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
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 Fall '08
 POOLE
 Calculus, Cos, Mathematical Series, dx, dθ

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