91Ó°ÊÓ

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

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

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

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

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 $\mathrm{ÒÏ}(x,y,z)=kx.$

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

Let's find the mass of the solid:

$$\begin{gathered} M={∭}_R\mathrm{ÒÏ}(x,y,z)dxdydz \\ M={∫}_{x=a_1}^{a_2}{∫}_{y=b_1}^{b_2}{∫}_{z=c_1}^{c_2}kdxdydz \\ M=k(a_2−a_1)(b_2−b_1)(c_2−c_1) \end{gathered}$$

Now, the center of mass is,

$$\begin{gathered} \stackrel{¯}{x}=\frac{M_{yz}}{M}=\frac{{∭}_{T}x\mathrm{ÒÏ}(x,y,z)dxdydz}{∭\mathrm{ÒÏ}(x,y,z)dxdydz} \\ \stackrel{¯}{x}=\frac{1}{M}{∫}_{x=a_1}^{a_2}{∫}_{y=b_1}^{b_2}{∫}_{z=c_1}^{c_2}\left(kx\right)dxdydz \\ \stackrel{¯}{x}=\left(\frac{k}{2M}\right)(a_2^2−a_1^2)(b_2−b_1)(c_2−c_1) \\ \stackrel{¯}{x}=\left(\frac{{\displaystyle k}}{{\displaystyle 2\left(k(a_2−a_1)(b_2−b_1)(c_2−c_1)\right)}}\right)(a_2^2−a_1^2)(b_2−b_1)(c_2−c_1) \\ \stackrel{¯}{x}=\frac{a_1+a_2}{2} \end{gathered}$$

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