Solutions_Test3_B

# Solutions_Test3_B - MATH 251 sectiﬁ Test#3 Variant B Fall...

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Unformatted text preview: MATH 251 sectiﬁ Test #3, Variant B, Fall 2010 . Name ................ ............ ......... 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 % Ill). 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: mzasinqﬁcosd, yzasingbsind, z=acos¢, 0§¢§7n 036\$27r or m=ucosv, y=usinv, z=\/a2—u2, Oguga, Ogvg27r. 2. For sphere with the given above ﬁrst parametric presentation we have the following expression for the normal vector and its magnitude r¢ X r9 = (12(sin2 gbcosd, sin2 ¢sin 0, sinqﬁcos qﬁ), |r¢> X [‘9' = a2 sin d). 3. Some formulas for trigonometric functions: 1— _‘ 2.. 1 r‘ 2 2sinacosa = sin(2a), sin2 oz = w, 00820 = Your score: 1. Calculate the line integral / F - dr7 Where F = (312312 + 21:2)i + (2:53:11 + 22)j + (:L'2 + 2yz)k C and the curve C consists of the line segments connecting the points P(1, 0, O) with Q(1, 1, 3) and then Q(1,1,3) with R(1,2,1). Hint: show that the vector ﬁeld F is conservative and use its properties. i J ‘4 q ‘ . (Uri); = 3x 97 2*} :<2% ’23) XX,” 2 “46) 573% Zxﬁ-HZ X2t2’ﬂ‘ : 4 (7717/ 0> 773M szl: as X32+2X2 ﬁ-ﬁ + l) 0;?) =X§+Xé+giz+glié) ¥j=2x33+et Mazzgéy/Hb lj= 2 M,z>:g/¢[email protected] 31% ~‘— Xzf-Qr-E .2/(3) Q 976:0 350 lid?” = £(R>ufﬁP>;£0’2iq>_,&4'0}v) >— 22; + z +c v(0+o+o+c):@ \ ,5? fF" “:3 (31341319) Zﬂﬁti’ 1+6?) «9/2) 390% x \ = O J (2Mt1+3+igﬁ)cﬂ=k3 +2+ +Z¢ezwss+ [email protected] 0 0 C1 Came (L {/3 fmmmeiLn‘w 06 Z: R5572=<0ﬂfzy C K X;4 L” 3347* oéféy 3. it for the vector-ﬁeld F(:L’, = (my + y)i + y3j and the curve C’ that is the circle :52 + y2 = 4 oriented in a counterclockwise direction. (This means that you have to compute both sides in the Green’s theorem equality and SllOW that they are equal.) .grecws C 8-; Pag~m§g W! (m1 ‘StvﬂN/‘cm’wl Cam in xy-ﬂwu, Mia?) a mar/2w y 7621 WWW/{m Hm mm“ MHz/1mg D. 7730‘ . . CleWj =ﬂ%~g§)d4 In 1% cm géywlwi’ﬁ "' a? \$22231: 3096;21-e’éﬂb’ﬁﬂiﬁ’ﬂfmv t~ .\ , x” _ +l2t‘x+‘+8§4»3%2aa+dk Sllﬁﬁwwj‘ We (12?“ ‘9 “25"” (i Q if)st ° : dgeggmfwttmwé WWHS BM “6‘3 5‘” 45;, t .' “fsz ,4. -- ,l__. 47") Van); [#7r9))d%:-—og((x + )) 0M 0 01 g* 2 AV \ Jaw) 0M : “- (((y‘unﬂ—l Owl WW = '5 Mrﬁjmwd r119 1) I) g; 4 é -— a £5— : '* 4T Mm - no Dav i 2 Wow Bi A. Use Green’s Theorem to evaluate the integral + e’”)dm + (.732 — ln(1 + y))dy where C is positively oriented curve that consists of the line segments from (0,0) to (7r, 0) and the curve y=siii\$,0§x§7r. WEE.) 5 (Problem 9 pp. 835) Find the surface area of the part of the sphere 1:2 + 3/2 + Z2 = a2 that lies inside the cylinder 3:2 + y2 : am and above the guy—plane. ’ i? 55: ﬁz—XZ-jl +3“? 55 Sf/iﬂa [1260/2 Wefw Var I) 6. (Problem 10, pp. 925) Evaluate the surface integral // 33" zdS, Where S is part of the S sphere \$2 + 3/2 + 2:2 = 1 that lies above the cone z : V3032 + 212). Whimedmﬁd' sﬂwfe W W X2+'\{2‘+ 3A%+\9?):/y Kg”: F-ui‘: 4‘” ,2 3: (problem 16, Section 14.6, pp. 913, modiﬁed) Find an equation of the tangent plane at the point (1, —2, l) to the parametric surface ac 2 U2, y = v - (42, z : v2. /)=vL m “J7 42 2 O> "a: -— 0 ’7 z “ 2 VI} (2%) 2"?) > wt. {lifoa it: <02 A '1) ﬁr<09 4) iv) v ) ) ijﬁz<4>4al7 “7+ 77"” p’I‘Z/Q W :3 avm‘w WW kw mﬂﬁﬁﬁ 753W ['4 ~22 0 “ﬁn/(47W? WﬁWﬁ‘fﬁPpﬁaM 9? 46H) +4/ +L)+Z{%—I>"/O K ix—Zw‘b/Ii’4 +%-—I>O [EB O 8. Calculate the work done by the force ﬁeld F = (33“ + y2)i + (yy + Z2)j + (22 + 1:2)k when a particle moves under its inﬂuence around the edge of the part of the plane 22: +3; + z = 1 that lies in the ﬁrst octant, in the counterclockwise direction as viewed from above. Hint: Use the Stokes Theorem. SllWl 4: stﬁerv—p 7 X, 9444/), Dal ‘ a?) . r.— (—L) £62 ~4va H4 +9) 0%: FOOD/{(1 ~3x ﬁﬁM 77’ 44X 96’ '2. re ' -x, ale Axi— 2’5Xll-2X>”L[l”27<> pix w w l a x 2 6. "‘ “H2 Q’ZxXHX)’ 04074“? ll‘l’l‘” i “1’4 W “We” 0 922% +6)? 0 . V; 5/ a'JCF—WW~~<%'%i+%“>:v(§wi+é) gm — .2 units ﬂows with a 9. (Problem 35, Section 14.7, p. 926) A ﬂuid of density p(9;,y,z) — velocity v = (y,1,z). Find the rate of ﬂow upward through the surface of the paraboloid z = 9 — (11:2 + 312) / 4 which is within the cylinder 1'2 + y2 = 36. Hint: Use the fact that the ﬂow rate is given by the surface integral // F ' (is, where F 2 [JV and S is the surface. S E=12<ﬂ)4)%> 21: x72 )9 1;’7 ﬁg dS= <~2m~25n>0ik 53. 4. i J),§x2+11£363 6‘ g» L7: 9. L =<1-:3—L- DEV) of 5w) :4 7 z (magmas V+ 18w +13 ’95) “(for @ 10. (Problem 11, 14.10, p.937) Use the divergence Theorem to calculate the surface integral // F - (is, Where F : yezi + 3/2 j + ewyk and S is the surface of the solid bounded by the 5' cylinder 302 +112 : 9 and the planes 2 = 0 and z 2 y M 3. 11 11. (BONUS PROBLEM, no partial credit) (problem 12, pp. 931) Calculate the line integral / F - dr, Where F : (12y, %:1:3,:13y) and the curve C is the intersection of the hyperbolic C paraboloid z : y2 ~ 2:2 with the cylinder :52 +1112 : 1 oriented counterclockwise Viewed from above. 3», {L .2 .11. X; w *2“ :7: [PM ...
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Solutions_Test3_B - MATH 251 sectiﬁ Test#3 Variant B Fall...

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