steele (pss669) – HW 14 (Last Quest) – cepparo – (55660)1Thisprint-outshouldhave23questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

1.Calculate the value of the double integralI= 128 cu. unitscorrect2.I= 126 cu. units3.I= 129 cu. units4.I= 127 cu. units5.I= 130 cu. unitsExplanation:The double integralI==A2xsin(x+y)dxdywhenAis the rectangle(x, y) : 0≤x≤π2,0≤y≤π2.1.I=-2π2.I= 2π3.I=πcorrect

00210.0 pointsI=A2xsin(x+y)dxdywhenAis the rectangle(x, y) : 0≤x≤π2,0≤y≤π2.1.I=-2π2.I= 2π3.I=πcorrect

steele (pss669) – HW 14 (Last Quest) – cepparo – (55660)24.I= (4-π)5.I=-2(4-π)6.I=-(4-π)Explanation:By treatingIas an iterated integral, inte-grating first with respect tox, we see that2xsin(x+y)dx=-2xcos(x+y)+ 2cos(x+y)dx= 2-xcos(x+y) + sin(x+y).Thusπ/202xsin(x+y)dx=-πcosπ2+y+ 2 sinπ2+y-2 sin(y).In this case,I=-πsinπ2+y-2 cosπ2+y+ 2 cos(y)π/20.Consequently,I=π.00310.0 pointsFind the volume of the solid lying under thecircular paraboloidz=x2+y2and above therectangleR= [-1,1]×[-6,6].1.2802.4443.296correct4.2225.420Explanation:The volume is given by the double integralV=6-61-1(x2+y2)dx dyover the rectangular regionR= [-1,1]×[-6,6] ={(x, y) :-1≤x≤1,-6≤y≤Integrating each term separately, we see thatV=6-61-1x2dx dy+6-61-1y2dx dy= 8+288.Consequently,volume = 296.00410.0 pointsEvaluate the double integralI=3220ex-ydxdy .1.I=e-3-e-2-e-1-12.I=e-3-e-2+e-1+ 13.I=e-3+e-2-e-1+ 14.I=e-3-e-2-e-1+ 1correctExplanation:After integration with respect tox,I=32ex-y20dy=32(e2-y-e-y)dy .But then, after integrating next with respecttoywe see thatI=-e2-y+e-y32=-e-1+e-3-(-1 +e-2).

steele (pss669) – HW 14 (Last Quest) – cepparo – (55660)3Consequently,I=e-3-e-2-e-1+ 1.00510.0 pointsEvaluate the iterated integralI=31202(x+y)2dx dy .1.I=12ln342.I= ln343.I= 2 ln344.I= ln955.I=12ln956.I= 2 ln95correctExplanation:Integrating the inner integral with respecttoxkeepingyfixed, we see that202(x+y)2dx=-2x+y20= 21y-12 +y.In this caseI= 2311y-12 +ydy= 2 ln(y)-ln(2 +y)31.Consequently,I= 2 ln(3)(1 + 2)(2 + 3)= 2 ln95.A={(x, y) : 1≤x≤4,0≤y≤4}.1.I= 6 4 ln 4-1342.I= 6163ln 4 +1343.I= 6163ln 4-134correct4.I= 6163ln 4-1325.I= 6316ln 4-134Explanation:Since the area of the rectangleAis 12, theaverage value offoverAis given byI=112A6(x+y) lnx dxdy.The integral with respect toycan be carriedout immediately, whereas the integral with re-spect toxwould require integration by parts,00610.0 pointsFind the average value,I, of the functionf(x, y) = 6(x+y) lnxover the rectangle