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calc hw 6

calc hw 6 - lawrence(cdl678 Homework 6 ODELL(56280 This...

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lawrence (cdl678) – Homework 6 – ODELL – (56280) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The shaded region in lies inside the polar curve r = 3 cos θ and outside the polar curve r = cos θ . Which of the following integrals gives the area of this region? 1. I = 1 2 integraldisplay π/ 2 0 2 cos θ dθ 2. I = 1 2 integraldisplay π/ 2 π/ 2 8 cos 2 θ dθ correct 3. I = 1 2 integraldisplay π/ 2 0 8 cos 2 θ dθ 4. I = integraldisplay π/ 2 π/ 2 2 cos θ dθ 5. I = 1 2 integraldisplay π/ 2 π/ 2 2 cos θ dθ 6. I = integraldisplay π/ 2 π/ 2 8 cos 2 θ dθ Explanation: As the graphs show, the polar curves inter- sect when 3 cos θ = cos θ , i.e. at θ = ± π/ 2. Thus the area of the shaded region is given by I = 1 2 integraldisplay π/ 2 π/ 2 braceleftBig (3 cos θ ) 2 (cos θ ) 2 bracerightBig dθ . Consequently, I = 1 2 integraldisplay π/ 2 π/ 2 8 cos 2 θ dθ . keywords: area, polar cooordinates, definite integral, circle, 002 10.0 points Find the area of the region bounded by the polar curve r = 4 ln θ + 3 as well as the rays θ = 1 and θ = e . 1. area = 1 2 (3 e + 1) correct 2. area = 1 4 (3 e + 2) 3. area = 3 e + 2 4. area = 3 e + 1 5. area = 1 4 (3 e + 1) 6. area = 1 2 (3 e + 2) Explanation: The area of the region bounded by the graph of the polar function r = f ( θ ) as well as the rays θ = θ 0 , θ 1 is given by the integral A = 1 2 integraldisplay θ 1 θ 0 f ( θ ) 2 dθ . When f ( θ ) = 4 ln θ + 3 , θ 0 = 1 , θ 1 = e ,

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lawrence (cdl678) – Homework 6 – ODELL – (56280) 2 therefore, the area of the enclosed region is thus given by the integral A = 1 2 integraldisplay e 1 ( 4 ln θ + 3) 2 = 1 2 integraldisplay e 1 (4 ln θ + 3) dθ . To evaluate this last integral we use Integra- tion by Parts, for then A = 1 2 bracketleftBig 4 θ ln θ + 3 θ bracketrightBig e 1 1 2 integraldisplay e 1 4 = 1 2 bracketleftBig 4 θ ln θ 1 θ bracketrightBig e 1 . Consequently, area = A = 1 2 (3 e + 1) . 003 10.0 points Find the area of the region enclosed by the graph of the polar function r = 2 sin θ . 1. area = 7 2 π 2. area = 17 2 π 3. area = 5 2 π 4. area = 9 2 π correct 5. area = 3 π Explanation: The area of a region bounded by the graph of the polar function r = f ( θ ) and the rays θ = θ 0 , θ 1 is given by the integral A = 1 2 integraldisplay θ 1 θ 0 f ( θ ) 2 dθ . On the other hand, the graph of r = 2 sin θ is the cardioid similar to the one shown in so in this case we can take θ 0 = 0 and θ 1 = 2 π . Thus the area of the region enclosed by the graph is given by the integral A = 1 2 integraldisplay 2 π 0 (2 sin θ ) 2 dθ . Now (2 sin θ ) 2 = 4 2 sin θ + sin 2 θ = 9 2 2 sin θ 1 2 cos 2 θ , since sin 2 θ = 1 2 parenleftBig 1 cos 2 θ parenrightBig . But then, A = 1 2 integraldisplay 2 π 0 parenleftbigg 9 2 2 sin θ 1 2 cos 2 θ parenrightbigg = 1 2 bracketleftBig 9 2 θ + 2 cos θ 1 4 sin 2 θ bracketrightBig 2 π 0 .
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calc hw 6 - lawrence(cdl678 Homework 6 ODELL(56280 This...

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