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In Exercises 61–64, let R be the rectangular solid defined by

R=(x,y,z)|0a1xa2,0b1yb2,0c2zc2.

Assume that the density at each point in R is proportional to the distance of the point from the yz-plane.

(a) Without using calculus, explain why the y- and z-coordinates of the center of mass are y¯=b1+b22andz¯=c1+c22,respectively.

(b) Use an appropriate integral expression to find the x-coordinate of the center of mass.

Short Answer

Expert verified

Part (a) The y- and z-coordinates of the center of mass are y¯=b1+b22andz¯=c1+c22,respectively because it is given that density at each point in R is proportional to the distance of the point from the yz-plane, so y-and z-coordinates will be the same as the center of mass coordinates and the x-coordinate of the center of the mass should be zero.

Part (b) The x-coordinate of the center of mass isx¯=a1+a22.

Step by step solution

01

Part (a) Step 1. Given Information. 

The given rectangular solid is defined byR=(x,y,z)|0a1xa2,0b1yb2,0c2zc2.

02

Part (a) Step 2. Explanation. 

As we know the center of mass is the midpoint of the coordinates, so the center of mass coordinates are:

(x¯,y¯,z¯)=a1+a22,b1+b22,c1+c22

It is given that density at each point in R is proportional to the distance of the point from the yz-plane, so y-and z-coordinates will be the same as the center of mass coordinates and the x-coordinate of the center of the mass should be zero.

Thus, the y- and z-coordinates of the center of mass areb1+b22,c1+c22.

03

Part (b) Step 1. Find the x-coordinate of the center of mass.  

We have to use an appropriate integral expression to find the x-coordinate of the center of mass.

Now, the density at each point in Ris proportional to the distance of the point from the yz-plane, so ρ(x,y,z)=kx.

The x-coordinate of the center of mass isx¯=MyzM.

Let's find the mass of the solid:

M=Rρ(x,y,z)dxdydzM=x=a1a2y=b1b2z=c1c2kdxdydzM=k(a2a1)(b2b1)(c2c1)

Now, the center of mass is,

x¯=MyzM=Txρ(x,y,z)dxdydzρ(x,y,z)dxdydzx¯=1Mx=a1a2y=b1b2z=c1c2kxdxdydzx¯=k2M(a22a12)(b2b1)(c2c1)x¯=k2k(a2a1)(b2b1)(c2c1)(a22a12)(b2b1)(c2c1)x¯=a1+a22

Thus, the x-coordinate of the center of mass isx=a1+a22.

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