Consequently I 3 2 \u03c0 2 011 100points By changing to polar coordinates evaluate

Consequently i 3 2 π 2 011 100points by changing to

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Consequently,I=32π2.01110.0pointsBy changing to polar coordinates evaluatethe integralI=integraldisplay integraldisplayRe-x2-y2dxdywhenRis the region in thexy-plane boundedby the graph ofx=radicalbig9-y2and they-axis.1.I=π(1-e-9)2.I=12π(1-e-3)3.I=π(1-e-3)4.I=12π(1-e-9)correct5.I=14π(1-e-9)6.I=14π(1-e-3)Explanation:In polar cooordinates,Ris the setbraceleftBig(r, θ) : 0r3,-π2θπ2bracerightBig,
byrne (hcb539) – Homework 11 – spice – (54070)7whileI=integraldisplay integraldisplayRe-r2(rdrdθ)=integraldisplay integraldisplayRre-r2drdθ ,sincex2+y2=r2. But thenI=integraldisplay30integraldisplayπ/2-π/2re-r2drdθ=πintegraldisplay30re-r2dr .The presence of the termrnow allows this lastintegral to be evaluated by the subsitutionu=r2. For thenI=12πbracketleftBig-e-ubracketrightBig90=12π(1-e-9).01210.0pointsThe solid shown inis bounded by the paraboloidz= 1-x2-y2and thexy-plane.Find the volume of thissolid.1.volume =14π2.volume =12πcorrect3.volume =184.volume =145.volume =126.volume =18πExplanation:Theparaboloidintersectsthexy-planewhenz= 0,i.e., whenx2+y2-1 = 0.Thus the solid lies below the graph ofz= 1-x2-y2and above the diskD=braceleftBig(x, y) :x2+y21bracerightBig,so its volume is given by the integralV=integraldisplay integraldisplayD(1-x2-y2)dxdy .In polar coordinates this becomesV=integraldisplay10integraldisplay2π0r(1-r2)dθdr= 2πintegraldisplay10(r-r3)dr= 2πbracketleftBig12r2-r44bracketrightBig10.Consequently,volume =V=12π.01310.0pointsThe solid shown in
byrne (hcb539) – Homework 11 – spice – (54070)8lies inside the spherex2+y2+z2= 25and outside the cylinderx2+y2= 16.Find the volume of the part of this solid lyingabove thexy-plane.1.volume = 18πcorrect2.volume = 27π3.volume = 94.volume = 185.volume = 276.volume = 9πExplanation:From directly overhead the solid is similartoxy54θrIn Cartesian coordinates this is the annulusR=braceleftBig(x, y) : 16x2+y225bracerightBig.Thus the volume of the solid above thexy-plane is given by the integralV=integraldisplay integraldisplayR(25-x2-y2)1/2dxdy .To evaluateVwe change to polar coordi-nates. NowR=braceleftBig(r, θ) : 4r5,0θ2πbracerightBig,so that after changing coordinates the integralbecomesV=integraldisplay54integraldisplay2π0radicalbig25-r2rdrdθ= 2πintegraldisplay54rradicalbig25-r2dr=πbracketleftBig-23(25-u)3/2bracketrightBig2516,using the substitutionu=r2. Consequently,volume =V= 18π.01410.0pointsA swimming pool is circular with a 36 ft diam-eter.The depth is constant along east-westlines and increases linearly from 1 ft at thesouth end to 7 ft at the north end.

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