91Ó°ÊÓ

The given rectangular solid is defined by

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

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

Now, the moment of the inertia about the x-axis is,

$$\begin{gathered} I_x={∭}_T(y^2+z^2)\mathrm{ÒÏ}(x,y,z)dV \\ {I}_x={∫}_{x=a_1}^{a_2}{∫}_{y=b_1}^{b_2}{∫}_{z=c_1}^{c_2}(y^2+z^2)\left(k\right)dxdydz \\ {I}_x=={k{∫}_{x=a_1}^{a_2}{∫}_{y=b_1}^{b_2}\overline{)\left(y^{2}z+\frac{z^3}{3}\right)}}_{z=c_1}^{c_2}dxdy \\ {I}_x==k{∫}_{x=a_1}^{a_2}{∫}_{y=b_1}^{b_2}{y}^2(c_2−c_1)+\left(\frac{1}{3}\right)(c_2^3−c_1^3)dxdy \end{gathered}$$

Integrate with respect to 'y'

$$\begin{gathered} I_x={{∫}_{x=a_1}^{a_2}(c_2−c_1)\overline{)\left(\frac{{\displaystyle y^3}}{{\displaystyle 3}}\right)}}_{y=b_1}^{b_2}dx+{{∫}_{x=a_1}^{a_2}\left(\frac{{\displaystyle 1}}{{\displaystyle 3}}\right)(c_2^3−c_1^3)\overline{)\left(y\right)}}_{y=b_1}^{b_2}dx \\ {I}_x={\frac{{\displaystyle (c_2−c_1)(b_2^3−b_1^3)}}{{\displaystyle 3}}\left(x\right)|}_{x=a_1}^{a_2}+{\frac{{\displaystyle (c_2^3−c_1^3)(b_2−b_1)}}{{\displaystyle 3}}\left(x\right)|}_{x=a_1}^{a_2} \\ {I}_x=\frac{{\displaystyle (c_2−c_1)(b_2^3−b_1^3)(a_2−a_1)}}{{\displaystyle 3}}+\frac{{\displaystyle (c_2^3−c_1^3)(b_2−b_1)(a_2−a_1)}}{{\displaystyle 3}} \\ {I}_x=k\left(\frac{{\displaystyle (a_2−a_1)(b_2−b_1)(c_2−c_1)}}{{\displaystyle 3}}\right)(b_2^2+b_1^2+b_1{b}_2+c_2^2+c_1^2+c_1{c}_2) \end{gathered}$$

To find the radius of gyration about the x-axis, we have to find the mass of the rectangular solid:

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

So, the radius of gyration is:

$$\begin{gathered} R_x=\sqrt{\frac{I_x}{m}} \\ {R}_x=\sqrt{\frac{k\left(\frac{(a_2−a_1)(b_2−b_1)(c_2−c_1)}{3}\right)(b_2^2+b_1^2+b_1{b}_2+c_2^2+c_1^2+c_1{c}_2)}{k(a_2−a_1)(b_2−b_1)(c_2−c_1)}} \\ {R}_x=\sqrt{\frac{(b_2^2+b_1^2+b_1{b}_2+c_2^2+c_1^2+c_1{c}_2)}{3}} \end{gathered}$$