9322 The lemniscate with polar equation r 2 4 cos2 \u03b8 is shown in Fig 9315 We

9322 the lemniscate with polar equation r 2 4 cos2 θ

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9.3.22: The lemniscate with polar equation r 2 = 4 cos2 θ is shown in Fig. 9.3.15. We will find the area of the right-hand loop. First, r = 0 when cos 2 θ = 0, so that θ is an odd integral multiple of π/ 4. Hence the area of the right-hand loop is A = 1 2 π/ 4 π/ 4 4 cos2 θ dθ = π/ 4 0 4 cos2 θ dθ = 2 sin 2 θ π/ 4 0 = 2 . 9.3.23: Given: The polar equation r 2 = 4 sin θ . One way to construct its graph is first to construct the Cartesian graph y = 4 sin x , which is shown next for 0 x 2 π . 3 6 x -4 -2 2 4 y Then construct the Cartesian graph of y = ± 4 sin x , shown next, also for 0 x 2 π . Note that there is no graph for π < x < 2 π but two graphs for 0 x π . 966
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3 6 x -4 -2 2 4 y Finally, use the last graph to construct the polar graph r 2 = 4 sin θ by sketching both r = 4 sin θ and r = 4 sin θ . As θ varies from 0 to π , r begins at 0, increases to a maximum r = 2 when θ = π/ 2, then decreases to 0 as θ runs through the values from π/ 2 to π . Meanwhile, r sweeps out the mirror image of the previous curve, and thus we obtain the “double oval” shown in the next figure. -1 1 x -2 -1 1 2 y The area of the upper loop is therefore A = 1 2 π 0 4 sin θ dθ = 2 cos θ π 0 = 2 ( 2) = 4 . 9.3.24: The area of one loop of this rose is A = 1 2 π/ 12 π/ 12 36 cos 2 6 θ dθ = 36 π/ 12 0 1 2 (1 + cos 12 θ ) = 18 θ + 1 12 sin 12 θ π/ 12 0 = 3 π 2 . 9.3.25: See the graph below. 967
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-1 1 x -1 1 2 y Find the area A of the region both inside the circle r = 2 sin θ and outside the circle r = 1. The circles cross where 2 sin θ = 1, thus where θ = π/ 6 and where θ = 5 π/ 6. So A = 1 2 5 π/ 6 π/ 6 (4 sin 2 θ 1) = 1 2 5 π/ 6 π/ 6 2(1 cos 2 θ ) 1 = 1 2 θ sin 2 θ 5 π/ 6 π/ 6 = 1 2 5 π 6 π 6 + 3 2 + 3 2 = 2 π + 3 3 6 1 . 913222954981 . 9.3.26: See the graph below. -2 -1 1 2 3 4 x -2 -1 1 2 y The circles r = 4 cos θ and r = 2 cross where 4 cos θ = 2, thus where θ = π/ 3 and where θ = π/ 3. To find the area A of their intersection, we double the area of its top half; thus A = π/ 3 0 4 + π/ 2 π/ 3 16 cos 2 θ dθ = 4 π 3 + 4 π/ 2 π/ 3 2(1 + cos 2 θ ) = 4 π 3 + 4 2 θ + sin 2 θ π/ 2 π/ 3 = 4 π 3 + 4 π 2 π 3 3 2 = 4 π 3 + 4 π 8 π 3 2 3 = 8 π 6 3 3 4 . 91347879 . 9.3.27: See the figure below. 968
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-0.5 0.5 x -0.5 1 y The two circles r = cos θ and r = 3 sin θ cross where cos θ = 3 sin θ ; tan θ = 3 3 ; θ = π 6 ; they also cross at the pole. The region inside both is divided by the ray θ = π/ 6 into a lower region of area B and an upper region of area C . Thus we find the area of the region inside both circles to be A = B + C . Now B = 1 2 π/ 6 0 3 sin 2 θ dθ = 3 4 π/ 6 0 (1 cos 2 θ ) = 3 4 θ sin θ cos θ π/ 6 0 = 3 4 π 6 3 4 and C = 1 2 π/ 2 π/ 6 cos 2 θ dθ = 1 4 π/ 2 π/ 6 (1 + cos 2 θ ) = 1 4 θ + sin θ cos θ π/ 2 π/ 6 = 1 4 π 3 3 4 . Therefore A = B + C = 5 π 6 3 24 0 . 221485767606.
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