Q. 64 - Calculus

Question

In Exercises 61–64, let R be the rectangular solid defined by $\mathfrak{R}=\left\{(x,y,z)|0\le a_1\le x\le a_2, 0\le b_1\le y\le b_2, 0\le c_2\le z\le c_2\right\}.$

Assuming that the density at each point in R is proportional to the distance of the point from the yz-plane, find the first moment of inertia about the x-axis and the radius of gyration about the x-axis.

Step-by-Step Solution

Answer

The first moment of inertia about the x-axis is $I_x=k\left(\frac{(a_2-a_1)(b_2-b_1)(c_2-c_1)}{3}\right)(b_2^2+b_1^2+b_1{b}_2+c_2^2+c_1^2+c_1{c}_2)$ and the radius of gyration about the x-axis is $R_x=\sqrt{\frac{(b_2^2+b_1^2+b_1{b}_2+c_2^2+c_1^2+c_1{c}_2)}{3}}.$

Step 1. Given Information.

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

Step 2. Find the first moment of inertia about the x- axis.

It is given that the density at each point in R is proportional to the distance of the point from the yz-plane, so $\rho (x,y,z)=kx.$

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

$$\begin{gathered} I_x={\iiint }_T(y^2+z^2)\rho (x,y,z)dV \\ {I}_x={\int }_{x=a_1}^{a_2}{\int }_{y=b_1}^{b_2}{\int }_{z=c_1}^{c_2}(y^2+z^2)\left(kx\right)dxdydz \\ {I}_x={k\overline{)\left(\frac{x^2}{2}\right)}}_{x=a_1}^{a_2}\times {{\int }_{z=c_1}^{c_2}\overline{)\left(\frac{y^3}{3}\right)}}_{b_1}^{b_2}+{\int }_{x=a_1}^{a_2}{\overline{)z^2\left(y\right)}}_{b_1}^{b_2}dz \end{gathered}$$

Integrate with respect to 'y'

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

Step 3. Find the radius of the gyration.

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

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

So, the radius of gyration is:

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