Calc III - enclosed is then A = R 2 π 1 2 r 2 dθ = R 2 π 1 2(1-sin θ 2 dθ = R 2 π 1 2(1-2sin θ sin 2 θ dθ = R 2 π 1 2(1-2sin θ 1-cos2 θ

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Quiz 2 (Solution) 1. Sketch the region in the plane consisting of points whose polar coordinates satisfy the give conditions: (1): 1 < r 3, - π/ 2 θ < π/ 6 (2): r = θ + π/ 2, θ 0 Figure 1: Graph of 1 < r 3, - π/ 2 θ < π/ 6
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5 10 15 30 210 60 240 90 270 120 300 150 330 180 0 Figure 2: Graph of r = θ + π/ 2, θ 0. 2. Sketch the curve and find the area it encloses: r = 1 - sin θ Sol: We know that θ (0 , 2 π ) ( Note : Actually, we may use any interval ( a,a + 2 π )) , which leads to the same result. The area
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Unformatted text preview: enclosed is then: A = R 2 π 1 2 r 2 dθ = R 2 π 1 2 (1-sin θ ) 2 dθ = R 2 π 1 2 (1-2sin θ + sin 2 θ ) dθ = R 2 π 1 2 (1-2sin θ + 1-cos2 θ 2 ) dθ = 1 2 ( θ + 2cos θ + θ 2-sin2 θ 4 ) | 2 π = 3 2 π 0.5 1 1.5 2 30 210 60 240 90 270 120 300 150 330 180 Figure 3: Graph of r = 1-sin θ ....
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This note was uploaded on 02/23/2010 for the course MATH 221 taught by Professor Staff during the Spring '08 term at Tulane.

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Calc III - enclosed is then A = R 2 π 1 2 r 2 dθ = R 2 π 1 2(1-sin θ 2 dθ = R 2 π 1 2(1-2sin θ sin 2 θ dθ = R 2 π 1 2(1-2sin θ 1-cos2 θ

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