HW10-solutions - sipes(cms5336 HW10 schultz(56190 This print-out should have 16 questions Multiple-choice questions may continue on the next column or

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.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).
B.For all continuous functionsf,1020f(x, y)dx dy=2010f(x, y)dx dy.
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 surfacef(x, y)dxdy,which in turn can be written as the repeatedintegral309x208xy dydx.
Find the volume of the solid under this graphover the regionA.Af(x, y)dxdy,which in turn can be written as the repeatedintegral309x208xy dydx.

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