Explanation Given: The density is ρ ( x , y , z ) = k and the region is Q : x a + y b + z c = 1 , ( a , b , c > 0 ) , 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 : x a + y b + z c = 1 x a + y b + z c = 1 z c = 1 − x a − y b z = c ( 1 − x a − y b ) So, z varies from 0 to c ( 1 − x a − y b ) Now on xy -plane z = 0 . So, x a + y b + ( 0 ) c = 1 x a = 1 − y b x = a ( 1 − y b ) So, x varies from 0 to a ( 1 − y b ) . Now, put x = 0 , z = 0 . So, ( 0 ) a + y b + ( 0 ) c = 1 y b = 1 y = b So, y varies from 0 to b . 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 d v = k ∫ 0 b ∫ 0 a ( 1 − y b ) ∫ 0 c ( 1 − x a − y b ) d z d x d y = k ∫ 0 b ∫ 0 a ( 1 − y b ) [ z ] 0 c ( 1 − x a − y b ) d y d x Further solve and get, m = k ∫ 0 b ∫ 0 a ( 1 − y b ) [ c ( 1 − x a − y b ) ] d x d y = k c ∫ 0 b ∫ 0 a ( 1 − y b ) [ ( 1 − y b ) − ( x a ) ] d x d y = k c ∫ 0 b [ ( 1 − y b ) x − x 2 2 a ] 0 a ( 1 − y b ) d y = k c ∫ 0 b [ { ( 1 − y b ) a ( 1 − y b ) } − a 2 ( 1 − y b ) 2 2 a ] d y Further solve and get, m = k a c ∫ 0 b [ ( 1 − y b ) 2 − ( 1 − y b ) 2 2 ] d y = k a c 2 ∫ 0 b [ ( 1 − y b ) 2 ] d y = k a c 2 [ ( − b ) ( 1 − y b ) 2 3 ] 0 b

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Chapter 14.6, Problem 40E

To determine

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

Q:xa+yb+zc=1 ,(a,b,c>0) of the density ρ(x,y,z)=k bounded by the graphs of the

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

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Chapter 14.6, Problem 39E

Chapter 14.6, Problem 41E

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Chapter 14 Solutions

EBK CALCULUS: EARLY TRANSCENDENTAL FUNC

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 : x a + y b + z c = 1 , ( a , b , c > 0 ) of the density ρ ( x , y , z ) = k 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:xa+yb+zc=1 ,(a,b,c>0) of the density ρ(x,y,z)=k 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 : x a + y b + z c = 1 , ( a , b , c &gt; 0 ) of the density ρ ( x , y , z ) = k 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:xa+yb+zc=1 ,(a,b,c>0) of the density ρ(x,y,z)=k 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 : x a + y b + z c = 1 , ( a , b , c > 0 ) of the density ρ ( x , y , z ) = k bounded by the graphs of the equations x = 0 , y = 0 , z = 0 .

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

Q:xa+yb+zc=1 ,(a,b,c>0) of the density ρ(x,y,z)=k bounded by the graphs of the

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