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 a mass of the rectangular solid is

$\stackrel{¯}{x}=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2},\stackrel{¯}{y}=\frac{m_1{b}_1+m_2{b}_2}{m_1+m_2},\stackrel{¯}{z}=\frac{m_1{c}_1+m_2{c}_2}{m_1+m_2}$where the $m_1\mathrm{and}{m}_2$are the masses of the body.

Since the density is uniform $m_1=m_2=k.$

Thus, the center of the masses are$\stackrel{¯}{x}=\frac{a_1+a_2}{2},\stackrel{¯}{y}=\frac{b_1+b_2}{2},\stackrel{¯}{z}=\frac{c_1+c_2}{2}.$

It is given that the density of R is uniform throughout, so$\mathrm{ÒÏ}(x,y,z)=k.$

To find the center of a mass using the integral expression, let's first find the mass,

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

Now, the center of the mass of the rectangular solid is

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

By proceeding with the center of a mass,

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

Now,

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

Hence proved, the center of mass is$\left(\frac{a_1+a_2}{2},\frac{b_1+b_2}{2},\frac{c_1+c_2}{2}\right).$