Explanation Given information: Mass density is δ ( x , y ) = x 2 and region is bounded by y = 6 − x , x = 0 , y = 0 Formula used: The center of mass is the point P C M = ( X C M , Y C M ) , where X C M = M y M , Y C M = M x M Where, M is mass of the solid, M ∬ D δ ( x , y ) d A M x = ∬ D y δ ( x , y ) d A M y = ∬ D x δ ( x , y ) d A Calculation: First find M x , M y M x = ∬ D y δ ( x , y ) d A = ∫ 0 6 ∫ 0 6 − x y x 2 d y d x As δ ( x , y ) = x 2 and region is bounded by y = 6 − x , x = 0 , y = 0 Thus, 0 ≤ x ≤ 6 , 0 ≤ y ≤ 6 − x Now to integrate first evaluate inner integral treating x as constant M x = ∫ 0 6 x 2 ( ∫ 0 6 − x y d y ) d x = ∫ 0 6 x 2 [ y 2 2 ] 0 6 − x d x = ∫ 0 6 x 2 [ ( 6 − x ) 2 2 ] d x = ∫ 0 6 x 2 2 ( 36 + x 2 − 2 x ) d x = ∫ 0 6 ( 18 x 2 + x 4 2 − 6 x 3 ) d x = ( 18 x 3 3 + x 5 10 − 6 x 4 4 ) 0 6 = ( 6 x 3 + x 5 10 − 3 x 4 2 ) 0 6 = ( 6 ( 6 ) 3 + ( 6 ) 5 10 − 3 ( 6 ) 4 2 ) = ( 1296 + 3888 5 − 1944 ) ⇒ M

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

4th Edition

ISBN: 9781319312886

Author: Rogawski, Jon, Adams, Colin, FRANZOSA, Robert

Publisher: W. H. Freeman

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Chapter 15.5, Problem 25E

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Find center of mass where density is δ(x,y)=x2 and region is bounded by y=6−x,x=0,y=0.

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Chapter 15.5, Problem 24E

Chapter 15.5, Problem 26E

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Chapter 15 Solutions

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

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Find center of mass where density is δ ( x , y ) = x 2 and region is bounded by y = 6 − x , x = 0 , y = 0 .

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Find center of mass where density is δ(x,y)=x2 and region is bounded by y=6−x,x=0,y=0.

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Chapter 15 Solutions