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Homework 12

# Homework 12 - Vincent(jmv777 m HW12 w Gilbert —(57495 1...

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Unformatted text preview: Vincent (jmv777) m HW12 w Gilbert — (57495) 1 This print—out should have 20 questions. is the volume of the solid below the graph of Multiple—choice questions may continue on f having the rectangle the next column or page m find all choices before answering. A = {my} ; 0 g 1; g 4, 0 g y g 4} 001 10‘0 pOiﬂtS for its base. Thus the solid is the wedge The graph of the function 2 : ftyay) m 7_\$ is the plane shown in and so its voiuine is the area of trapezoidal face multiplied by the thickness of the wedge. Consequently, 'I = 80 cu. units , Determine the value of the double integral ' I m f / any date! A H ) y keywords: over the region _ 002 10f) points A m {(m,y):ogazgzi, egg/54} Find the value of the integral in the say—plane by first identifying it as'the 2 volume of a solid below the graph of f. I 2 f f (a: y) day 0 1. I = 82611. units when 2. I m 80 cu. units correct ﬂax, y) m 455: ~ 53323;. 3. I = 81 cu. units 40 _ 1. I = 8+ 3:!) 4. I z: 79 cu. units 2.1 e 4. m5 2 5. I m 830:1. units 9 y Explanation: 3 I __ 8 40 The double integral - w "H "gay correct I m j [A meme 4. — 8+£§y2 L...‘ l Vincent (jaw???) — HWIZ —— Gilbert w (57495) 3 Consequently, ‘ III!!!- 005 10.0 points Evaluate the double integral 1+3:2 In dd [[41+3JZ 33y when m {(ay) 0<w31, ogygl} 1 1. I 2 "2‘71" 1 2. I m 67f E" 3. I 2 gr 4. I = ":7? 1 5. I n gt” correct Explanation: Since A = {(3:,y):0§23\$1, Ogygl} is a rectangle with sides parallel to the coor- dinate axes, the double integral can be repre» sented as the iterated integral 111+\$2 Im- dd. fe<fe Maggy Now 1 2 1+3: 1 1 3 3- d a [ m . . [0 1+3}? 3: 1 y? t+3m10 Thus Consequently, 1 Izm— 7T . 806 10.0 points Calculate the value of the double integral I = // 2msin(cc+y)dmdy A . when A is the rectangle {(m,y):05mg%, Ogygg}. 1. I = iﬁﬁ+2(ﬂ“l) 2.1 a iaﬂ~2(\/§mi) 3. I = 2(\/§m1)+—:w(2—V§) 4.1 = Ell—ﬂx/é—Zb/E—l-l) 5.1 x 2(\/§+1)m»:«1r(2+\/§) 1 205 — 1) ~ _7r(2 ~ ﬂ) correct . I 6 4 H Explanation: By treating I as an iterated integrai, inte~ grating first with respect to as, we see that / 2m sin(x + y) 013: m —22: cosh: + y) + 2/ cos(:c +y)dac = 2( may eosm + y) + sin(.’1: + 11)) after integration by parts. Thus WM . 1 7r /0 2m51n(:c+y) dzt = w~~2~7rc30e(~&w +31) . 75" . +2s1n(1 +y) — Zsmy. vincent (jaw???) w HWIB m Gilbert w (57495) since cos2u 2 Ecoszu.m1. Consequentiy, 1 1 77/6 I = 2/ y[u+wsin2u] dy 0 2 0 1 i z 2 hf: [W2] , 6 4 2 e 009 10.0 points Evaluate the iterated integral I m f:{fyﬂ 9ydm}dy. . 6 1. I m — 5 2. I r: 3 9 3. I = — 5 12 4. I 2 m» 5 3 5. I m ~5- correct Explanation: AS an interated integral, integrating ﬁrst with respect to x, we see that .I m £1[9xy]:§dy 3 m 9/ (213/2~y2)dy- 0 Consequently, [l CHICO 010 10.0 points Find the value of the integral I m f/A{2(cc—2)-—3y}dmdy when A is the region {(w.y):osys:cw2. 29:4}. 7 1. I = w 3 2. I = 2 8 3. I = _. 3 5 4. I m — 3 5. I m georrect Expianation: The integral can be written as the repeated integrai I m fmmmaamamagywy} day. Now . x—Z [0 {2&3 -~ 2) m em 3 2 = Plaid)?” “2’9 if? Consequently I = [24-;(x-2)2dm = [éedﬁ]. and so 4 I m — 3 011 10.0 points Evaluate the double integral 89 I: f/D m2+1dmdy Vincent {jmv777) — HVV 12 —— Gilbert * (57495) . 7 Thus ' Explanation: 7 The volume of the solid under the graph of I = f {in 7 m In 3;} d3; f is given by the double integral . i ' 7 = :ﬁln’?——[azln:cwm]l. V f/Af(az,y)dmdz, Consequently, which in turn can be written as the repeated integral V 2 6—1137. 013 10.0 points f: (fay/m 82129013,!) disc. N OW the inner integral is equal to m ﬂing) x 893g [4ey2]0 z: 4m(4 w 3:9). rI‘he graph of over the bounded region A in the ﬁrst qued— Thus rant enclosed by 2 W Vm4/02m(4—x2)am=[menaﬁﬂd and the m, y~axes is the surface y Consequently, Volume = 16 cu. units. 014 10.0 points Evaluate the double integral Im /L(3m+2y)dxdy when A is the region enclosed by the graphs Find the volume of the solid under this graph Of :2: = 1, m m y m 1’ y m 1' over the region A. 1. Volume 2 32 cu. units , 1. I = Scarred; 2. Volume m80‘u. units 2. I = g, 3. Volume 221604. units correct 3. I m 3 4. Volume 3 4cu. units 4_ I = 2 .- {xv-Volume m g on units 5. I = Z 3 3 vincent (jinx/777) w HW12 w Gilbert M (57495) 9 _ Thus 7? [2/ (27rm233msinsc)dm U 2 71' = [21m —— a: + cosmL). ' Consequently, I I=7r2w . 016 10.0 noints Find the volume of the solid in the ﬁrst octant bounded by the cylinders \$2+y2=16, y2+z2 16. ll Hint: in the first octant the cylinders are shown in 128 1. volume = Mg— cu. units correct 2. volume at 44 cu. units 124 3. volume m —-3— cu. units 116 4. volume : —— cu. units 3 5. volume = 40 cu. units Explanation: As the ﬁgure shows, the solid in the first octant bounded by the cylinders x2+y2 m :6, y2+32 16 is the soiid below the graph of z = 16 —— yg above that part of the circte 2:2 + 3,12 = 16 lying in the first quadrant of the xy-{plane Thus the volume of the solid is given by the double integra} V m // x/16my2dzcdy A Where A is the region in the ﬁrst quadrant of the w—y plane bounded by the quarter—circle {(x,y):0_<_'xg x/l6—y2,{)§yg4}, and so V can be represented as the iterated integral V w [)4{/0W Wdac}dy. In this case, 4 16—- 2 V2/[93x/16my3L) ydy 0 4 =/ (16~92)dy- G Consequently, "42 — cu. units . 0 3 1 3]4 128 [mi/‘39 017 10.0 points Vincent (jmv???) w HVV12 - Gilbert — (57495) 11 4. I m /:( 25 f(.’13, y)dm)dy ‘x/g 5.1 = /.25(/Gﬂ f(.r,y)d:c)dy o n w e. I z] (/ f{m,y)d\$)dy 0 5 Explanation: The region of integration is the set of all points hayhﬁﬁySmKOSmg5} in the plane bounded by the ar—axis and the graphs of This is the shaded region in (not drawn to scale). Integration is taken ﬁrst with respect to y for ﬁxed :3 along the dashed vertical line. To change the order of integration, first ﬁx y in the interval [0, 25] and let CC vary along the solid horizontal line. Then m varies from V5 to 5. Hence, after changing the order of integration, keywords: double integral, reverse order inte— gration, parabola, 019 10.0 points Evaluate the double integral I = ff 43/8932 dmdy A when A is the region in the ﬁrst quadrant bounded by the graphs of N. 5...! H CD 9" >-...( ll (e m» 1) correct 4. I = 2(em l) 5.1 m 4(e+r) Explanation: 2 Since the function ﬁx) = em cannot be integrated directly, we have to represent I as a repeated integral, integrating ﬁrst with respect to y, where the region of integration is similar to the shaded region in Thus as a repeated integral fol/f 1/2 4318552 rig) d1: and so 1 Im fog+ Consequently, 2y y2 dy Vincent ﬁrm/T777) — HVV 12 w Gilbért ~ (M495) 1 = [1n(9+y2)]0. l3 ...
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Homework 12 - Vincent(jmv777 m HW12 w Gilbert —(57495 1...

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