# To determine d we look at the where the plane z 6

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To determineDwe look at the where theplanez= 6 intersects the paraboloidz= 8-2x2-2y2.This intersection occurs when6 = 8-2x2-2y2,i.e., when2 = 2(x2+y2).Thus the surfaces intersect in the circle{(x, y,6) :x2+y2= 1}.SoDis the darker-shaded region in thexy-plane bounded by the unit circlex2+y2= 1shown in
byrne (hcb539) – Homework 11 – spice – (54070)11zyxandI= 2integraldisplay integraldisplayx2+y21(1-x2-y2)dxdy .BecauseDis circular and the integranddepends only on the distance of (x, y) fromthe origin, we change from Cartesian to polarcoordinates using the transformationsx=rcosθandy=rsinθ. ThenI= 2integraldisplay2π0integraldisplay10(1-r2)r drdθ= 2integraldisplay2π0bracketleftBigr22-r44bracketrightBig10= 2integraldisplay2π014= 2·π2.Consequently, the solid hasvolume =π.keywords:volume of solid, double integral,change coordinates, paraboloid, plane, coor-dinate transformation, curve of intersection,01810.0pointsEvaluate the iterated integralI=integraldisplay10integraldisplay1-y20cos(πx2+πy2)dxdyby converting to polar coordinates.1.I=142.I=14π3.I=12π4.I=125.I= 0correctExplanation:In Cartesian coordinates the region of inte-gration isbraceleftBig(x, y) : 0xradicalbig1-x2,0y1bracerightBig,which is the shaded region inxy1θrThe graphic shows that in polar coordinatesthe region of integration isbraceleftBig(r, θ) : 0r1,0θπ/2bracerightBig.Thus in polar coordinatesI=integraldisplay10integraldisplayπ/20cos(πr2) (rdθdr)=12πintegraldisplay10rcos(πr2)dr .Consequently,I=14bracketleftBigsin(πr2)bracketrightBig10= 0.
byrne (hcb539) – Homework 11 – spice – (54070)1201910.0pointsBy changing to polar coordinates evaluatethe integralI=integraldisplay integraldisplayRradicalbigx2+y2dxdywhenRis the regionbraceleftBig(x, y) : 16x2+y225,y0bracerightBigin thexy-plane.1.I=583π2.I=523π3.I=553π4.I=613πcorrect5.I=643πExplanation:In polar cooordinates,R=braceleftBig(r, θ) : 4r5,0θπbracerightBig,whileI=integraldisplay integraldisplayRr(rdrdθ) =integraldisplay integraldisplayRr2drdθ ,sinceradicalbigx2+y2=r. But thenI=integraldisplay54parenleftbiggintegraldisplayπ0r2parenrightbiggdr=πintegraldisplay54r2dr .Consequently,I=13bracketleftBigr3bracketrightBig54=613π.