   Chapter 14, Problem 59RE

Chapter
Section
Textbook Problem

Center of Mass In Exercises 59 and 60, 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 x ¯ using ρ ( x , y , z ) = k . Q : x + y + z = 10 ,   x = 0 ,     y = 0 ,     z = 0

To determine

To calculate: The mass and the indicated coordinate x¯ of center of mass of solid region Q with density given as (ρ=k). The solid region Q is bounded by the graph of the equation Q:x+y+z=10,x=0,y=0,z=0.

Explanation

Given:

The density of solid region Q is ρ(x,y,z)=k and the solid is bounded by the graph of equation,

Q:x+y+z=10,x=0,y=0,z=0.

Formula used:

Mass (m) of a solid region Q having density ρ is,

m=Qρ(x,y,z)dV

First moment about yz plane is,

Myz=Qxρ(x,y,z)dV

x Coordinate (x¯) of center of mass of a solid region Q is,

x¯=Myzm

Calculation:

A 3-D plot is made with the help of maple software,

The limits of x,y and z can be calculated with the help of this 3-D plot.

Take the provided equation of the solid Q:x+y+z=10,x=0,y=0,z=0 into consideration.

The upper limit of x can be calculated from the above equation as follows:

x+y+z=10

If value of x and y are zero, then,

x+0+0=10x=10

Thus, the upper limit of x is 10 and bounds of x is 0x10.

Take the xy plane into consideration.

In this case, since the value of z=0, the equation of the line will be given as,

y=10x

Also, as z depends on both x and y, the equation of z contain both x and y

x+y+z=10

On subtracting x+y from both sides, we get,

z=10xy

The lower limit of x,y,z is provided in the question, i.e.,

x=0,y=0,z=0

On applying the formula for the mass (m) of solid body Q, we get,

m=Qρ(x,y,z)dV=010010x010xykdzdydx=k010010x[z]010xydydx=k010010x[(10xy)0]dydx

Next, on integrating with respect to y, we get,

k010010x[(10xy)0]dydx=k010[10yxyy22]010xdx=ky010[10xy2]010xdx=k(10x)010[10x(10x)20]dx=k(10x)010[(10x)(10x)20]dx

Simplifying further,

k010010x[(10xy)0]dydx=k(10x)010[(10x)2]dx=k2010(10x)2dx=k2010(100+x220x)dx

Further, on integrating with respect to x, we get,

k2010(100+x220x)dx=k2[100x+x3320x22]010=k2[100(10)+103310(10)20]=k2[1000+100031000]=5003k

Therefore, mass (m) of the solid body Q is calculated to be 5003k

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