16 1 4 π 1 e 16 020100 points by changing to polar

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160=14π(1e16).02010.0 pointsBy changing to polar coordinates evaluatethe integralI=Rex2y2dxdywhenRis the region in thexy-plane boundedby the graph ofx=9y2and they-axis.1.I=14π(1e9)2.I=π(1e3)3.I=14π(1e3)4.I=12π(1e3)5.I=π(1e9)6.I=12π(1e9)correct
macias (ygm97) – Homework 13 – staron – (53940)13Explanation:In polar cooordinates,Ris the set(r,θ) : 0r3,π2θπ2,whileI=Rer2(rdrdθ)=Rrer2drdθ,sincex2+y2=r2. But thenI=30π/2π/2rer2drdθ=π30rer2dr .The presence of the termrnow allows this lastintegral to be evaluated by the subsitutionu=r2. For thenI=12πeu90=12π(1e9).02110.0 pointsThe solid shown inis bounded by the paraboloidz= 212(x2+y2)and thexy-plane.Find the volume of thissolid.1.volume = 4πcorrect2.volume = 2π3.volume = 14.volume = 25.volume =π6.volume = 4Explanation:Theparaboloidintersectsthexy-planewhenz= 0,i.e., whenx2+y24 = 0.Thus the solid lies below the graph ofz= 212(x2+y2)and above the diskD=(x, y) :x2+y24,so its volume is given by the integralV=D212(x2+y2)dxdy .In polar coordinates this becomesV=12202π0r(4r2)dθdr=π20(4rr3)dr=π2r2r4420.Consequently,volume =V= 4π.022
10.0 points
macias (ygm97) – Homework 13 – staron – (53940)14The planez= 3and the paraboloidz= 52x22y2enclose a solid as shown inzyxUse polar coordinates to determine the vol-ume of this solid.1.volume = 2π2.volume =52π3.volume = 3π4.volume =32π5.volume =πcorrectExplanation:From the figure, we see that paraboloidforms the upper boundary of the solid whilethe lower one is formed by the plane.Thusthe volume of the solid is given by the integralI=D((52x22y2)3)dxdywith domain of integration,D, in thexy-plane.To determineDwe look at the where theplanez= 3 intersects the paraboloidz= 52x22y2.This intersection occurs when3 = 52x22y2,i.e., when2 = 2(x2+y2).Thus the surfaces intersect in the circle{(x, y,3) :x2+y2= 1}.SoDis the darker-shaded region in thexy-plane bounded by the unit circlex2+y2= 1shown inzyxandI= 2x2+y21(1x2y2)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= 22π010(1r2)r drdθ= 22π0r22r4410dθ= 22π014dθ= 2·π2.Consequently, the solid hasvolume =π.
macias (ygm97) – Homework 13 – staron – (53940)15keywords:volume of solid, double integral,change coordinates, paraboloid, plane, coor-dinate transformation, curve of intersection,02310.0 pointsA cylindrical drill with radius 2 inches isused to bore a hole through the center of asphere of radius 4 inches leaving the solidshown inFind the volume of this solid.1.volume = 163 cu. ins2.volume = 243πcu. ins3.volume = 323πcu. inscorrect4.volume = 323 cu. ins5.volume = 243 cu. ins6.volume = 163πcu. insExplanation:In Cartesian coordinates this is the annulusR=(x, y) : 4x2+y216.Thus the volume of the solid is given by theintegralV= 2R(16x2y2)1/2dxdy;notice that the factor 2 is needed because theFrom directly overhead the solid looks sim-ilar toxy42θr

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