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Unformatted text preview: MATH 234 (A. Assadi) - Spring 2008 - FINAL Tuesday, May 13, 2008 NAME (print): Instructions: 1. 2. Write your name on each page. Circle the name of your TA: WU ' TONEJ C . Closed book. Closed notes. Calculators allowed. No laptops. 2 index cards of size 3” x 5” with formulas allowed. . Answer your questions on the exam paper. There are some extra pages in the back if you need them. . Show all your work. Partial credit is given only if your work is clear. Enclose your ﬁnal answers clearly in a rectangle. . Time allowed: 120 minutes. 1 cos29 , 1—cos20 ‘ "“1 _ c0s29 = ~+——- sm26 = “—— /sin“amda: = —W + n 1/sin”_2amcdav 2 1 2 na n sec2 0 = 1 + tanz 6’ C802 0 = 1 + COtZ 6 n cos”—1 ax sin an: n — 1 n—2 sin(Aﬂ:B) = sinAcosB icosAsinB /cos (”“12 = T + n /COS (1536111: cos(A :l: B) = cosAcosB q: sinAsinB tanA :l: tanB cos(a + b)a: cos(a'—— b)a: /sma:ccosbxd:c=— 2(a+b) — 201-1)) /e‘” cos b3: d2: = a2 + b2 6113 a2+b2 tan(AiB) = mg _ . sin(a—-b)x sin(a+b)a: smaazsmbxdx: ~——~———~— , _ 2(a—b) 2(a+b) / da: =1tan_1 ax—b fwsmosbxdx=w+w mums-b :5 , b 2(a—b) 2(a—I—b) / dx = ax+b / sin ax cos axdx = _co:2ax —-————-——— -—— a xx/aa:+b x/Eln x/ax+b b+\/5 1 da: 1 a: /cot‘a:1:da: = ~—1n|sinax| / = —tan_1— a a2+x2 a a 1 d3: 1 x a /tanaxdx = Elnlsecaxl : ——1 /a2—a:2 2ancc—a / dx ~1t ax / dx :lln a: 1+cosax—a an 2 ' x(ax+b) ‘b_ ax—I—b / dx -——lcota\$ / dx __ so + tan" 1a: l—cosax _ a 2 2 22 " 2 2 2 __’3' _ 1 (a 3:3) 2a ((1 +23) 2a a /tan2axdcc= Etanaw—x /—__2 2 :1n(x‘+ W) 1 ”a +3: /cot2amdx=———cotax—az «/ 2 2 ,/ 2 2 a /—a +2; dsz—aln —-————a+ (1 +x 1 :c a: /secaa:d_x= Elnlsecam+tanaacl /_.._d_m___.— lIn a+ a2+x2 /cscaa:dm— 11n|cs¢acv+cotax} Clix/a2+\$2 _ a IE 2 f a 1 / d2: ___\/a2+:c2 /sec2axdx: ~tanax m2x/a2 + x2 (123: a 1 / da: —sin 1 E /csc2 axda: = ~Ecotam ‘/a2_\$2 — a 2 2 lnazcdzc=xlnax~x /\/a2—~m2da:—gx/a2—m24—9—2—sm’ 3:— / a eazdm= l6“ / d3: ——1~1n a+\/a2—x2 a 1/ 2_ 2 i an: as a a: a an e (acosbx+bsinb:z:) dx' ——-—-——~—~— = sin‘1 (a: _ a) / van—932 a / e” sinbxdm': (a sin bx — b cos bx) /\/a2x+\$2dw=§a2 (n+2—I—— 2 21n(:+\/a2+x2) 2 fmdm:gm_glnlx+mu NAME (print): Problem 1. The plane ﬁns—F ﬁy+3z = 11 cuts the cone Z2 = 1:2 +3;2 in an ellipse. Using the method of Lagrange multipliers, ﬁnd the greatest and the smallest values that the function f (as, y, z) = my + 22 takes on the ellipse. Write the equations that you need to solve in the box below. Put your ﬁnal answer in the two boxes at the bottom. Greatest value: [: Smallest value: S NAME (print): Problem 2. (a) Compute the gradient of the function f (cc, 3/, z) = yzemy + cos z — 2x2. (b) Let F = 3562 i + sinyj + (43:2: ~ g) k. Find cur1(Vf + F). (c) Now let G = exyzi + y cos zj ~ 33,22 k. Compute div (G + curl(Vf + F)). NAME (print): Problem 3. Consider the annular ring A = {1 g \$2 + y2 S 4} and let M = y(cos(a:y) ~— 1), and N = 03(1 + cos(a:y)). (a) Sketch the region A and the vector ﬁeld F = M i + N j at the points Where the boundary of A intersects the coordinate axes. Make sure that your vectors have proper length! (b) Compute the circulation MM+N® 6A> of F around the boundary of A. HINT: Green’s theorem might be useful. , NAME (print): Problem 4. Consider the sphere x2 +y2 +22 = a2 and the cylinder \$2 +3;2 = 0?. Let —a g hl < hg S a.‘ Show that the surface area of the part of the sphere between the planes z = hl and z = hg equals the surface area of the part of the cylinder between the same twoplanes. In other words, the surface area of the sphere over any vertical span is equal to the surface area of the circumscribing cylinder over that same span. (This was already known to Archimedes!) NAME (print): Problem 5. Consider the surface S , which is the upper half of the torus (see ﬁgure) and Whose parameterization is - ‘ r(u,v) z ((a+bcosu)coSv, (a+bcosu) sinv,bsinu), 0 g u g 7r; 0 < v < 27r ) Z 0 < b < a. Let F : <y, —a:,ea‘2 cos(y3)>. Compute y f/curlF - nda, . S Where n is the upward normal. ac HINT: What is the boundary of the surface? NAME (print): Problem 6. Compute the outward ﬂux of F = 4xyi — 2y2j +zk across the boundary of the region between the plane 2 = O and the paraboloid z = 4 — m2 — y2, that is, across the boundary of the set D:{(x,y,z): O<z<4—\$2~y2}. ...
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