# HW10-solutions - sipes cms5336 HW10 schultz 56190 This...

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sipes (cms5336) – HW10 – schultz – (56190)1Thisprint-outshouldhave16questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0 pointsEvaluate the iterated integralI=10yy6y dx dy .1.I= 22.I=453.I=654.I=855.I=25correctExplanation:As an interated integral, integrating firstwith respect tox, we see thatI=10[ 6xy]yydy= 610y3/2y2dy .Consequently,I= 625y5/213y310=25.00210.0 pointsWhich, if any, of the following are correct?A.For all continuous functionsg,10y0g(x, y)dx dy=101xg(x, y)dy dx.B.For all continuous functionsf,1020f(x, y)dx dy=2010f(x, y)dx dy.1.both of them2.neither of them3.A onlycorrect4.B onlyExplanation:A.True:correct reversal of the order ofintegration when integrating over the uppertriangle in the square [0,1]×[0,1].B. False: incorrect reversal of the order ofintegration when integrating over a rectangle.00310.0 pointsEvaluate the iterated integralI=3π/20cos(θ)04esin(θ)dr dθ.1.I= 02.I= 4e3.I= 4(e1)4.I=e45.I=1e46.I= 41e1correctExplanation:After simple integrationcos(θ)04esin(θ)dr=4r esin(θ)cos(θ)0= 4 cos(θ)esin(θ).
sipes (cms5336) – HW10 – schultz – (56190)2In this case,I=3π/204 cos(θ)esin(θ)dθ=4esin(θ)3π/20.Consequently,I= 41e1.00410.0 pointsEvaluate the double integralI=D4y(x2+ 1)2dxdywhenDis the region(x, y) : 0x1,0yxin thexy-plane.1.I= 12.I= 23.I= ln 24.I= 4 ln 25.I= 2 ln 26.I=12correctExplanation:As an iterated integral, integrating firstwith respect toy, we see thatI=10x04y(x2+ 1)2dydx .Nowx04y(x2+ 1)2dy=2y2(x2+ 1)2x0= 2x(x2+ 1)2.In this case,I= 210x(x2+ 1)2dx=1x2+ 110.Consequently,I=12.00510.0 pointsEvaluate the double integralI=D2xcos(y)dxdywhenDis the bounded region enclosed by thegraphs ofy= 0,y=x2,x= 3.1.I= 2 (cos(9)1)2.I= 9sin(9)3.I= sin(9)14.I= cos(9)95.I= 2 (9cos(9))6.I= 2 (sin(9)1)7.I= 2 (9sin(9))8.I= 1cos(9)correctExplanation:After integration with respect toywe seethatI=302xsin(y)x20dx=302xsin(x2)dx=cos(x2)30,
sipes (cms5336) – HW10 – schultz – (56190)3usingsubstitutioninthesecondintegral.Consequently,I= 1cos(9).00610.0 pointsThe graph off(x, y) = 8xyover the bounded regionAin the first quad-rant enclosed byy=9x2and thex, y-axes is the surfaceFind the volume of the solid under this graphover the regionA.1.Volume = 162 cu. units2.Volume = 54 cu. units3.Volume =814cu. units4.Volume =812cu. units5.Volume = 81 cu. unitscorrectExplanation:The volume of the solid under the graph offis given by the double integralV=Af(x, y)dxdy,which in turn can be written as the repeatedintegral309x208xy dydx.

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