Io eI 10 Multiple Integrals Dmav 140.1 INTRODUCTION ortant topic in science and technology due to its large Integration is a very imp gy. The number of applications in different branches of science and technolo; students are already familiar with the integration containing single variable. But. the integration with several variables (two or more) are also used in different areas, such as, to find surface area, volume ofa solid, moment of inertia, ce gravity, etc. In this chapter, we shall discuss about the line integral, double i and triple integral of the functions containing two or three variables. ntre of 10.2 LINE INTEGRAL The line integrals are integrals of functions defined over curves. The curv be plane curves or curves in space. Let x=$(r) and y =y/(t) be the parame! and Y are functions defined in an interval |, in containing the curve C. D Let {02 = ff ns fps byrne by = Bo Where by < fy So In be any division of [a, B), see and let &, be any point in the interval ie A). Let x, = 9(t,), y, =yU,), and let ¢, y Now, we consider the sum S = Ef (GG, WG) Or ~ * = Ef (HE,), WAS) Ae, Re 4s, is the length of the interval [x, > "1 es may acurve C, where ? tric equation of dina A). Let f be a function defines rt) 303egrals (Ch 10) nof D, is infinite then the maxim — ju ds to 0, then the sum § lengt > 2 tends to gy tty Multiple ber of divisions. Ifthe num > (called norm) ten nes ol ‘ i sae atthe choice of the points ¢,, we denote this ny! w S Limi 4, symbol ; MOY ty i x, y)/dx= li (OE fe y)dx, i.e. [re y) lim © FOE). WE)) ay and this integral is called line integral of f(x, y) along the curve ¢. ifthe functions f+ @,y are continuous and @ possesses a continuoy, "then the above integral surely exists. derivate if x=o(t) and y= y(n,asts bbe the parametric equation of the : uve ¢ b ten [fs pace [ F000. vO) 70 IS a the curve Cbe y=O(x), aS x50 then b J fox yav=f F(x, 02) 6(x) dx. 3 Also, if Properties of line integrals In the above discussion, we have seen that the line integral can be r Ih a S ‘educed to an x alee a basic properties of the line integral are almost similar (a) Ifkisa constant, then [afte nae Kf 00a (b) Eee pabasive integrable functions on C, then f+ g is also integrable [tea f fac ef oa c ic (©) If f(x, y)20 for all x and y, then ah [te y)dx 20. a integrable on C, then || is also integrable on C; and I(x, y)dx|s (e) Ifan are ABis J ee composed of two arcs AC and CB, then een Jferd=f sanaref f(s yee a ded t0 a finite number of arcs. :Line Integral (Sec. 10.2) | 305 PLE 10.1 Evaluate the integral i) (2? dx + xydy) taken along c the line segment from (1, 0) to (0, 1), i) (ii) the first quadrant of the circle x = cos f, y = sin 1, joining the same points. (WBUT 2005) solution (i) The equation of the line joining (1, 0) and (0, 1) isx + y= 1 o fear tan = fie nae c 1 [ox +y= Land dx + dy=0] % 1 S j (2x? - x)dxr=-= 1 6 (ii) Inthe first quadrant of the circle x = cos f, y= sin ¢, joining the points, (1, 0) and (0, 1) the ¢ varies from 0 to 2/2. l2 Thus, j (x? dx + xydy) | (—cos” t sin t + cos” sin 1) dt = 0. c 5 EXAMPLE 10.2 Evaluate J (2 +y) dr, where Cis the are of the parabola c y= 4 ax between (0, 0) and (a, 2a). Solution First Method Then We express the curve as a single valued function of y, be. X= 47 integrand becomes a function of y and y varies from 0 to 2a (Figure 10.1). A (2a) (0, 0)| Figure 10.1? = 4ax. de 4 J (e +y arf ‘(a y c 0 (16a?“if eli) 2a | 16a? al] _t 64a® | 16a*] 7 2a| 16a? 6 am | 3° Second Method Let x = af and y = 2at be the parametric equation of y? = 4ax. Between (0. 0) and (a, 2a), t varies from 0 to 1. 1 fe + y)de= f(a 44a?) 2a0 c 0 1 ee) 7 = 2a" f (+40?) dr=20°. eet |=—a 0 6 3 EXAMPLE 10.3 Find the value of J (ay + y?) dx +x? dy, taken in the c clockwise sense along the closed curve C formed by y = x? and y = x Solution The curve C consists of two arcs OCB (y = x) and BAO (y = °) as shown? Figure 10,2. Figure 10.2 Cformed by y = x? and y = x J {ay + y*) det? dy) ¢ =| (Gay + rdvvat ay | (Oy + y2)dve x’ a) oce aIntegral (Sec. 10.2) | 307 reel koy Pe afueenaee asj+f (08 +24) det x2 2x dx} 0 0 -[arae] (x4 +32) de . 1 0 4 Compute j (2°y dx + y? x dy), where C is the rectangle i EXAMPLE 10.4 putes | y’ xdy, ectangle in the xy-plane bounded by x= 0, x=a,y=0, y=b. Solution Here the curve C is the combination of the lines OA, AB, BD and DO as shown in Figure 10.3. y=b Bia, b) heza ek y=0 A(a, 0) Figure 10.3 The rectangle bounded by x = 0, x= a, y=0, y= b. Me l,l. s+. pene the line OA, y = 0 and x varies from 0 to a. Aen the line AB, x = a and y varies from 0 to b. Along the line BD, y = b and x varies from a to 0. the line DO, x = 0 and y varies from 4 to 0. j (2? y det y*x dy) = j (2? 0dr +0.x.0) bn 0 Poy =0,dy=O1 =0,308 | Multiple Integrals (Ch. 10) i J ctydreyxdn= [0+ y’-a-dy) AB . r 0 ayers yx) | (x2 -b- dx +0) BD - [ey =b,dy=9) 5 J etydr+y'ed)=[ O+y? 0-d)=0 Fe , [7x =0,dr=0] 3 3 Hence [ccrartytran=0r S740 td 8 3 aha 2 =” wy? - a”)! 3 ¢ a’), EXAMPLE 10.5 Show that J (y dx — x dy) = 247, where zis the arc of the T eycloid x = 2 (t— sin 1), y = 2 (1 — cos 4), joining the points (0, 0) and 47 0 shown in Figure 10.4. | Solution In this case, f varies from 0 to 27. ) as (0, 0) Br. 0) Figure 10.4 Cyloid. Ng =] (2(1-c lo ©0861) 2(1 ~ cos 1) ~ 2 (¢ ~ sin 1) (2 sin 1)) OO —il ™ Line Integral (Sec. 10.2) | 309 a é| {4(1-cost)? = 4sin (1 sin) dt 0 an a4] [2=2cos¢=rsin 1] dr 0 24[21—2sin +1 C08 1 — sin)” 24[4n +27] =24z. nd the val ot | 2 c i 106 Fi ju i y dx + xy’ dy) taken in the clockwise ww the hexagon whose vertices are (+3a, ()), (+2a, + v3a) Figure 10.5 Hexagon. of the lines of the curve Care: AB: y=—V3a BC: y - V3x+3V3a=0 . CD: y + V3x-3N3a=0 DE: y= Ba EF : y- V3x-3V3a=0 _ FAzy+3x+3V3a=0. ombination of the lines AF, FE, EP. pe, CB, BA310 | Multiple Integrals (Ch. 10) Hence, J (x°y de + xy? dy) -|-Jo# (V3x + 3V3a) de “fe / (+ i + oo v3a f x? (V3x + 3y3a) de +f y? ( sid ta 0 ty V3a x? de 3a 0 +] | 2? (-V3x + 3V3a) de + (-4 sal (f a BJ] [fee (fir pars fy (a+*)5] 2a ., Bax de a =? (Bx- Nia dc+4 fy *(5 30) +2Via f 2 de 2a ws (Sinn Fp) —38 4 3 10.3 DOUBLE INTEGRAL The single integral is eval over a region R. Let f(x, ») be “ag “sled “14h and single valued function of x and y withi subdivided into n by a closed curve C. Let the luated over a line, while double integral isKe ee Double Integral (Sec. 10.3) | 314 AX, Figure 10.6 An elementary are AA, It is obvious that if n >< then AA, -> 0. The limit, lim Ef (xj.¥) MA ne i=] - may or may not exist. If jim z f (%;, y;) AA; exists then we say that f(x, y) is integrable over the region R and this limit is called the double integral of f(x, y) over R. The double integral of f(x, y) over R, is denoted by SJ fi R [fj fone asim F129 R ne i=l utation of double integral over R, we generally consider b-region or. rectangular grid. The rectangular ing R by lines parallel to the coordinate axes an elementary rectangular region is Ax, Ay, y) dA, i.e. To simply the comp the sub-region AA, as a rectangular sul sub-region are obtained by subdividi as shown in Figure 10.6. The area of the double integral SJ fx, y)dA is given by R nye (al =! ff flay) dx dy = lim z E fl 9d a6 Ay). R Properties of Double Integral 1. Iff(x, y) be integrable, 2. Iffis integrable in R, then || is also integrable in R. 3. Iffand g are integrable in R, then /* 8 fe and fig, where | g constant c, are also integrable. me Sr y) dx dy =k Sf. (x, y)drdy, 5. ff tats yy gts snide = JJ raneet fferae it is integrable on any sub-rectangle of R. | 2c for some where & is a constant312 | Multiple Inte grals (Ch 10) 6, Iffand gare integrable in R and if f2 g in R, then ff fix, y) dx dy ff g(x, y) dx dy. R R ¢ subdivided into two regions R, and R. >. Ifthe region R may b > then, Ly) ded) -ff f(x, y) dx dv+ff F(x, fre y) dx dy " pd x, y) dx dy, 40.3.1 Iterated or Repeated Integrals definition of double integral, that the computation ofd integral depends on the evaluation of a limit. But, practically determinatig this limit is not easy. It is, howe possible to evaluate a double integral a region by two successive single integrations. Suppose that the region R be defined as cSySd We observed, in the and g(y) SxS h(y) (see Figure 10.7). Oo Figure 10.7 Strips for repeated integrals. Let the equations of the curves ABC and i CD. a pe A be respectively x = Ay) Now, for each fixed y, f(x, y) is an integrable function of x, then this i defines a function of y say GQ) in cS y 4 “1 e repeated (ii) j F(x, y) dx exists for each value of yin eS y $d then the repeat ! ; integral : : i , yodv dy j yf f(x, y) dy exists and is equal to ffre y « abt Ch. 10) ee 314 | Multiple Imegrals ( e double integral it iggested that, first of all dray I W the To calculat i ‘ he range of X andy. Now, check the integrand /(x, y) it "ekion p itis gy and determine t oo pb wort. x then use {J f(x,y) dx dy a) yf fla ya R e * ay . Other ig b d f(xy)ae dy= | arf f(xy) dy. R meee to integrat use the relation {J Ifa double integral exists, th being equal. However, if the ay or may not exist. repeated integrals exist and be equal then double stence of one or both of the repeated integ integral f the double integral. Brals is ng Js exist but are unequal, the double integral ca mn hen the two repeated integrals car ‘annot double integral does not exist the CXist c Tepe ated Note: without integrals m: Again, if one or both may not exist, ice. the exi guarantee for the existence 0! If the two repeated integral exist. 1 he Deccary? 1 EXAMPLE 10.7. Show that J arf Eaeey* 7, =f Me P lo ox ty" th Solution i i ae ; oxty? =f, Filet uns [ae ana! i 1 . here ay=[ (orn t-1Jar=a — jze%e-ffr 2y? , erty do — Pe) arai—2y en eo? 2 RHS= fay x -y 1 0 (eae f{1-2yan" L]ay-0 Thus, cat 2-2 1 lo spr ty =0= | i I —y? ty lo doxt+ ye EXAMPLE 10.8 Prove that Walia we fa f FN) 7 O(x+ yy v (x+y)_ \ } — _Pouble Integral (See. 10.3) | 315 solution \ 1 yey FaY aya ({ te o(xty) oL(xt yy Crt yyt dy -| x rey I =|-—; + | = — | (x-y)? (ty) |, +x)? , . LHS = fa f= iy ee ea[ ol) = feet lakes I+z], Pj 2 2 : 2y lee a rd Letactal-e Gt» Ga (i+ yy 1 ‘md us 1 rus=f df 2% ar | et 0 | Jo(x+ yy o (1+ y) ade ; ee eo 2 2 ete te fal othe laa ieee 0 ‘0 (x+y) EXAMPLE 10.9 Evaluate the double integral 1 pd [frre ofr Solution ‘limits of x are 1 and 2 and that of y are 0 and 1. We write the double Js {: [fe +0) ast fo + xy) de [taken y as constant] 1 Fesin (x + y) dx dy. ald paid EXAMPLE 10.10 Find the value of J i} 7 fo Jo Solution Here the limits of x are 0 and 2/2 and that of y are O and aA. Cp en a ay a 4 Beets if" lz J sin x+y) as] 0 on -[ [cos (/2+ y) +c0s ¥]4y la =| (sin y + cos y) dy 0 =[-cos y+sin yf" =(-cos E+sin +coso-sind — 4 4 1 2 1-051. 2—————si sr srr Double Integral (Sec. 10.3) | 317 710.12 Evaluate Sf xy dx dy where R is the positive quadrant of R ty=a. of x2 +? =a? is shown in on of integration, i.e. the first quadrant awe O} (0, 0) 10.9 First quadrant of 2+ =a’.tegrals (Ch. 10) 318 We consider an elementary strip parallel to the x-axis. From Pi see that the strip extends from the y-ay Bune .€.x = 0, to the circle x2 4 09 a = qi which x= a? —y°. Hence the limits of x are x = 0 to x= : y entire region, the strips start with the minimum value of y, i.e. the y. maximum value of y, i.e. the point where the curve intersects the limits of y are therefore from y= 0 to y=a Thus, the given integral is pet [Loans f For Axis tg y-axis, [PLE 10. Fah 0.13 Evaluate j xy (x + y) dx dy over the area bound by = ‘ (WBUT Solution The region of integration is shown in Figure 10.10 yDouble Imegral (See. 10.3) | 319 we integrate first with respect to x and then wir, y. Therefore, we draw a e rip (BC) parallel to x-axis, At B, the value of x is y and at C, x Jy. ie. the of y are y and vy. The limits of y are then 0 and 1 sofa piven integral is equal to 1 vi j J xy (V+ y) de dy = er jim oni ; Hence the gi 0 if sa ’ ‘ ] J Me ad y y lay of 3 2 hs, Pee, a 2egliben 94 sy] D iregiam ig (ign Ws, be 1 3 =—+---= ite NGiS6 EXAMPLE 10.14 Evaluate Jer — y? dx dy over the triangle formed by the straight lines y = 0, x= 1 and y= x. (WBUT 2003, 2005 Solution The triangle bounded by the given lines are shown in Figure 10.11. We int first w.rt. y, so we draw a strip CD parallel to y-axis. At C, y = 0 and at D, y = x. ie. y varies from 0 to x. Also, x varies from 0 to 1. y =x ‘B(1, 1) Do ea 0 oA ’ Figure 10.11 Region bounded by y= 0, x= 1 and y= x. Hence the given integral equals to fi [er ~y dydx= five Be ee a 2 dX Ta. y w+ 2y' sin! fu5 | eae ero Re 1 es feu 2 0 2 @ EXAMPLE 10.15 Evaluate J © dr dy, E is bounded by x=2,y=x,xy=1, 9) ey Solution The region £ is shown in Figure 10.12. We integrate first w.r.t. y, so we draw a strip DF parallel to = At D, xy = 1, ie. y= I/x and at F, y =x. Therefore, the limits of y are — 1 and x. The minimum and maximum values of x within Z are | and 2. Thus, the ia ofxare and 2. Figure 10.12 Region bounded by x = 2, y= xand x¥= 1: tell 5 de dy = -[’ JouDouble Integral (Sec. 10.3) | 321 EXAMPLE 10.16 By changing the order of integration, evaluate 2a 3a-y j J (x? + y*) dr dy. 0 0 yi4a Solution Here the order of integration is first w.r.t. x and then w.r.t. y. The limits of x are x=y7/4a and x =3a-y and that of y are y = 0 and y = 2a. That is, the region of integration is bounded by x =)7/4 or y? = dav; x=3a~yorx + y=3a; y=Oand y=2a, which is shown in Figure 10.13. To integrate first w.r.t. y, we draw a strip parallel to y-axis. The strip starts at y = 0 but ends at two different curves. $0, we divide the region into two parts R, and R, by the line AC. The coordinates of A, B and C are (a, 0), (3a, 0) and (a, 2a). In region Rj, y varies from Oto Jax; and x varies from 0 to a. In region R,, y varies from 0 to 3a —x and x varies from a to 3a. Figure 10.13 Region of integration is shaded by horizontal lines. Thus, the given integral 2a 3a-y = 7 j J (2 +y*)aedy= ff (2 +y)drdy+ ff (2 + y?) de dy 0d y2740 Rh R =f fre +a der) tae +y?) dy dx a + vax ay y Bans . ay le Slee] ax f [eye] dx iMultiple Integrals (Ch. 10) _ a a yas 2 1 =f (ave e+ F(a" ‘Jace J [: -3+5 Gea a EXAMPLE 10.1 4 Changing the order of integration, prove that Ux yd) alt fa +f 70 __ == (#0), d+y Solution (+a) 4 Here the region of integration R is R[osxstxsys4] a # ie. bounded by the curves x = 0, x = 1, y= x, xy = 1. The region is sho Figure 10.14. The region is divided into two sub-regions R, and R,. y c iB k y% Fe D(O, 1) R, A (1, 1) xy=1 Oo x Figure 10.14 Shaded area represents region of integ! or He region Ry: x varies from 0 to y and y varies from 0 to 1. In region Rs * varies from 0 to I/y and y varies from 1 {9, Thus the given integral is equal to My i atom era Lf,Double Integral (Sec. 10.3) | 323 1 ne | asmenttan 3 ; el + | y = tan 25 [tan y], la Giuton” @d0 [by putting y aA 14 tan? 6 sec? ( *)+J Bea 0 24 sec! 8 es La bd ate mn 0da= a | cos 0 z j si ae) (l=cos 28) di 0 ' x nid =f +t[o- i028 42 zal oh sie ther. B on24y 0204 EXAMPLE 10.18 Prove that f [fr «| du= [fo (x-u) du, 0 [Jo 0 where /(x) is a continuous function. Solution Here the region of integration R is R:[0stsu;0susx] This is shown in Figure 10.15. t * (x0) Figure 10.15 Region of integration shown by horizontal line. _ Weintegrate first w.r.t. u, so we draw a strip parallel to u-axis. The strip starts '=1and ends at u = x, f varies from O to x. LbCh. 10) 24 | Multiple Invegrals ( oe: =f rola af PO (x= dt F 5 = [rene 0 40.4 CHANGE OF VARIABLE IN DOUBLE INTEGRATION Like integration of single valued function, the method of substitution js also user, eful technique in double integrals. Let the integral be i f(x, y) de dy, Supa x=O (GV) VEY y) be the transformation (or substitution) for the Variables xand y from the region Rin xy-plane to anew region R’ in wv-plane. Assume that the functions @and yare continuous and map the region R to the region R’, If the functions ¢and yhave continuous first order derivatives, then Jf fonaa= Jf, fee) yeu, | 4 ae R R where J (#0) is the Jacobian defined by ax ay 9%, y) _|du du}, O(u,v) |ax oy av av ee a. a he positive aaa Freez +y ly over the positive q! (WBUT 2006)Change of Variable in Double Integration (Sec. 10.4) | 325 Oo hen i ax ay — 9%, Y) _| Or 00 a(r,@) | dy oy or 00 _|cos@ — —rsing ~ | sin@ rcos 0 In the first quadrant of the circle x?+? = I, r varies from 0 to | and @varies from 0 to 772. Also, dx dy =| J | dr d0 =r dr dd. qe 2 ay? 1 parr [y_ 72 jj = de dy = | j r dO dr l¢x ty lo Jo Vier a2 1 =. =| dof to dr 0 o\V 1+r Substituting 2 = cos 2 g. Then 2r dr =-2 sin dd When r=0 then ¢= 2/4 and when r= I then @= 0. _@[° |1-cos2¢ ~ 2 Sais\ 1+ 00s 26 (Gs ® 21. == -2sin ¢ cos 9 di ar cos @ misshee (-sin 29) dd ai ms g it (1—cos 29) do _af, sin2e)"" “5/9 al 0 x ft) Fie-2, m4 2. 8 EXAMPLE 10.29 Show that ffNie-e dx dy taken the upper half of Me circle 2 +52 _ 207 = 0 is 2 ae EXAMPLE 10.28 Show that ff __dedydz_1 [ me ;] integration being taken over the volume bound () led by the coordinate planes and the plane ae (WBUT 2005) Solution As in previous example, th li -x- 0 to Ss rare 0 to 1. P le limits of z are 0 to 1 — x - y; y are aif ee dx dy dz “| Los plek-y a G+ytesy? ai J, fare, Ier-y dy &__ Triple ime _ nt| gral (Sec. 10.5) | 335 | 1 > I FS ldye 2ivtyeiy g (0% Vai 2 -|-Fe+teas rit x) | ‘| heat 5] = > log2-—=-—| log e-2 |. poe Z loge 3] EXAMPLE 10.29 Evaluate Sf 2° dx dy dz over the region defined by z > 0. r+p+rsa’, (WBUT 2004) Solution The region of integration is the up| per half of the sphere x24 y2 4 #2 [since z> 0), We first integrate w.rst, 2, where z varies from 0 to integrate over xy plane, rrpag (obtained by Then we where z = 0. The region on xy plane is the circle substituting z = 0). [lea ac-ff ie 8 lei xe [where R is the region 2 +2 = @ inl, es 1 Te a ahd, y= Sf ; ; Now, we Substitute x ; lecircle x2 +52 = 42 4 2 _ y2yua dx dy =rcos @, y=rsin @ in the region of integration is the In this region r varies from 0 to a and @ varies from 0 to 2 oy % J = Ax, y) 9r, 6) J cia? ~ x? ~ y?)¥? dedy k3336 | Multiple Integrals (Ch. 10) 6 —— -[. [ye y? rdrdo ine ) J “aol (a? =r?) rdr ve 1 (=r aml a= 24-2, 3 ae 5/2 bo 5 _2n a’ _2za (Qin xa 3.5 15 1 Note: Jf f(x,y, 2) dx dy dz over the region defined by the sphere x? +24 2 =@ is equivalent to mi foe [a2 eB [«lpael pes f(x, y, z) dz jj f(x, y, 2) dx dy dz over the first octant of the sphere x+y equivalent to F(x, y, 2) dz. 2 fezeRt=2 dy J I fel, ‘Similarly, fj f(x, y) dx dy where the field of integration is the circle = @ is equivalent to the interval fia afer Flay) d. EXERCISES Short Answer Questions we (Section A) L eee] (xdv+ y dy) taken slong he ine seanExercises | 337 —— dx dy is equal to i if \ rdrd@ is equal to .. "Jo Jasin® x 5, The value of lhe dx dy, where R:{|x|S1,1S y $2} is ween 6. The value of the integral J (2 + y”) dx dy over the rectangle k R:{0SxSL1S yS2} is. 7. The value of | (y dx+xdy), where C is the curve x = cos f, y = sin f, c Osrsz/2 is 1 2 3 8 The value of [ j J peaeayde is equal to... o Jo Jo 2 x2 9. The j J xy dy dx is equal to ... 0 0 Te 10. The value of f ie xy dy dx is... jo Ix (Section B) 1. Find the value of fj e”'* dxdy, if the domain of integration is the r triangle bounded by the straight lines y= x, y= 0 and x= I. 2. Evaluate the following integrals: 2 ely (@) fi Jiasx+ nae ida () j i z [(e-p? +y? ]dvdy 4 1 (©) J, Jrc-naa». [Dea (a) i i vty 1 e ‘ : (e) | J x(x? + 9") dy dx 0 Jo (f) IJ ae dy where R:{|x|S 1,15 y $2} RY 8lz © > 1 pl 5. If f(x, y=aytx” then the value of f j f(x, y) dr dy is 0 40 (a) 2 D4 © fete 6. j j x dy dx is equal to 0 Jo L (a) 3 J ©) & i ee 1. The value of [ if yet (a) Va wie (c) 3 J dx dy dz is 0 (0) 6 (d) none of these (b) 6 (d) none of these (b) 81x (d) none of these Ez. (b) 35 (d) none of these 1 () 5 (d) none of these 2 ) 5 (d) none of theseExercises | 341 4. ucisthecirclex=c0s A.y=sin 0505 1/2 then j (xdy + y dx) is h © equal to b) 1 a) 0 ( a -l (d) none of these : rhe value of ff (y-2x) de dy, Ri{1S. x5 2,35 y<5) is . k @2 (b) 4 (0 (d) 1 10. [ fe dx dy is equal to (ae (b) (e~1) (© (e'-)(e-1) (d) none of these i ear 11. The value of f j ) xyz} dx dy dz, is equal to 0 Jo Jo me ah (@ 15 ) 1 Os (d) none of these 2. fi J, fi arayac is equal to (@O- (b) a(c—b) @ 0 (b) 0 (d) none of these a Long Answer cuore. Jite- yy? dx + (x— y)' dy} =3za", taken along the circleANSWERS ~~ Short Answer Questions (Section A) ate ™ Kae) Binal 2 “ xa’ 5. 0 8 2 & 5 9 72.0 8 oe 3 (8 () Soe? ()0 ae g) ws (c) 2 20 (c) we(3) @ pbAnswers | 347 Woe te SO y) de ay? cs af Oy) de to ote +] ayf P(x, y) de 0 P10 2 12. 7. (b) 8 (Section c) . (b) 3. (b) 4. (c) 5 8. (a) 9. (a) 10. (a) 13. (a) ta