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211-2005&2006-3-F10-July2006

211-2005&2006-3-F10-July2006 - Kuwait University...

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Unformatted text preview: Kuwait University Math. 211 July 30, 2006 Dept. of Math. & Comp. Sc. Final Exam Duration 2 hours Calculators and mobile phones are not allowed Answer the following questions. Each question is worth 4 points. 1. Test the absolute convergence, conditional convergence, or divergence of the series on n2 2( Willi) cos2n. n=1 1? 2. Find the interval of convergence of the power series CO 2 imp-,5 + 2)n_ 3. Use Lagrange multipliers to find the extreme values of flx,y) = 2:2 +322 —4x— 2y subject to the constraint 1:2 + y2 = 20. S 2 4. Reverse the order of integration to evaluate the integral J. J- ysinx7dxdy. 0 w 5. Find the value ofthe triple integral ofthe functionflx,y,z) = nyz over the region in the first octant bounded by the cylinder 2 = 2 — %x2 and the planes 2 = 0, y = x andy = 0. 2 «(E—7:7 9v“); 6. Use cylindrical coordinates to evaluate! J. I 2zdzajldx. O 0 Q 7' Let ?'(x=y,2) = x2 tan—'2? - XCOSyz-F +ye’227c’, a. Find div F'. b. Find curl F. 8. Evaluate the line integral dex + (x +y)a§J+xzdz c whereCisgiven byx : e“, y : 22‘, z = e’; 0 S t g 1. 9. Show that the line integral (4,3) L ”)(3x2 — 630061): + (—3x2 + 23nd}; is independent of path, and evaluate it by finding the potential function. 10. Use Green’s theorem to evaluate §(e" — @2de + (cosy + —§~x3 )dy, where C is the circle C x2+y2=4. OO 99 Kuwait University Math. 211 July 30, 2006 Dept. of Math. 3.: Comp. Sc. Final Exam Solution 1 ”2 1. (1—11—13)??? cosZn g (1—111—3) n n The series i 1 1. 111—3 *1 conver es b the Root Test ' n g Y 1 1 lim” (1—3—3)n2 =lim (l—E)n :e—ln3:l<1 n—voo n 11—‘00 fl 3 So the given series is absolutely convergent 1n n . un+1 la: + 2| . 2. If “a.” = (—1)"— (I + 2)”, then 11111 = , so by the Ratio Test, the n 3” rel—mo it” 3 series converges when 1:1: + 2[ < 3 4:} -5 < m < 1. When :1: z: —5, the series becomes 2 132 which diverges by I. T, B. C. T L. C T. When a:_ — 1 the series becomes 71. 71:1 00 Inn :1 [“1]?! —nw which converges conditionally. So the interval of convergence is (— 5 1] 11,: 3. g(m,y) 2332+y2-20 f: :Agm, fy=Agy =>233—4:2A:c, 2y—2:2/\y :1:=2y &$2+y2:20 y=12=m=i4 =>(—4,—2)1(4,2) f (—4, —2) = 40 is a L. Max. f (4.2) = 0 is a L. Min. 3' 2| 2 1'3 2 1 4. j/ysinm7dxdy :— // sine? (ydyjdz: = %/a:5 sinx7da: : _E [cosfl]: 0 ‘3/5 0 0 o l — cos 128) l E(1—C0527) = fi( 2.2 2 :1: (2:1:yz) )mdzdyd :,,(2//Iy 2x )2dyd33 = % 0 0 0R3 8 (”R-mt ll NIH H 4:. l tu—II—l H a: + 31H 8 ._$ to II 2 %/ (41$3 — 2m5 + £557) (112; 0 (1).») mm 2 V4 2? x/Q—wg—yg 2 4—3: ‘ g 2 6' f f / ZZdzdydx :/ j (9 “ $2 — yg) dyda: = [d9 / (9 -— 7'2) rd?“ 0 U 0 0 0 ‘ —-) _ 2 2 —1 3 2 a —zz Ta dwF _6w (3: tan z)+3y (~xcosy)+a(ye ) = 23: tan—1 2 + 21:3; sin y2 —— 21,12“?22 —> —> —> z j k: —* 3 (9 (9 7. z _ _ W b and F 6.1." 6y 4% (cc2 tan 12;) (—mcos yz) ye ’22 2 = (8—22) fl + (1:32) :1? — (cosy2) ? 1 8. [scrim + (a: + y) dy + :2:de 2 f6” (—e’tdt) + (8" + (32‘) (Qegtdt) + 6—38‘ (etdt) C 0 1 H ./(—8—2t + 33‘ + Qadtjdt 0 _ 1 _ [%e 2t+3€‘+%e‘“]fl= %e 2+%E4+3€—4 5M EN 9. = —6$ = ~— => [Mobs + Ndy is indepencient of path 0 .5};— 6‘3: 6 Roma—i=3x2-Bmy and 3—i—w3$2+2y => f($}y)=m3—~3332y+y2+0 {4.3) de + Ndy = [933 — 33:23; + yz] Eff)” = —71 + 3 z (—152) ’ 10. %(em — §y3) d3: + (cosy + $313) dy = // [6% (cosy + gma) — 5% (em — §y3)] dA C R 2172 =//2[$2+y2)dA=2//r2rdrd9 R 0 CI = 2W [irflfi = 2(2w)(4) =- , \ me CO ...
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