142p23s11solution[1]

B 3 0 0 3 3 kg cm kg cm 0 27 y 1 x2 1 1 1

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Unformatted text preview: π (1 − 3 x2 9 ) ∆x (1 − ) dx = 2π ⋅ ∫ (1 − ) dx = 4π x2 9 x2 9 0 kg mass ≈ ∑ δ∆x = ∑ ( 3 + 2 x 2 ) ∆x ⋅ cm = kg cm (a) 3 ( 3 + 2 x ) dx = 27 ∫ xδ dx = ∫ ( 3x + 2 x ) dx = 2 cm since x= ( 27 δ dx ∫ mass = ∫ 2 0 3 y = 0. (b) 3 0 0 3 3 kg ⋅cm kg ) = cm . 0 2.7 y = 1 − x2 1 1 1 ∫ xδ A dx = ∫ x ⋅ 2 1 − x dx = ∫ x ⋅ 1 − x dx x= ∫ δ A dx ∫ 2 1 − x dx ∫ 1 − x dx Ax x 0 2 0 1 0 1 x 2 0 1 2 0 2 0 ∆x For the integral in the numerator, let u = 1 − x 2 , then − 1 du = x dx . 2 When x = 0, u = 1. And when x = 1, u = 0. Then, ∫ 1 0 0 x 1 − x 2 dx = ∫ − 1 u du = 2 1 1 2 ∫ 1 u du = 1 . 3 0 For the integral in the denominator, let x = sin θ with − π ≤ θ ≤ π . 2 2 Then dx = cos θ dθ . When x = 0, θ = sin −1 ( 0 ) = 0 . And when x = 1, θ = sin −1 (1) = π . 2 Then, ∫ 1 0 π π 2 2 1 − x 2 dx = ∫ cos θ 1 − sin 2 θ dθ = ∫ cos 2 θ dθ = So, x = 0 1 3 π 4 = 4 . 3π 0 π 1 2 ( sin θ cos θ + θ ) 0 2 =π 4 2.8 2 n work ≈ ∑ work on slice at...
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This note was uploaded on 02/04/2014 for the course MAT 132 taught by Professor Poole during the Fall '08 term at SUNY Stony Brook.

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