   Chapter 14.6, Problem 38E

Chapter
Section
Textbook Problem

Center of Mass In Exercises 37-40, find the mass and the indicated coordinate of the center of mass of the solid region Q of density ρ bounded by the graphs of the equations.Find y ¯ using ρ ( x , y , z ) = k y Q : 3 x + 3 y + 5 z = 15 ,     x = 0 ,     y = 0 ,     z = 0

To determine

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.

Explanation

Given:

The provided density is ρ(x,y,z)=ky and the provided region is Q:3x+3y+5z=15,x=0,y=0,z=0.

Formula used:

Total mass of the solid region is represented by m=Qρ(x,y,z)dv and the first moment about the xz-plane is represented by Mxz=Qyρ(x,y,z) dv and y¯=Mxzm;

where, ρ(x,y,z) is the density.

Calculation:

Consider the provided solid region, Q:3x+3y+5z=15

Solve for z,

3x+3y+5z=155z=153x3yz=33x53y5

So that, z varies from 0 to 33x53y5

Now on xy-plane z=0.

Now solve for y,

3x+3y+5(0)=153y=153xy=5x

So, y varies from 0 to 5x.

On x-axis, y=0,z=0. So,

3x+3(0)+5(0)=153x=15x=5

So, x varies from 0 to 5.

Total mass of the solid region m=Qρ(x,y,z)dv.

Therefore,

m=Qρ(x,y,z)dv=Qkydv=k05y05x033y53x5dzdydx=k05y05x[z]033y53x5dydx

Further solve and get,

m=k05y05x(33y53x5)dydx=k0505x[(33x5)y3y25]dydx=k05[(33x5)(y22)y35]05xdx=k05[(33x5)(5x)22(5x)35]dx

That is,

m=k05[35(5x)(5x)22(5x)35]dx=k05[(5x)35(321)]dx=k1005(5

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