Solutions_Test3_A - So M'Hme MATH 251 sect Test#3 Variant A...

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Unformatted text preview: So M'Hme MATH 251 sect' , Test #3, Variant A, Fall 2010 Name ...................... i ... E ... i .. m ../.... ........... Closed book, no notes, calculators OK. Justify your answers by supporting every step of your reasoning, simply, show the whole work. Remember: you get a full credit for a complete answer. However, you may loose more credits if you are misstating the formula, or using wrong limits of integration, or fail to sketch the domain (when asked) than making simple technical error or a misprint. There are 10 problems each 10 pts for total 100 pts. The bonus problem is extra 5 pts (so you can score 105 % !!!). Here are some useful formulas: 1. One can use the following two parametric representation of a sphere with a radius a and center at the origin: xzasinq‘wosfi, yzasinqfisinfl, zzacosd, Og¢g7n 036$27r or xzucosv, y=usinv, z=\/a2—u2, 0<u<a, 0<v<27r. 2. For sphere with parametric presentation as given above we have the following expression for the normal vector and it magnitude r¢ >< r0 = (12(sin2 g1'>cos(?,sin2 gbsinfi, sinq5cos (15), |r¢ X rel = a2 sin Q5. 3. Some formulas for trigonometric functions: 1 — 2 1 2 2sinacosa : sin(2a), sin2 a = w, cos2 a = W. Your score: P#1 P#2 P#3 P#4 P#5 P#6 P#7 P#8 P#9 P#10 P#11 Total I mevrf’A 1 1. Calculate the line 1erntga1/F-r,d wher e=F (2:632y+21)+(3m2y2+2yz)j+(y2 +2.1 z)k aneclth uerv Cc 0n81 188t oefth hen seegmnts 011n ecietngtheop1018nt8,,1,P(100)WthQ(11,3) :ndthm 11CQ(11.,1,,3)WthR(211) Hint. sohwthatthe eev ct0 fieildFs ceons rievatv and 181: srp oepr erti.es :7 1‘ J K 01 11:: Y 129% 9*. b 172,.2,,11—21,11,%6112> x1313 3’8‘712/3‘ 121L271} 1: 4'71”)” 1 ygs E 17, WWW/114v; Vet/7w 1M4 W I; _. w: ‘— I' /\’2 ::fi(24,4) 180,017):22.l+2'+1+/:/[”+”+”/. =412—H 2 C: Vm’l'an‘i' A 3. Verify Green’s for the vector—field F(a:, y) : $3i + (my + ac) j and the curve C that is the circle as? + 3/2 2 4 oriented in a counterclockwise direction. (This means that you have to compute both sides in the Green’s theorem equality and show that they are equal.) 3253M+ [fl-MM?) : £7323- £50110»— KQ H)d'4_ U 0 c F 61 DD ‘ = but 9-st + _ . $315M ‘21:“; 23% (m/ 0991/ (MM/€05 ’ ¥~ \mbfi OAJCAZF "} 7mm? , 2r 553%; SL8 03% (43M)+(4ani&n++2w$t)émijdck ‘3 '7 (E0 ; 9 6 21% C O 4 34$ 0 2g) 70 /\ 9’ Zr =5i—WQL’WfiWWWWt Hmmyaw-z 859”” f , D f Zr 2 maid/ts (72 US [if mZtM/t :E : 3610939) Amrzédrzzgwix/ o . A: a q - 4. Use Green’s Theorem to find the work done by the force F(x,y) = (x2 + my)i + 3512ij in moving a particle from the origin along x-axis to (1, 0), then along the line segment to (0,2) and then back to the origin along the y—axis. 6. (Problem 10, Section 14.7, pp. 925) Evaluate the surface integral // ygzdS, where S is part of the sphere 11:2 + 3/2 + 22 = 2 that S lies above the cone z = «372 + yr". 2. . 33+ ‘71 + x243}: 2 X2” ; 4- g’lafls : wan-fig Ma Ilium? 5L 8mg 0% 1*:de o W 9”? L,.___J [-4 i 147? I 1 23!“?10 :Zgth'J-Té W..— i ZR 7. (p1ob1e1n 16, Section 14.6, pp. 913 modified) Find the equation of the tangent plane at point (1, —2, 1) to the parametric surface 111,1;1112271; “i=4 M V: 2— M-l=-—Z m v1=4 j v=l 27 a: (oil/1) 4) 0) V$Y\=<2—V) -4Mv) ~4VIV> 17;: 4 OJ ‘Zv’ 2V) fvr u 111th ”(x Dvfljn) +4621)» 8. Calculate the work done by the force field F = (33” + z2)i + (yy + 222) j + (2" + y2)k when a particle moves under its influence around the edge of the part of the plane m+2y +z = 1 that lies in the first octant, in the counterclockwise direction as viewed from above. Hint: Use the Stokes Theorem. 9: "Vfl .oi/WEQ x @310me :2/5 @«x) j(4~X)»~—4j{4—><lv/fll)< l/m, A 9. (Problem 36, Section 14.7, pp. 926) A fluid has density 15 and velocity field v :2 —yi + asj + 2zk. Find the rate of flow outward trough the upper half of the sphere x2 + y2 + 22 = 16. Hint: Use the fact that the flow rate is given by the surface integral // F ‘ dS, where F : {JV and 5’ is the surface. S mm: Km: "F: (sew—1L) S 577% Pmm‘n‘c X=4Sm (fan? 3=4£m€ «9M5 %:4®@ 059.4.- 2r mm? {0 A? g; 12; 31> gxg =16< 5»sz 60919) $13030 é‘mfl) 8”?”059? gsf§(*4£§fl(ffiw) 45"”‘lm’355 8603‘? a; (jg ._. [(46.69 ((#19qu WWW; J‘W‘MW ° 5 J) {‘fiM/Jmfib awn/903 £60360 ‘ 004 2 .3 m ~ 3 Hm/Jraw‘m w; 474 = 960 (K. 9% H51 *1 W :9 .Oarg A6553 WW) =MQD J gwfiegmejf age“: 2), 2J( 0: 0 0 i: [280W 1 10 M 10. (Problem 42, Review 14, pp. 940) Use the divergence Theorem to calculate the surface integral // F ~ dS, where S F : w3i + y3j + zk and S is the surface of the solid bounded by the cylinder 902 + y2 = 1, above the plane 2 = 0 and below the plane 2 = y + 3. 3542/ V: { 597)éo9=€><z+71.443 (952 $393 £(E’0(§= [aflal’gfifl/ = [(((3xl+3yz+)>ail/ V l/ ugh/7;: 5x2+ 372+4 W3 15MWeM:Jg{¢axz+s~,z+.wl .: 637 75 Mr WWW/445 (fl: aér‘i—j} #ng W 699.5% Dtl/‘S'thmfi’ =5 lérfirflm‘wm) r/Mr Q 0 (9 4 =j£<3rflfifg§ 97(19‘4‘ (3r3+r)3\24r jfll‘f‘ = gritJZVQ/ozrs 0 0. Bo 3 532751.} _ (1521’ 11 11 (BONUS PROBLEIVI no pa1tia101ed1t) (piobbm 12 pp 931) Calculate the1ineinteg1a1 </ 1-(11‘.\\11e1c F — (7722/ 333319317) and the cune C is the intelsection oi the h§ pe1bol1c 1) .7) . , .- . . . . 13511 ahoioid 3 : y“ — 3:“ w1th the cyhnder x2 + y2 = 1 oriented counterciockase as Viewed 110m above. M7MZ72777 737 71433;? @1177762167/7/177077172277771 JW 7:“ pi}; Cfi 7mm%%ngw7m C: 2:1fi%wa&%d $5agkgwywfimsfi~é&m1msfi;> 1 "1/1 31(1)” j<w51<11241 1wgg~tjé1w+§tbf> fi/H>d71' [7/1 :S(7pi%#i+iwfif+4€gffl:f)fif” 0271’" \“M” ”émw [(729174221‘1—1(7111D2LW 1? 0 171/ “:35 1147711 277' 17W 1{7+£m977+aw21)d1“ 0 _ ...
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