ha (lvh262) – HomeWork 15.3 – karakurt – (56295) 1This print-out should have 25 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsEvaluate the integralI =Z10Zx20(x + 5y) dy dx .1. I =11202. I =34correct3. I =13204. I =7205. I =920Explanation:After integration with respect to y we seethatI =Z10xy +52y2x20dx=Z10x3+52x4dx=14x4+12x510.Consequentl y,I =14+12=34.002 10.0 pointsWhich, if any, of the foll owing are correct?A. For all continuous functions f,Z10Z20f(x, y) dx dy =Z20Z10f(x, y) dx dy.B. For all continuous functions g,Z10Zy0g(x, y) dx dy =Z10Zx0g(x, y) dy dx.1. both of them2. A only3. B only4. neither of them correctExplanation:A. False: incorrect reversal of the order ofintegration when integrating over a rectangle.B. False: incorrect reversal of the order ofintegration when integrating over the uppertriangle in the square [0, 1] × [ 0, 1].003 10.0 pointsEvaluate the double integralI =Z1−1Zyy2(6x − y) dx dy .1. I = −4152. I =215correct3. I = −7154. I =135. I = −115Explanation:Treating I as an iterated integral, integrat-ing first with respect to x with y fixed, we seethatI =Z1−1Zyy2(6x − y) dx dy=Z1−13x2− xy yy2dy .ha (lvh262) – HomeWork 15.3 – karakurt – (56295) 2ThusI =Z1−13(y2− y4) dy −Z1−1(y2− y3) dy=3y33−y55−y33−y441−1.Consequentl y,I =215.004 10.0 pointsFind the value of the double integralI =Z ZA(6x − y) dxdywhen A is the regionn(x, y) : y ≤ x ≤√y, 0 ≤ y ≤ 1o.1. I =11152. I =8153. I =19304. I =565. I =1330correctExplanation:The integral can be writt en as the repeatedintegralI =Z10"Z√yy(6x − y) dx#dy.NowZ√yy(6x − y) dx =h3x2− xyi√yy= 3y − y3/2− 2y2.But thenI =Z10(3y − y3/2− 2y2) dx=h32y2−25y5/2−23y3i10.Consequentl y,I =1330.005 10.0 pointsEvaluate the double integralI =Z ZD8x3y2dxdywhenD =n(x, y) : 0 ≤ x ≤ 1, −x ≤ y ≤ xo.1. I =232. I =473. I = 04. I =1621correct5. I =89Explanation:The double integral can be rewritten as therepeated integralI =Z10Zx−x8x3y2dydx ,integrating first with respect to y. NowZx−x8x3y2dy =h83x3y3ix−x=163x6.ha (lvh262) – HomeWork 15.3 – karakurt – (56295) 3Consequentl y,I =163Z10x6dx =1621.006 10.0 pointsEvaluate the double integralI =Z ZD4y(x2+ 1)2dxdywhen D is the regionn(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤√xoin t he xy-plane.1. I = ln 22. I = 4 ln 23. I = 24. I = 2 ln 25. I = 16. I =12correctExplanation:As an iterated integral, integrating firstwith respect to y, we see thatI =Z10nZ√x04y(x2+ 1)2dyodx .NowZ√x04y(x2+ 1)2dy=h2y2(x2+ 1)2i√x0= 2x(x2+ 1)2.In this case,I = 2Z10x(x2+ 1)2dx =h−1x2+ 1i10.Consequentl y,I =12.007 10.0 pointsEvaluate the double integralI =Z ZDx4y − 3xy2dAwhenD =n(x, y) : 0 ≤ x ≤ 1, −x ≤ y ≤ xo.1. I = −152. I = 03. I = −25correct4. I = −455. I = −35Explanation:The double integral can be rewritten as therepeated integralI =Z10Zx−xnx4y − 3xy2odydx ,integrating first with respect to y. NowZx−xnx4y − 3xy2ody=h12x4y2− xy3ix−x= −2x4,((−x)2= x2remember!). Consequently,I = −2Z10x4dx = −25.ha (lvh262) – HomeWork 15.3 – karakurt – (56295) 4008 10.0 pointsEvaluate the double integralI =Z ZD2x cos(y) dxdywhen D is the bounded region enclosed by thegraphs ofy = 0 , y = x2, x = 1 .1. I = sin(1) − 12. I = 1 − cos(1) correct3. I = 2 (1 − sin(1))4. I = 2 (1 − cos(1))5. I = 2 (sin(1) − 1)6. I = 2 (cos(1) − 1)7. I = cos(1) − 18. I = 1 − sin(1)Explanation:After integration with respect to y we seethatI =Z10h2x sin(y)ix20dx=Z102x sin(x2) dx =h−cos(x2)i10,using substitution in t he second integral .Consequentl y,I = 1 − cos(1) .009 10.0 pointsEvaluate the double integralI =Z ZD(3x + 4) dAwhen D is the bo unded region enclosed byy = x and y = x2.1. I =1112correct2. I =343. I =7124. I =5125. I =14Explanation:The area of integration D is the shadedregion in the figureTo determine the limits of integration, there-fore, we have first to find the points of int er-section of the line y = x and the parabolay = x2. These occur when x2= x, i.e., whenx = 0 and x = 1. T hus the double integralcan be written as a repeated integralI =Z10Zxx2(3x + 4)dydx,integrating first with respect to y. After i nte-gration this inner integral becomesh(3x + 4)yixx2= (3x + 4)(x − x2)= 4x − x2− 3x3.ThusI =Z10(4x − x2− 3x3) dx=2x2−13x3−34x410.ha (lvh262) – HomeWork 15.3 – karakurt – (56295) 5Consequentl y,I =1112.010 10.0 pointsFind the volume, V , of the solid under thegraph of the functionf(x, y) =1x + y + 1and over the triangular region A enclosed bythe graphs ofx = 1, x = 5, x + y = 4, y + 1 = 0 .1. V = 5 + ln(5)2. V = 5 − ln(5)3. V = 6 − ln(5)4. V = 4 + ln(5)5. V = 4 − ln(5) correctExplanation:The volume of the solid is the value of thedouble integralI =Z ZAf(x, y) dydx .As the region of integration, A, is given by(1, −1)(5, −1)(1, 3)(x, −1)(x, 4 − x)(not drawn to scale) the double integral canbe written as the repeated integralI =Z51Z4−x−11x + y + 1dydx ,integrating first with respect to y from y = −1to y = 4 − x. Now the inner integral is equaltohln(x + y + 1)i4−x−1= ln(5) − ln(x).ThusI =Z51(ln(5) − ln(x)) dx= 4 ln(5) −hx ln(x) − xi51.Consequentl y,V = 4 − ln(5) .011 10.0 pointsEvaluate the double integralI =Z ZD(2x − y) dAwhen D is the region enclosed by the graphsofx = 1, x + y = 1, y + 1 = 0.1. I =732. I = 23. I = 34. I =53correct5. I =83Explanation:ha (lvh262) – HomeWork 15.3 – karakurt – (56295) 6The graphs ofx = 1, x + y = 1, y + 1 = 0are straight lines intersecting at the points(1, 0), (1, −1), (2, −1) .Thus the region of i ntegration is the shadedtriangular region(1, −1)(2, −1)(1, 0)(x, −1)(x, 1 − x)so I can be written as the repeated integralI =Z21Z1−x−1(2x − y) dydx ,integrating first with respect to y from y = −1to y = 1 − x. Now the inner integral is equaltoh2xy −12y2i1−x−1= 5x −52x2.ThusI =Z21n5x −52x2odx =h52x2−56x3i21.Consequentl y,I =53.012 10.0 pointsThe solid shown inzyxis bounded by the paraboloidz = 4 − x2− y2,and the coordinates planes as well as theplanesx = 1 , y = 1 .Find the volume of this solid.1. volume =732. volume =103correct3. volume =834. volume = 35. volume =113Explanation:The volume of the sol id is given by thedouble integralV =Z ZDz dxdy=Z ZD(4 − x2− y2) dxdywhere D is its base as a region in
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