Explanation Given: The density is ρ ( x , y , z ) = k y and the region is Q : 3 x + 3 y + 5 z = 15 , x = 0 , y = 0 , z = 0 . Formula used: Total mass of the solid region is given by m = ∭ Q ρ ( x , y , z ) d v and the first moment about the xz -plane is given by M x z = ∭ Q y ρ ( x , y , z ) d v and y ¯ = M x z m ; where, ρ ( x , y , z ) is the density. Calculation: Here, Q : 3 x + 3 y + 5 z = 15 3 x + 3 y + 5 z = 15 5 z = 15 − 3 x − 3 y z = 3 − 3 x 5 − 3 y 5 So, z varies from 0 to 3 − 3 x 5 − 3 y 5 Now on xy -plane z = 0 . So, 3 x + 3 y + 5 ( 0 ) = 15 3 y = 15 − 3 x y = 5 − x So, y varies from 0 to 5 − x . On x -axis, y = 0 , z = 0 . So, 3 x + 3 ( 0 ) + 5 ( 0 ) = 15 3 x = 15 x = 5 So, x varies from 0 to 5 . Total mass of the solid region m = ∭ Q ρ ( x , y , z ) d v . That is, m = ∭ Q ρ ( x , y , z ) d v = ∭ Q k y d v = k ∫ 0 5 y ∫ 0 5 − x ∫ 0 3 − 3 y 5 − 3 x 5 d z d y d x = k ∫ 0 5 y ∫ 0 5 − x [ z ] 0 3 − 3 y 5 − 3 x 5 d y d x Further solve and get, m = k ∫ 0 5 y ∫ 0 5 − x ( 3 − 3 y 5 − 3 x 5 ) d y d x = k ∫ 0 5 ∫ 0 5 − x [ ( 3 − 3 x 5 ) y − 3 y 2 5 ] d y d x = k ∫ 0 5 [ ( 3 − 3 x 5 ) ( y 2 2 ) − y 3 5 ] 0 5 − x d x = k ∫ 0 5 [ ( 3 − 3 x 5 ) ( 5 − x ) 2 2 − ( 5 − x ) 3 5 ] d x Further solve and get, m = k ∫ 0 5 [ 3 5 ( 5 − x ) ( 5 − x ) 2 2 − ( 5 − x ) 3 5 ] d x = k ∫ 0 5 [ ( 5 − x ) 3 5 ( 3 2 − 1 ) ] d x = k 10 ∫ 0 5 ( 5 − x ) 3 d x <

Calculus: Early Transcendental Functions (MindTap Course List)

6th Edition

ISBN: 9781285774770

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 14.6, Problem 38E

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To calculate: The mass and the value y¯ of the center of the mass of the solid region

Q:3x+3y+5z=15 of the density ρ(x,y,z)=ky bounded by the graphs of the equations

x=0,y=0,z=0.

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Chapter 14.6, Problem 37E

Chapter 14.6, Problem 39E

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Calculus: Early Transcendental Functions (MindTap Course List)

Ch. 14.1 - Evaluate the iterated integral: 0433cosrdrd Ch. 14.1 - Prob. 1E Ch. 14.1 - Evaluate the integral: xx2yxdy Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Evaluate the integral: 04x2x2ydy Ch. 14.1 - Evaluate the integral: x3x(x2+3y2)dy Ch. 14.1 - Evaluate the integral: eyyylnxxdx;y0 Ch. 14.1 - Evaluate the integral: 1y21y2(x2+y2)dx Ch. 14.1 - Evaluate the integral: 0x2yeyxdy

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The mass and the value y ¯ of the center of the mass of the solid region Q : 3 x + 3 y + 5 z = 15 of the density ρ ( x , y , z ) = k y bounded by the graphs of the equations x = 0 , y = 0 , z = 0 .

Solution Summary: The author explains the formula used to calculate the mass of the solid region and the moment of inertia.

Videos

To calculate: The mass and the value y¯ of the center of the mass of the solid region

Q:3x+3y+5z=15 of the density ρ(x,y,z)=ky bounded by the graphs of the equations

x=0,y=0,z=0.

Want to see the full answer?

Chapter 14 Solutions

The mass and the value y ¯ of the center of the mass of the solid region Q : 3 x + 3 y + 5 z = 15 of the density ρ ( x , y , z ) = k y bounded by the graphs of the equations x = 0 , y = 0 , z = 0 .

Solution Summary: The author explains the formula used to calculate the mass of the solid region and the moment of inertia.

Videos

To calculate: The mass and the value y¯ of the center of the mass of the solid region Q:3x+3y+5z=15 of the density ρ(x,y,z)=ky bounded by the graphs of the equations x=0,y=0,z=0.

Want to see the full answer?

Chapter 14 Solutions

To calculate: The mass and the value y¯ of the center of the mass of the solid region

Q:3x+3y+5z=15 of the density ρ(x,y,z)=ky bounded by the graphs of the equations

x=0,y=0,z=0.