tapia (jat4858) – HW07 – clark – (52990) 1This print-out should have 24 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsThe shaded region inlies inside the polar curve r = 2 cos θ andoutside the polar curve r = cos θ. Determinethe area of this region?1. area = 3π2+ 12. area = 3π3. area =34π2+ 14. area =32π2+ 15. area =34π correct6. area =32πExplanation:As the graphs show, the polar curves inter-sect when2 cos θ = cos θ ,i.e. at θ = ±π/2. Thus the area of the shadedregion is given byI =12Zπ/2−π/2n(2 cos θ)2−(cos θ)2odθ=32Zπ/2−π/2cos2θ dθ .To evaluate this last integral we use the doubleangle formulacos2θ =12(1 + cos 2θ) .For thenZπ/2−π/2cos2θ dθ =12Zπ/2−π/2(1 + cos 2θ) dθ=12hθ +12sin 2θiπ/2−π/2=12π .Consequently, the shaded region hasarea =34π.Are there any other ways of calculating thi sarea without using integration?keywords: a rea, polar cooordinat es, definiteintegral, circle,002 10.0 pointsFind the area of the region bo unded by thepolar curver =√2 ln θ + 3as well as the rays θ = 1 and θ = e.1. a rea =12(3e − 1) correct2. a rea =12(3e + 2)3. a rea = 3e − 14. a rea =14(3e + 2)5. a rea = 3e + 26. a rea =14(3e − 1)tapia (jat4858) – HW07 – clark – (52990) 2Explanation:The area of the region bounded by thegraph of the polar function r = f(θ) as wellas the rays θ = θ0, θ1is given by the integralA =12Zθ1θ0f(θ)2dθ .Whenf(θ) =√2 ln θ + 3 , θ0= 1 , θ1= e ,therefore, the area of the enclosed region isthus given by the integralA =12Ze1(√2 ln θ + 3)2dθ=12Ze1(2 ln θ + 3) dθ .To evaluate this last integral we use Integra-tion by Parts, for thenA =12h2θ ln θ + 3θie1−12Ze12 dθ=12h2θ ln θ + 1θie1.Consequently,area = A =12(3e −1).003 10.0 pointsFind the area of the regi on enclosed by thegraph of t he pola r functionr = 2 − cos θ .1. area =72π2. area =192π3. area = 3π4. area =92π correct5. a rea = 4πExplanation:The area of the region bounded by thegraph of the polar function r = f(θ) andthe rays θ = θ0, θ1is given by the integralA =12Zθ1θ0f(θ)2dθ .On the other hand, the graph ofr = 2 −cos θis the cardioid similar to the one shown inso in this case we can take θ0= 0 and θ1= 2π.Thus the area of the region enclosed by thegraph is given by the integralA =12Z2π0(2 − cos θ)2dθ .Now(2 − cos θ)2= 4 − 4 cos θ + cos2θ=92− 4 cos θ +12cos 2θ ,sincecos2θ =121 + cos 2θ.But then,A =12Z2π092− 4 cos θ +12cos 2θdθ=12h92θ − 4 sin θ +14sin 2θi2π0.tapia (jat4858) – HW07 – clark – (52990) 3Consequently,area = A =92π.keywords: polar graph, area, cardioid004 10.0 pointsFind the area of the shaded-region insidethe polar curver = cos θshown inφ2φ1xywhen φ1= π/4 and φ2= π/3.1. area =148π +18√32. area =748π +116(2 −√3)3. area =748π +116(2 +√3) correct4. area =148π +116(2 −√3)5. area =748π +18√36. area =148π +116(2 +√3)Explanation:The area of the region bounded by thegraph of the polar function r = f(θ) andthe rays θ1and θ2is given by the integralA =12Zθ2θ1f(θ)2dθ.Whenf(θ) = cos θ , θ1= −φ1, θ2= φ2,therefore,A =12Zπ/3−π/4cos2θ dθ .On the other hand,cos2θ =12(1 + cos 2θ) .ThusA =14Zπ/3−π/4(1 + cos 2θ) dθ=14hθ +12sin 2θiπ/3−π/4.Consequently,area =748π +116(2 +√3) .keywords: polar graph, polar integral, doubleangle,005 10.0 pointsFind the area of the shaded regiontapia (jat4858) – HW07 – clark – (52990) 4inside the graph ofr = 1 + 2 cos θ .1. area = π +3√34correct2. area =12π +3√343. area = π −3√344. area =π25. area =32π6. area = π7. area =12π −3√34Explanation:The area of a region bounded by the graphof the polar function r = f(θ) and the raysθ = θ0, θ1is given by the integralA =12Zθ1θ0f(θ)2dθ .In this question the function isr = 1 + 2 cos θ ;on the other hand, to determine the rays θ =θ0and θ = θ1bounding the shaded region,note that as θ ranges from 0 to 2π/3 thegraph(i) begins on the x-axis when θ = 0 at r = 3,(ii) crosses the y-axis w hen θ =π2at r = 1,(iii) first passes though the origin whenr = 0, i.e., when θ = 2π/ 3.Thus the area of the shaded region is given bythe integralA =12Z2π/30(1 + 2 cos θ)2dθ .But(1 + 2 cos θ)2= 1 + 4 cos θ + 4 cos2θ= 3 + 4 cos θ + 2 cos 2θ ,sincecos2θ =121 + cos 2θ.HenceA =12Z2π/303 + 4 cos θ + 2 cos 2θdθ=12h3θ + 4 sin θ + sin 2θi2π/30.Consequently,area = A = π +3√34.keywords: polar graph, area, cardioid, polarintegral006 10.0 pointsFind the area of the shaded region inspecified by the graphs of the circlesr = 2 cos θ , r = 2 sin θ .1. a rea = 2π4+ 12. a rea = 13. a rea = 2π4. a rea =π2− 1tapia (jat4858) – HW07 – clark – (52990) 55. area =π2+ 1 correct6. area = πExplanation:The area of a region bounded by the graphof pol a r function r = f(θ) between the raysθ = θ0, θ1is given by the integralA =12Zθ1θ0f(θ)2dθ .Now the graph of r = 2 cos θ is a circlecentered on the x-axis, while that of r =2 sin θ is a circle centered on the y-axi s; inaddition, both pass through the origin andhave the same radius. So the circles intersectalso at θ = π/4, as the figure shows.The area of the shaded region can thus becomputed as the differenceI1− I2= 2Zππ/4sin2θ d θ− 2Zπ/2π/4cos2θ dθof the area of the resp ective shaded regions inNowI1=Zππ/4(1 − cos 2θ) dθ=hθ −12sin 2θiππ/4=3π4+12,whileI2=Zπ/2π/4(1 + cos 2θ) dθ=hθ +12sin 2θiπ/2π/4=π4−12.Consequently, the shaded region hasarea =π2+ 1.keywords: definite integral, area betweencurves, polar area, circle,007 10.0 pointsWhich one of the following integrals givesthe arc length of the portion shown as solidblue in the graphof the polar curver = 1 − 2 cos θ .1.12Zπ/20(1 − 2 cos θ)2dθ2.Zπ0√5 − 4 cos θ dθ3.12Zπ/3−π/3(1 − 2 cos θ)2dθ4.Z2π/30√5 − 4 cos θ dθ5.12Z2π/30(1 − 2 cos θ)2dθ6.12Zπ0(1 − 2 cos θ)2dθtapia (jat4858) – HW07 – clark – (52990) 67.Zπ/20√5 −4 cos θ dθ8.Zπ/3−π/3√5 − 4 cos θ dθ correctExplanation:The arc length of a pol ar curve r = f (θ)between θ = θ0and θ = θ1is g iven by theintegralL =Zθ1θ0qr2+ f′(θ)2dθ .Now whenr = 1 − 2 cos θ .its graph will(i) pass through the origin when r = 0, i.e.,at θ = −π/3, π/3,(ii) cross the x-axis also at θ = 0, π,(iii) and cross the y-axis also at θ =π/ 2,