lewis (scl876) – HW 15.3 – radin – (55315) 1This print-out should have 11 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsEvaluate the double i ntegralI =Z ZD4y(x2+ 1)2dxdywhen D is the regionn(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤√xoin the xy-plane.1. I = 4 ln 22. I = 2 ln 23. I = 14. I = ln 25. I = 26. 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 =12.002 10.0 pointsEvaluate the double i ntegralI =Z ZDx3y − 6xy2dAwhenD =n(x, y) : 0 ≤ x ≤ 1, −x ≤ y ≤ xo.1. I = −352. I = −153. I = −254. I = −45correct5. I = −1Explanation:The double integral can be rewritten as therepeated integralI =Z10Zx−xnx3y − 6xy2odydx ,integrating first with respect to y. NowZx−xnx3y − 6xy2ody=h12x3y2− 2xy3ix−x= −4x4,((−x)2= x2remember!). Consequently,I = −4Z10x4dx = −45.003 10.0 pointslewis (scl876) – HW 15.3 – radin – (55315) 2Evaluate the double i ntegralI =Z ZD2x sin(y) dxdywhen D is the bounded region enclosed by thegraphs ofy = 0 , y = x2, x = 1 .1. I = 2 (sin(1) − 1)2. I = 2 (1 − cos(1))3. I = 1 − cos(1)4. I = sin(1) − 15. I = 1 − sin(1) correct6. I = cos(1) − 17. I = 2 (1 − sin(1))8. I = 2 (cos(1) − 1)Explanation:After integrat ion with respect to y we seethatI = −Z10h2x cos(y)ix20dx= 2Z10x(1 − cos( x2)) dx=hx2− sin(x2)i10,using substitution in t he second integral .Consequentl y,I = 1 − sin(1) .004 10.0 pointsThe graph off(x, y) =1x + y + 4over the triangular region A enclosed by thegraphs ofx = 1, x + y = 2, y + 4 = 0is the surfaceFind the volume V of the solid under thisgraph and over the regio n A.1. V = 6 − ln 62. V = 5 − ln 6 correct3. V = 5 + ln 64. V = 6 + ln 65. V = 7 − ln 6Explanation:As the region o f integration is given by(1, −4)(6, −4)(1, 1)(x, −4)(x, 2 − x)(not drawn to scale) the double integral canbe written as the r epeated integralI =Z61Z2−x−41x + y + 4dydx ,lewis (scl876) – HW 15.3 – radin – (55315) 3integrating first with respect t o y from y = −4to y = 2 − x. Now the inner integral is equaltohln(x + y + 4 )i2−x−4= ln 6 − ln x.ThusI =Z61nln 6 − ln xodx= 5 ln 6 −hx ln x − xi61.Consequentl y,V = 5 − ln 6 .005 10.0 pointsDetermine the volume of the solid lyingunder the graph ofz = 2x + 3yand above the bounded region enclosed by thegraphs of y2= x and y = x2.1. volume =11202. volume =13203. volume =17204. volume =34correct5. volume =920Explanation:The volume of the solid lying under thegraph of z = f (x, y) and above the region Ain the xy-plane is given by the double integralV =Z ZAf(x, y) dxdy .Thus when f(x, y) = 2x + 3y and A is thebounded region enclosed by the graphs of y2=x and y = x2, the volume of the solid is gi venby the double integralV =Z10Z√xx2(2x + 3y) dydx ,since the g r aphs of y2= x and y = x2in-tersect at (0, 0) and ( 1, 1). But then afterintegration with respect to y,V =Z10h2xy +32y2i√xx2dx=Z10n2(x3/2− x3) +32(x − x4)odx=h225x5/2−14x4+3212x2−15x5i10.Consequentl y,I =310+920=34.006 10.0 pointsFind the volume of t he solid in the firstoctant bounded by t he cylindersx2+ y2= 9 , y2+ z2= 9 .Hint: in t he first oct ant the cylinders areshown inxyz33lewis (scl876) – HW 15.3 – radin – (55315) 41. volume = 20 cu. units2. volume = 22 cu. units3. volume = 18 cu. units correct4. volume = 19 cu. units5. volume = 21 cu. unitsExplanation:As the figure shows, the solid in the firstoctant bounded by t he cylindersx2+ y2= 9 , y2+ z2= 9is the solid below the graph ofz =p9 − y2above that part of the ci rclex2+ y2= 9lying in the first quadrant of the xy-plane.Thus the volume of the solid is given by thedouble integralV =Z ZAp9 − y2dx dywhere A is t he region in the first quadrant ofthe x-y pl ane bounded by the quarter-circlen(x, y) : 0 ≤ x ≤p9 − y2, 0 ≤ y ≤ 3o,and so V can be represented a s the it eratedintegralV =Z30Z√9−y20p9 − y2dx dy .In this case,V =Z30hxp9 − y2i√9−y20dy=Z30(9 − y2) dy .Consequentl y,V =9y −13y330= 18 cu. units.007 10.0 pointsEvaluate the double i ntegralI =Z ZD3(4 + x + y)3/2dAwhen D is the regi on in the first quadrantbounded byy = 3x + 12, x = 12as well as the x and y-axes.1. I = 132. I = 143. I = 12 correct4. I = 155. I = 16Explanation:The regio n of integration D is the regionshaded in124y = 3x + 12the vertical line interior to the regi on showingthat we should integrate first with respect toy. Then I becomes the repeated integralI =Z120Z3x+1203(4 + x + y)3/2dydx.lewis (scl876) – HW 15.3 – radin – (55315) 5Now the inner integral ish−6(4 + x + y)1/2i3x+120=3(4 + x)1/2,and soI =Z1203(4 + x)1/2dx =h6(4 + x)1/2i120.Consequentl y,I = 12 .008 10.0 pointsReverse the order of integration in the inte-gralI =Z42 Z1√x/2f(x, y) dy!dx,but make no attempt to evaluate either inte-gral.1. I =Z1√2/2 Z4y22f(x, y) dx!dy cor-rect2. I =Z1√2/2 Z2y22f(x, y) dx!dy3. I =Z√2/20 Z4x22f(x, y) dx!dy4. I =Z√2/20Z24y2f(x, y) dxdy5. I =Z1√2/2Z22y2f(x, y) dxdyExplanation:The region o f integration is similar to theone in the figure2 41√22This shaded region is enclosed by the graphsof 4 y2= x, y = 1 , and x = 2. To changethe order of integration, first fix y. Then,as t he graph shows, x varies from 2 to 4y2.To cover the region of integration, therefore,y must vary from√2/2 to 1. Hence, afterchanging the order of integration,I =Z1√2/2 Z4y22f(x, y) dx!dy.009 10.0 pointsReverse the order of integration in the inte-gralI =Zln 50Zex1f(x, y) dydx ,but make no attempt to evaluate either inte-gral.1. I =Z51Zln y0f(x, y) dxdy2. I =Z51Zln 5ln yf(x, y) dxdy correct3. I =Z50Z5eyf(x, y) dxdy4. I =Zln 50Z5eyf(x, y) dxdy5. I =Z50Zey5f(x, y) dxdylewis (scl876) – HW 15.3 – radin – (55315) 66. I =Zln 50Zey1f(x, y) dxdyExplanation:The regio n of integration is t he set of allpointsn(x, y) : 1 ≤ y ≤ ex, 0 ≤ x ≤ ln 5oin the plane bounded by the graph of y = exand the the linesy = 1 , y = 5 , x = ln 5 .This i s the shaded region inxy15ln 5Integration is taken first wi th respect to y forfixed x along the solid vertical line.To change the order of integration, now fixy and let x var y alo ng the solid horizontal lineinxy15ln 5(not drawn to