Explanation Given: Equation of density is, ρ = k x y Lamina is bounded by graphs y = 2 x , y = 2 x 3 , x ≥ 0 , y ≥ 0 Formula used: Mass of lamina is given as; m = ∫ 0 1 ∫ 2 x 3 2 x ρ d y d x Mass Moment( x ) is given as; M x = ∫ 0 1 ∫ 2 x 3 2 x ρ y d y d x Mass Moment( y ) is given as; M y = ∫ 0 1 ∫ 2 x 3 2 x ρ x d y d x Centre of Mass = ( x ¯ , y ¯ ) = ( M y m , M x m ) Calculation: Take the lamina bounded by y = 2 x , y = 2 x 3 , x ≥ 0 , y ≥ 0 into consideration. The mass of lamina is calculated as follows: m = ∫ 0 1 ∫ 2 x 3 2 x ρ d y d x = ∫ 0 1 ∫ 2 x 3 2 x k x y d y d x = k ∫ 0 1 x [ y 2 2 ] 2 x 3 2 x d x = k 2 ∫ 0 1 x [ ( 2 x ) 2 − ( 2 x 3 ) 2 ] d x = 2 k ∫ 0 1 x 3 − x 7 d x = 2 k [ x 4 4 − x 8 8 ] 0 1 = 2 k [ 1 4 4 − 1 8 8 − 0 ] = k 4 Therefore, the mass of lamina, m = k 4 Mass Moment( x ), M x is given as; M x = ∫ 0 1 ∫ 2 x 3 2 x ρ y d y d x = ∫ 0 1 ∫ 2 x 3 2 x k x y 2 d y d x = k ∫ 0 1 x [ y 3 3 ]

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ISBN: 9781337275576

Author: Larson, Ron, Edwards, Bruce

Publisher: Cengage Learning,

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Chapter 14, Problem 39RE

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To calculate: The mass and the center of mass of the lamina bounded by the graphs y=2x,y=2x3,x≥0,y≥0 for the density given as ρ=kxy.

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Chapter 14, Problem 38RE

Chapter 14, Problem 40RE

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The mass and the center of mass of the lamina bounded by the graphs y = 2 x , y = 2 x 3 , x ≥ 0 , y ≥ 0 for the density given as ρ = k x y .

Solution Summary: The author calculates the mass and center of mass of the provided lamina bounded by the graphs rho =kxy.

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To calculate: The mass and the center of mass of the lamina bounded by the graphs y=2x,y=2x3,x≥0,y≥0 for the density given as ρ=kxy.

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Chapter 14 Solutions