Q. 63

Question

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+b22 and z¯=c1+c22, respectively.

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

Step-by-Step Solution

Verified
Answer

Part (a) The y- and z-coordinates of the center of mass are y¯=b1+b22 and z¯=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 is x¯=a1+a22.

1Part (a) Step 1. Given Information.

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

2Part (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 are b1+b22,c1+c22.

3Part (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 is 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 is x¯=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 is x=a1+a22.