: = Flog + v2) Example: 4.10 Evaluate {4 [2 e”/x dydx Solution: The given integral is in correct form se aya = EY LG)-@le =fflxe® — xldx =f x(e* — 1) dx -[xe*- -@(e*- SI (by Bernoulli’s formula) 4(e* 4) -(e*- 8) -@- 1] =4e*—16-e4 +841 =3e'-7 Example: 4.11 Sketch roughly the region of integration for f° [* f(x,y) dydx Solution: Given Jo f* fey) dydx x varies from x = Oto x= 1 y varies from y = 0 toy =xSolution: Ke f= a dydx is the correct form x limit varies from x = 0 to x = y limit varies from y = Vax—x* to y = Va? —x? ie, y? = ax —x? to y? ie, y? +x? =ax to y?+x? x? +y? =ax isacircle with centre (2,0) and radius £ 2x? +y? =a? isacircle with centre (0,0) and radius a Exercise 4.1 Evaluate the following integrals Lf fy G? + y?) dye Ans: 262. {6 [G2 +y2) dedy Ans: 3.8 Sp ody Ans: = 4. f° iy) dxay Ans: 13 5. Ne +y) axay Ans: 2 10. {27 PG?) dyad 4.3 Double integration in Polar co-or Consider the integral i? S72 £7, dr do which isin polar form. This integral is bounded over the region by the straight line 0 = 0 8 = 8, and the curves r = 74,7 = 7;.To evaluate the integral, we first integrate with respect to r between the limits r = 7, and = r,(treating @ as a constant).The resulting expression is then integrated with respect to @ between the limits 8 = @, and @ = 0, Geometrically, AB and CD are the curves r = f,(@) and r = f,(9) bounded by the lines 8 = 6, and @ = , so that ABCD is the region of integration. PQ is a wedge of angular thickness 58. ‘Then f.? f(r, @)dr indicates that the integration is performed along PQ(i.c.. 7 varies , 8 constant) and the integration with respect to 02 fp, £7, dO means rotation of the strip PQ from AC to BD Problems based on double integration in Polar Co-ordinates Example: 4.13 sind Evaluate {./? f° rdodr Solution: sind Given 0"? 2°" rdbdr = 2? [2 rard@ (Correct form) = fale = 5" EFI fi? sin? odo eo do = ffl [2 — o]ao Example: 4.14 Evaluate fo fj"? rdrado Solution: ven J fo" rdrao =e = a0 =i |e ale [@—0)- (0-0)] Example: 4.15 Evaluate ff fo rdrd@ Solution: Given Jf ff rardo =f 2 | "ae 0 10Example: 4.16 Evaluate [07 {°° r2aedr re Jo Solution: sven (7/2, (209 2 Given [77 So" dear LS" rdrd® (correct form) nla penpeeos® Lrilaly 4 =p? [ee ("i oaa = 25007, cos? do in =£@) [0 costo do 162 4) 32 Seal - Example: 4.17 Evaluate fo” fo coso)7d0dr Solution: Given fea cosoy? 0dr = uo do Sr" 11 — a — cose)3]40 = §7/?[1 — (1 — 3c0s0 + 3cos*O — cos*6)}40 = £4"? [3cos0 + 3cos?0 — cos*4)|a0 2 [@sinoyy? Epone Uw_ 2 fasconeal alo =f [44—9n] Exercise 4.3 Evaluate the following integrals 1 cose ™4raa 2S" Sosna TAO m/% pcos20 Lento rardo 4,47? 20"? faz Tarde Ans: =n — 4) ‘ans: 22-2) an Ans: 222 4.4 Change of order of integration Change of order of integration is done to make the evaluation of integral easier The following are very important when the change of order of integration takes place 1, If the limits of the inner integral is a function of x (or function of y) then the first integration should be with respect to y (or with respect to x ) 2. Draw the region of integration by using the given limits 3. If the integration is first with respect to x keeping y as a constant then consider the horizontal strip and find the new limits accordingly 4. If the integration is first with respect to y keeping x a constant then consider the vertical strip and find the new limits accordingly 5. After find the new limits evaluate the inner integral first and then the outer integral Problems Example:4.18 Change the order of integration in fy’ {° f(x, y)dy dx Solution: Given y:x >a x04 The region is bounded by y = x,y = a,x =Oandx=a WDx axis limit represents the horizontal strip and y axis limit represents vertical x0oy y0>a By changing the order we get [ “fr Gi y)dx dy o Jo Example: 4.19 Change the order of integration {2 [* f(x,y)dy dx Solution: Given 413.By changing the order we get [ [senaray 0 Jy Example: 4.20 ‘Change the order of integration and hence evaluate it {¢' f° (x?-+y*)dy dx Solution: It is correct form, given order is dydx given y:x >a x0>%a the region is bounded by y = x,y = ax = Oandx =a x axis limit represent the horizontal strip y axis limit represents vertical path changed order is dx dy x0oy yi0>a Rots Dexdy = 82+ yx) ay ~h E+ y |r Example: 4.21 Change the order of integration for" [2,4 xy dy dx Solution: Itis correct form Given order is dydx Given y 23 Ve 14x: 09 4a The region is bounded by x? = 4ay , y? = dax x = Oandx=4a se" fp aya dy =" | Ey -[@t- I 6 ial, _taue* _ aye _ 1280" _ 4096 3 492 stat “a Example: 4.22 ae Change the order of integration of {2 {2 xydxdy and hence evaluate it Solution: It is correct form Given order is dxdy Given x:02 BF =y? yr O>b 415yx=0,x=% bey? 3 4% The region is bounded by x = 0,x = 2 /b?@=y? 354% = F yaOy Changed order is dydx Draw the vertical strip y 10-9 2Naraa xr03a LET waar = CPE Example: 4.23 ‘Change the order of integration and hence evaluate {3 {2 " xy dy dx Solution: It is correct form Given order is dydx 16Given y: x? 2—x xP 031 The region is bounded by = ‘Now divide the region in to two parts ie, Ri and Re Changed order is ddy Draw horizontal strip For Region Ri Limits are x: 0 + Jy yoot Si xy dxdy = fp exp? 2 Io dy = sion 2 ay For region Ro Limits are x: 0 2-y yi132 [ [PoweExample: 4.24 Change the order of integration in J. Sy dx dy and hence evaluates Solution: It is correct form xiyo2-y y0o1 The region is bounded by x= y ,x+y=2 ysO,yrl y ‘Now divide the region into two parts ie. Ry and Re Changed order is dydx Draw horizontal strip For region Ri Limits are x: 0 1 13For region Ro x12 y:0 92-x 2 pez _ pepe LR xy dy dx = ff ey dx =f ESE o]ar a p2x(a4z? 4x) =p ee ax 2 fax + x9 — 4x7) de Example: 4.25 Change the order of integration {.° f” Pay dx and hence evaluate it Solution: 19Itis correct form Given order is dy dx Given y: x > 00 x1 09% Changed order is dxdy Draw a horizontal strip oy 05 0 =[le%dy = —{e~* -e°] Example: 4.26 Change the order of integra m= iS ae Solution: Itis correct form Given order dady xiysa yi0>a The region is bounded by x = y,x =a y=0,y=a 20 dx dy and the evaluate itChanged order is dy dx Draw a vertical strip yi0ox :03a LGpa ds =x HO], ax Example: 4.27 Evaluate {* f°" xy dy dx by changing the order of integration Solution: Its correct form Given order dy dx Given y: 09 Vat =x? x10%a the region is bounded by y = 0, y = Va? — x x=0,x=a 1changed order dxdy Draw horizontal strip x1 0>Jat—y? yr0>a xy dy dx =fo fy" xy dx dy =o a dy =i y@?-y*) dy “yi aly Exercise: Change the order of integration and hence evaluate the following 1 RO Ge? +9) dy ae 2. fo JN x2 dy dx Ans: tat 3.8 BF dy dx (aay? 4A Son Jaraye ae dy Ans: 2 5. Rye = Ans: 2 6.4. Sy xy dxdy Ans: 22TS dydx ‘Ans: log4 — 1 5 8. fox? dy dx Ans: 24 9. Sy dx dy Ans: = log(1 + 2) 10. f* Sig ax dy Ans: 8-2 log 4 4.5 Area enclosed by plane curves (Cartesian coordinates) Area = f fdydx (or) Area=f fdxdy Example: 4.28 Find the area enclosed by the curves y=2x? and y? = 4x Solution: Area = f f dy dx yt 2x? + ave x 1031 Area =f. [N" dy dx = {Fax = fo Ve = 2) dx axY/2 axa] ae 8 0 “Pal 4 3 2 3 23Example: 4.29 Find the area between the parabola y? =4ax and x? = Solution: Area =f f dy dx y = 2ax 4a x10 4a = fi" (2 dydx 4a =f ois ax = {iver - 5 ax ; fe afere | yp, ia A = fa (402 - SS" 6a? Example: 4.30 ye bE Find the area of ellipse = + Solution: vnArea =4 ff dx dy 19094 bray? x02 bay yi 0 ab Oa Area =4f? fi ¥ dy dx =a “0 ber jo Example: 4.31 2 Evaluate ff xy dx dy over the positive quadrant of the circle x? + y’ Solution: x:0>/1-y? yi Oot 25Sfxy dxdy =f) f° xy dx dy “SET =iRG1 yy ay -ifa-y)yay =i fo- yay Example: 4.32 Find the smaller of the area bounded by y = 2— x and x? + y? =4 Solution: Area= ff dy dx yi 2-x+Vaox? x2032 =i GA dy dx = Kier ax = IVF ~ 2 — wax = [Vea + Esin4 (8) - 2x +E OF @ 445 =n —2 square unit 6Example: 4.33 Evaluate ff xy dxdy over the positive quadrant for which x + y <1 Solution: x: 031-y yrOot 0 pe Example: 4.34 Using doubl Solution: 7Area = ff dy dx yixtox x:001 Example: Evaluate ff (x? + y )dxdy where A is area bounded by the curves x”: x2. about x axis Solution: yr Ox? xt132 28Example: 4.36 Find the area enclosed by the curves y = x? andx + y —2 Solution: Given y=x? andx +y —2 =0 x oy1 f2 —2 Area = ff dy dx yixto2—x x: -231 SE Sa “dy dx = fly) ax 1 2° = [2 - x —x*)dx -b-#-41, Example: 4.37 Find by double integration the area lying between the parabola y=4x-x? and the line y=x Solution: Given y = 4x —x* andy = x 29y ate x ]0 Area= ff dy dx pix 4x —x? x09 3 LIE ay dx = ye ax = Gx =x? -d)dx = {2@x-x?)dx 1. Evaluate ff xdy dx over the region between the parabola y? = x and the lines x + y x=Oandx=1 + 2. Evaluate ff y® dxdy over the area of ellipse = +% =1 Ans: 22 3. Find the area between the curve y? = 4 —x and the line y2 =x Ans: = 4.6Area Enclosed by Plane Curves [Polar co-ordinates] Area= ff rdrd@ Problems Based on Area Enclosed by Plane Curves Polar Coordinates Example: 4.38 dr [A.U 2011][A.U 2014][A.U 2015] Find using double integral, the area of the cardi (1 + cos®) Solution: 30Area= ff rdrdo ‘The curve is symmetrical about the initial line @ varies from : 0 1 r varies from: 0 > a(1 + cos®) a(t003) gg Hence, required area = 2 fy" = fb ao = Jf'[a?C + cos0)? — 0]d0 =a? f"[1 + cos?0 + 2cos 0]d0 =a? fo [1 +422 + 2c050] a0 = =e +1 + cos20 + 4cos6] do = > JG[3 + cos20 + 4cos6] de == [ae-+ S32" + 4sino]” = [Gn+0+0)- (+040) 2 Fa’ square units Example: 4.39 Find the area of the cardioid r = a(1 — cos0) Solution: Area= ff rdrd0 ‘The curve is symmetrical about the initial line. 31© varies from : 0 1 varies from: 0 ~ a(1 — cos®) Hence, required area = 2 fO-" f"=90°°") rdrdo do = fi [r2|=gene"") ao = J'la*C1 - cos0)* — 0]d0 =a? f"[1 + cos*@ — 2cos @]d0 =a [1 + #5" — 2050] d0 oS [2 +1 +cos26 — 4cos6] de = > {.'[3 + cos20 — 4cosé] do sino)” = £[@n+0-0)-@+0-0)) a’mt square units. Example: 4.40 Find the area of a circle of radius ‘a’ by double integration in polar co-ordinates. Solution: Area= ff rdrd0 Onna The equation of circle with pole on the circle and diameter through the point as initial ine is 2acosd Area=2 X upper area SE (079° a0 = 4a? f2cos*eda = 4a? 5 +F = ma? square units 32Example: 4.41 Find the area of the lemniscates r’ 'cos20 by double integration. [A.U R-08] Solution: Area= ff rdrd0 O=x/2 X area of upper half of one loop. a4 rarao = 2860 = 2a fF cos26 d0 2 OS): = a square units, Example: 4.42 Find the area that lies inside the cardioid r = a(1 + cos@) and outside the circle r = a, by double integration. [A.U 2014] Solution: Area= ff rdrd0 ean Both the curves are symmetric about the initial line 33,Hence, the required area =2 fg52 [2° rdrdo ceo wo ee Pha =27 | foeesnt “ | ae =a? J2 [1 + cos?6 + 2cos @ — 1]d0 oleae + 2cos0] a0 © (a[1 + cos20 + 4cosé] do = £[(F+0+4)-@+0+0)] =< (r+ 8)square units. Example: 4.43, Find the area inside the circle r = asin@ and outside the cardioid r = a(1 — cos®) [A.U.Jan.2009] Solution: oxen oo ras Onn ono ca From the figure, we get 0 varies from : 0 > 5 r varies from: a(1 —cos@) — asin® The required area = ey epaoy FEES 2 pyayr=asin®) -8 EE 6 0 L2]p=a(1-cos8) _ pk fetsinta _ a= cosy? leer ees 34= [Isin?9 — 1 —cos*d + 2cos6] do =© [ff sintoaa — f2d0 — [3 cos*ad0 + 2 J? cosoda = £[-1015 + 2lsinalj] Fsintedo = [Fcos?edo -£L-(E-0)+20-0) =i 3] = fj sinodo = 2.1] Exercise: 4.6 1. Evaluate ff rsin6 drd0 over the cardioid r = a(1 — cos®) above the initial line. Ans: 2 3 2. Evaluate {f 25 over one loop of the lemniscates. r? = a?cos20 Ans:a (2-2) 3. Find by double integration the area bounded by the circles r = 2sin0 andr = 4sin@ Ans:3 1 4. Find the area outside r = 2acos® and inside r = a(1 + cos0) Ans: = 4.7 Triple Integrals Triple integration in cartesian co-ordinates is defined over a region R is defined by Wa fly. dxdydz or fff, fy.2dV or fff, flx,y,z)d0xy,2) Type I - Problems on Triple Integrals Example: 4.46 Evaluate ff) {s(x-+ + z)dzdydx Solution: 37= (oe + 24%) be =e 2 =atb+o Example: 4.47 Evaluate {fix fy" xdzdxdy Solution: So Spe So" xdzdxay = ff) bea 5-* axdy ap = fp Sex —») dxdy = Io Spel — x?) dxdy Example: 4.48 Evaluate J. fy” {o"” e'dxdydz Solution: SESE BE et dxayde = ff f* fe etdadydx = fee 1S dyaxe = foe"? ~ 1) dydx Sle? —yls* dx Ser* 1 4+x-e*) dx = fie-1+x-e%)dx = [e212 ef 38e-1+3-e-040-04+1 Example: 4.49 a Ne eT Evaluate fo fo" ” Y""* ” dadxdy Solutic GIF PO draxdy = IF aE axay SI fat =H dxdy = IN | (VERT x2 aay =f F (a? — y?) — x? + Example: 4.50 Evaluate f°?" {2 [2"" et **dzdydx Solution: $088 fF [2 ext etdzdydx = [08 fer He] dydx gon °F (2 — e+) dydx tog a [ez a re aL (E- el te 393 2ioga 4 ploga 2434 ate spoon gas (2 eas [velox Type:II Problem on Triple Integral if region is given Example: 4.51 x] Express the region x > 0,y = 0,z > 0,x? + y? + 2? < 1 by triple integration. Solution: For the given region, z varies from 0 to fT — x? —y? y varies from 0 to VI— x? x varies from 0 to 1 a ie So So my Se dadydx Example: 4.52 Evaluate fff x*yzdxdydz taken over the tetrahedron bounded by the planes x=0,y=0,2=Oand +2 4+7=1, Solution: Xyvyze Given $+24+2=1 Limits are, z varies from 0 to ¢(1-# y varies from 0 t0 b (1-2 x varies from 0 toa IW xyzdxayaa = fo OD sol [re=1-3]2 pa pbk 2 2k = Sf Phey (ie + - 22) dyax HES Ps? (yk? +8 — 2) aya aohe lS phe (+e “Se CH a ME pene (et 204 ave 232( =e ett 2 _ ware x? | wean | x peg ee Example: 4.53 Find the value of {ff xyzdxdydz through the positive spherical octant for which x+y ta? t=a* —2xdx = dt =a>t=0 Example: 4.54 Evaluate fff, (x + y + 2) dxdydz where D:1 {o* 2°" x dzdydx 3. Bvaluate ff Sp xyzdedyax (9v3 - 1) - log3| B4, Evaluate fy°8* J ("87 exty+dzdydx Ans:3 S-Evaluate fof 0°" &*¥*%dadydx Ans: [e* — 6c + Be" 6. Evaluate fff, (x + y +z) dxdydz where the region V is bounded by xty+z=a(a>0),x=0y=0,2 Ans: = 1. Evaluate J ne * wr Hy ps “Ans: 8. Evaluate f°, 7 70 + y +2) dxdydz ‘Ans: 0 4.8 Triple Integrals — Volume of Solids Volume = fff, dzdydx where V is the volume of the given surface. Example: 4.56 Find the volume of the sphere x? + y? + 2? = Solution: Volume = 8 X volume of the first octant z varies from 0 to fa? — x? — y? y varies from 0 to Va? — x? x varies from 0 to a F dadydx aap = pe Ze aydx Ig fa? =x? =? dydx =a fee |W aw) — y? dyax =8 fe Ley? + =8 J (0+ sin*1 — 0) dx = 4 {2@? —x2)Edx f(a? — x2) dx a4= cu.units. 3 Example: 4.57 Find the volume of the ellipsoid = Solution: Volume =8 X volume of the first octant z varies from 010 ¢/1- 5-4 y varies from 0 to |1 —2> x varies from 0 to a peop ehee dadydx a PY ya Be =8f fy “Izl) @ ™ dydx where k? = b? (1 — Pe lo HEL (lie = Pie ax saya be (1-3) ax = 2ber fi (1-4) dx = aber fe- 2] (0+ sin-*1 -0) ax = 2ben|a-2) AS= 2hen (a -$) = 2ben x 2 smane ‘1 = 2° cu.units. Example: 4.58 Find the volume of the tetrahedron bounded by the coordinate planes and = +5 + Solution: Volume = fff, dzdydx 2 varies from 0 to ¢ (1 — y varies from 0 to b (1 x varies from 0 to a v=f ped i (39) azayax =e = £5 abe =-F 0-1) == cu.units 6 Example: 4.59 Evaluate fff, dxdydz where V is the volume enclosed by the cylinder x + y? = 1 and the planes z = 0 and z = 2— x. AGSolution: In the positive octant, the limits are z varies from 0 to 2—x x varies from 0 to fT y2 y varies from —1 to 7 Sf dxdydz = 2 f°, 0° §2* azdxdy = 2°," Lal dxay =2 ft, 2 - w axdy =4 fit] ay - fl1-y"lay =A I= Hai] 8 nt] - Exercise: 4.8 1, Find the volume of the tetrahedron whose vertices are (0,0,0), (0,1,0), (1,0,0) and (0,0,1) 1 Ans: Zcu.units. 2. Evaluate fff, dzdxdy, where V is the volume enclosed by the cylinder x? + y? = 4 and the planes y +z = 4 and z= 0. Ans:/ 6rtcu. units 3. Find the volume of the region bounded by the paraboloid z = x? + y? and the plane z = 4 Ans:8ncu. units 4, Find the volume of the ellipsoid = + += = 1 by using triple integration Ans: 3 2ncu. units 5. Find the volume of the tetrahedron bounded by coordinate planes and the plane 2 Ans: 4cu. units xyy ptht a ay4.9 Change of variables in Double integral 4.9(a) Evaluation of double integrals by changing Cartesian to polar co- ordinates: Working rul Step:1 Check the given order whether it is correct or not, Step:2 Write the equations by using given limits. Step:3 By using the equations sketch the region of integration. Step:4 Replacement: put.x=reos@ ,y = rsinO , x? +? = 7? and dxdy = rdrdO Step:5 Find r limits(draw radial strip inside the region) and @ limits and evaluate the integral Example: 4.60 Change into polar co-ordinates and then evaluate f," f," e“"+¥ dydx [AU June 2011,Dec2005] Solution: =n (Cartesian form) O=-W22 (Rota fren} Given order dyad is in correct form, Given limits are y : 0 > 0 x: 0 > 0 Equations are y =0,y = 0,x = 0,x = 0 Replacement: Put x? + y? =r? , dydx = rdrd@ Limits: 4g1040, 0048 SSIS PM aya = (FIP er Substitution: Put r’ = 2B r= + e%d0 =!F@+1d0 (re = 0,e = 1) = 7 Je ae =7O5 Example: 4.61 Change into polar co-ordinates and then evaluate > Si za avax Solution: Given order dxdy is in correct form Given limits are x: y> a, ys 044 Equations are x =y,x=a,y=0,y=a Replacement: Put x= 2 dxdy = rdrdo 49Limits: 7:0 > =25 6:03 % Je Se a ayde = iirc 2 para = filrcosoly* do [e —0)a0 =a g 6 = 400); =a(Z-0) + Example: 4.62 Evaluate > S ixdy by changing into polar co-ordinates. [AU Apr 2009, May 2005,Nov 1998] Solution: Given order dxdy is in correct form Given limits arex y> a, y: 04a Equations are x =y,x=@,y=0,y= Replacement: Ve +y?, dxdy ~rdrdo Put x2 = eos’, x? +9" Limits: 0 > 35 8:0 > cost SoS a axdy = fe gore sorta rdrd@ we [Ecostof a i 50= fF 2ypcos*e — 0) do 36056 = £f§ 5 c0s*9 a0 cos = Se seco do = Llog(seco + tand))3 = £ flog (sec® + tan) — tog (seco + tand)] = © [log(v2 + 1) - log - 0] = “tog(v2 +1) Note: 2 1, x? +y? =1rcos’0 + sin? = 2. {Fcos*ed@ = J? sin?eda = 3. fecos*edo = fz sintodo =2x2x* 3. fZcos*ede = J? sin*ede = 2x2 x® 4, [Zcos*@sin*ed8 = +x? x= Example: 4,63 By changing into polar co-ordinates and evaluate J." f(x? + y2)dydx [AU Dec 1999, AU A/M 2011] Solution: o=a2 ar the circle J [Polar form} Given order dyad is in correet form. Given limits are y : 0 V2ax— x7, x: 0> 2a 51Equations are y = 0, y = V2ax—x?, x= 0,x=2a y? = ax —x? x? +y? —2ax = 0 isa circle with centre (a,0) and radius ‘a’. Replacement: Put x? + y? ,dxdy = rdrd@ Limits: r:0 > 2acos@ , 6:0 > 5 S29 fF (2 by? dyae = 3 2° 2? x rardo = Ee drdo E ppayzacose “GE, = 3 — 0) 40 = fat {2 cos*0 do m4at xxix Co (Fcosteda =2 «2x2 ) rat we Example: 4.64 ine is i 2 Vix By changing into polar co-ordinates and evaluate fy {> Fae ody Solution: O=x2 he circle. 0-22 polar form) Given order dixdy is in incorrect form. The correct form is dydx = [2 {)*™ =e dydx Given limits are y : 0» V2x— x? , x: 02 Equations are y = 0, y= V2R—X? x= 0, 4-2 52x 2x = Ois a circle with centre (1,0) and radius ‘1" Replacement: Put x =reosO, x? + y? = 1?, drdy = rdrdO Limits: r 0+ 2cos8 , 8:0 += 2080 reos8 dydx = J? J; a x rdrdo oh =[Z[rcoso]3"? do = §i2cos9 -0)d6 = 2 ff cos? dé =2x1x% fs fF cos? =2xixt — (« fZcosoao Example: 4.65 By changing into polar co-ordinates and evaluate fy {)° x? + ydydx Solution: o=n 1 the circle, Given order dxdy is in correct form, Given limits are y : 0» Va? a2, x: 0 a a? =x? x= 0,x Equations are y = 0, x? +y? =a? isa circle with centre (0,0) and radius ‘a’ Replacement: Para fx? ty? dydx = rdrdo 53. Put x? +y?Limits: r:0 >a , 0:0 += RR (P+ yFdydx = J? [> rx rdrdo = (3 ff arae Example: 4.66 Solution: drd0 Replacement: Put x? = r2cos?6, y? = r?sin20 2, dxdy = rdrdo Given the region is between the circles x? + y= x+y? 54Limits: r:a > b , 0:0 20 2 andy = JP [2 Peeters x yarag s?@xsin?@ 2m pb = fo fp ees x drdo = {2" 2 r¥cos?0 x sin®0 drdo ab =f," [FE] cos?e x sin?0 do = tet —a*) cos?0 x sin20 do ot a8) 92m 552 2 =) (° cos?8 x sin? d® WA 4 x fF eoste xsinto de ( (°" =4f2 ) = (bt at) x f2cos?0 x sin? do =(b* + f2cos?0sin2ed0 =! x2 x 5) (b' [3 cos*Osin?0d0 = 7 x>x >) at)xtxtx® aX2%9 nibt-at) Example: 4.67 fa? — x? —y? dydx by transforming into polar co-ordinates. [AU May 2011,June 2008,Nov 2007] Evaluate f° {2 Solution: o-x2 olar 1 the circle. Given order dydx is in correct form Given limits are y : 0-9 Va? =x?, x: 0a a? =x? ,.x=0,x=a Replacement 5s.= (2 +y?) =a? = 1? dydx = rdrdo ai Limits: r: 0 >a , 0:0 > Joo fH dye = fF fo Va? rerae = BUS Ve =F rarjda Substitution: Put a?—r?=t if r=0>t=a? —2rdr=dt ifr=a>t=0 rdr= —# 2 atra?+0 Zea J o0 dt, SGUp NaF rdryd0 = f3U2 Ve Sd0 = 50 F SZ Vide a0 = Use Bardo Evaluate the following by changing into polar co-ordinates. Via? Le RO aya 24 Kore +y*)dydx 563. fp fr" xy dxdy Ans: 2 SS aitay sm 5 fee PF A axdy 6 SQ? + y dyad Ans: 32° 7. fo RO (ty + y?)dxdy Ans: = 8 Lt + Dayar Ans: 2% <1 10.ff wap dy over the positive quadrant of the circle x? + y* Change of Variables in Triple Integral 4.9(b) Change of variables from Cartesian co- ordinates to cylindrical co — ordinates. To convert from Cartesian to cylindrical polar coordinates system we have the following transformation, x=rcosé aren * Hence the integral becomes II f(x, y,z) dzdydx = i f(r, 0,2) dzdrdo Example: 4.68 Find the volume of a solid bounded by the spherical surface x? + y? +z? = 4a? and the cylinder x? + y? — 2ay = 0. Solution: 57Cylindrical co — ordinates x=rcos@ y=rsino z= The equation of the sphere x? + y? +2? = 4a r2cos? 0 + 1? sin? 0 +22 r? +2? = 4a® And the cylinder x? +y? — 2ay = 0 x? ty? = 2ay r’cos* +r? sin? @= 2arsin@ r? = 2arsin @ r= 2asin0 Hence, the required volume, Volume = J f fax dy dz = Jf Jrdodrdz Va a4 fr? pease rdz.dr do a4? am? gar? ar do asin ® 4 0" [-S4a?— 2972] a0 Se" [-Ga? — 4a? sin? 0)? + 82°] 40 £ §(-8a? cos? 0 + 8a*) d0 534 gaa V2, = 5 8a? {C1 — cos? 0) do cubic units Example: 4.69 Find the volume of the portion of the cylinder x + y = 1 intercepted between the plane x = 0 and the paraboloid x? + y? = 4— Solution: Cylindrical co — ordinates x=rcos6 y=rsind z Given x? ty? = 1 cos? 0 +r? sin? @ = 1 rei r= dt Given x? ty? = 4-2 rcos? 0 +r? sin? 0 = 4-2 a 4-2 z=4-7 Hence the required volume Volume = ff frdz dr de = R" ff P razarao =" fo rad? arao = fo" f. r= 1?) drdo =f" f) G@r- 13) ar do do 2 4 =e" fe = &" [2@-2)- @-0] a tae = 2 [2n-0] = 3 weubic.units 59Example: 4.70 Find the volume bounded by the paraboloid x? + y? = az, and the cylinder x? + y? = 2ay and the plane z = 0 Solution: Cylindrical co — ordinates reos8 y=rsin@ z The equation of the sphere x? + y2+= az rcos? 0 +r? sin? 0 = az And the cylinder x? +y? = 2ay rcos? 6 +r? sin? @ = 2arsin® 2arsin 6 2asin Hence, the required volume, Volume = J f f dx dy dz = Jf frdodrdz 4a? x 2 f'!? sin*ede 4222 =4a9 x2 4.9(c)Change of variables from Cartesian Co — ordinates to spherical Polar Co - ordinates To convert from Cartesian to spherical polar co-ordinates system we have the following transformation =rsindcosp y= rsindsinp z= rcos0 60acy) j= 22 ar 8.@) sind Hence the integral becomes {i f(x,y,2) dedydx = i f(r, 0,2)r2sin0 drdody Example: 4.71 Evaluate f ff dx dy dz over the region bounded by the sphere xe+y2+22=1. Solution: Let us transform this integral in spherical polar co — ordinates by taking x=rsin@ cos y=rsin@ sin z=rcosO dx dy dz = (r? sin 8) dr de dp Hence ¢ varies from Oto 27 varies from 0 tom varies from 0 to 1 an pm pt = So" So So aaa Jo" ae] [{f'sino do] [J r* sin dr dodo ar] 4 =Qn-0) 141) ES ar = 40 fe ar Putr=sint ; dr= cost dt = t=0 rs1ate2 /2_sin? An Jo Fan cost dt = an (27 costae an fr? cost at = 4m J” sin? tat 61i} = a Example: 4.72 i= iz Evaluate oo" Sire es Solution: Given varies from 0 to 1 yvaries from Oto VI— x? zvaries from x? Fy? to 1 Let us transform this integral into spherical polar co — ordinates by using x=rsin8cos sin sin cos® dx dy dz = (r? sin) dr a6 a Let 2= x? Fy? = P= x+y? = 1? cos?@ =r? sin20 cos’ + 1? sin?0 sin? > cos? @= sin? @ [y cos®p + sin’ =1] 205 Let z=1 = rcos@=1 ares cos6 > r=secd The region of integration is common to the cone z? = x? + y? and the cylinder x? + y? = 1 bounded by the plane z = 1 in the positive octant. Limits ofr: r=0 to r=secé Limits of 6: @=0 to O=* Limits of: =0 to p= 2 = IP Gt sine drdadh — =f? 62/" °°" sin dr dB de w/2 (s/s [oo repeee n/2 af [sec?0 sind. = fr" [sine E] ao ag HL | a0 46 w/2 ait a pny + = fo fe/4 Esec Otan 0 dd ab = EG? ae | [G* secotano ao 62Example: 4,73 Evaluate { [ f (x? + y? + 2? dx dy dz taken over the region bounded by the volume enclosed by the sphere x? + y? + z’ Solution: Let us convert the given integral into spherical polar co — ordinates. x sinO cos = x? 2 sin? 8 cos? ysrsin@sing = y? =r?sin?@sin? > z rcos® = 27 =r* cos? 6 dx dy dz = (r? sin @) dr d0 dg SS [G+ y+ 2 )dxdydz = [0 E" Fr? ©? sin 0 do do dr) Limitsofr: r=0 to r=1 Limits of@: @=0 to @=7 Limits of: =0 to 6=2n SS SOP + y? + 2? dx dy dz = fT)" fpr? (e? sin 6 do de dr) = [Prt ar] [f'sin 0 a0] [ae] [E]; Ccosors 818" (3-0) a+) @r-0) (2) @e@n 63.