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HW5 - Math 191 6.5 6 a dx dy HW 5 Solutions = 1 y-1/2 dx dy...

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Math 191 HW 5 Solutions 6.5: 6 a. dx dy = 1 + y - 1 / 2 dx dy 2 = ( 1 + y - 1 / 2 ) 2 S = 2 π R 2 1 ( y + 2 y ) p 1 + (1 + y - 1 / 2 ) 2 dx . b. 3 3.2 3.4 3.6 3.8 4 4.2 4.4 4.6 4.8 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 x y y+2 sqrt(y)=x c. S 51 . 33. 6.5: 12 y = x 2 + 1 2 x = 2 y - 1 dx dy = 2; S = R d c 2 πx r 1 + dx dy 2 dy = R 2 1 2 π (2 y - 1) 1 + 4 dy = 2 π 5 R 2 1 (2 y - 1) dy = 2 π 5; Geometry formula: r 1 = 1 , r 2 = 3, slant height = p (2 - 1) 2 + (3 - 1) 2 = 5 Frustum surface area = π (1 + 3) 5 = 4 π 5 in agreement with the integral value. 6.5: 14 dy dx = 1 2 x - 1 / 2 dy dx 2 = 1 4 x S = R 15 / 4 3 / 4 2 π x q 1 + 1 4 x dx = 2 π R 15 / 4 3 / 4 q x + 1 4 dx = 2 π h 2 3 ( x + 1 4 ) 3 / 2 i 15 / 4 3 / 4 = 28 π 3 6.5: 36 dx dt = a (1 - cos t ) and dy dt = a sin t r ( dx dt ) 2 + dy dt 2 = p [ a (1 - cos t )] 2 + ( a sin t ) 2 = a 2 1 - cos t S = R 2 πyds = R 2 π 0 2 πa (1 - cos t ) · a 2 1 - cos t = 2 2 πa 2 R 2 π 0 (1 - cos t ) 3 / 2 dt . 6.5: 38 dx dt = h and dy dt = r r ( dx dt ) 2 + dy dt 2 = h 2 + r 2 S = R 2 πds = R 1 0 2 πrt h 2 + r 2 dt = 2 πr h 2 + r 2 R 1 0 tdt = 2 πrt h 2 + r 2 h t 2 2 i 1 0 = πr h 2 + r 2 . Check: slant height is h 2 + r 2 Area is πr h 2 + r 2 . 6.6: 7 The force required to haul up the rope is equal to the rope’s weight, which varies steadily and is

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