Question

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

$\mathfrak{R}=\left\{(x,y,z)|0\le a_1\le x\le a_2,0\le b_1\le y\le b_2,0\le c_2\le z\le c_2\right\}.$

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 $\overline{y}=\frac{b_1+b_2}{2}\text{and}\overline{z}=\frac{c_1+c_2}{2},$respectively.

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

Short answer

Part (a) The y- and z-coordinates of the center of mass are $\overline{y}=\frac{b_1+b_2}{2}\text{and}\overline{z}=\frac{c_1+c_2}{2},$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$\overline{x}=\frac{a_1+a_2}{2}.$

Step-by-step solution

Part (a) Step 1. Given Information. 

The given rectangular solid is defined by$\mathfrak{R}=\left\{(x,y,z)|0\le a_1\le x\le a_2,0\le b_1\le y\le b_2,0\le c_2\le z\le c_2\right\}.$

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:

$(\overline{x},\overline{y},\overline{z})=\left(\frac{a_1+a_2}{2},\frac{b_1+b_2}{2},\frac{c_1+c_2}{2}\right)$

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$\left(\frac{b_1+b_2}{2},\frac{c_1+c_2}{2}\right).$

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 $\rho (x,y,z)=kx.$

The x-coordinate of the center of mass is$\overline{x}=\frac{M_{yz}}{M}.$

Let's find the mass of the solid:

$$\begin{gathered} M={\iiint }_R\rho (x,y,z)dxdydz \\ M={\int }_{x=a_1}^{a_2}{\int }_{y=b_1}^{b_2}{\int }_{z=c_1}^{c_2}kdxdydz \\ M=k(a_2-a_1)(b_2-b_1)(c_2-c_1) \end{gathered}$$

Now, the center of mass is,

$$\begin{gathered} \overline{x}=\frac{M_{yz}}{M}=\frac{{\iiint }_{T}x\rho (x,y,z)dxdydz}{\iiint \rho (x,y,z)dxdydz} \\ \overline{x}=\frac{1}{M}{\int }_{x=a_1}^{a_2}{\int }_{y=b_1}^{b_2}{\int }_{z=c_1}^{c_2}\left(kx\right)dxdydz \\ \overline{x}=\left(\frac{k}{2M}\right)(a_2^2-a_1^2)(b_2-b_1)(c_2-c_1) \\ \overline{x}=\left(\frac{{\displaystyle k}}{{\displaystyle 2\left(k(a_2-a_1)(b_2-b_1)(c_2-c_1)\right)}}\right)(a_2^2-a_1^2)(b_2-b_1)(c_2-c_1) \\ \overline{x}=\frac{a_1+a_2}{2} \end{gathered}$$

Thus, the x-coordinate of the center of mass is$\overline{x}=\frac{a_1+a_2}{2}.$