Q. 58 - Calculus

Question

In Exercises 57–60, let R be the rectangular solid defined by

R = {(x, y, z) | 0 ≤ x ≤ 4, 0 ≤ y ≤ 3, 0 ≤ z ≤ 2}.

Assume that the density of R is uniform throughout, and find the moment of inertia about the x-axis and the radius of gyration about the x-axis.

Step-by-Step Solution

Answer

The moment of inertia about the x-axis is $I_x=104$ and the radius of gyration about the x-axis is $R_x\approx 2.0816.$

Step 1. Given Information.

The given rectangular solid is defined by $\mathfrak{R}=\left\{\right(x,y,z)\mid 0\le x\le 4, 0\le y\le 3, 0\le z\le 2\}.$

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

It is given that the density of R is uniform throughout, so $\rho (x,y,z)=1.$

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

$$\begin{gathered} I_x={\iiint }_{\mathrm{\Omega }}(y^2+z^2)\rho (x,y,z)dv \\ {I}_x={\int }_0^4{\int }_0^3{\int }_0^2(y^2+z^2)dzdydx \\ {I}_x={\int }_0^4{\int }_0^3\left[{\int }_0^2(y^2+z^2)dz\right]dydx \\ {I}_x={\int }_0^4{\int }_0^3\left[y^2{\int }_0^{2}dz+{\int }_0^2{z}^{2}dz\right]dydx \\ {I}_x={\int }_0^4{\int }_0^3\left[2y^2+\frac{8}{3}\right]dydx \\ {I}_x={\int }_0^4\left[2{\int }_0^3{y}^{2}dy+\frac{8}{3}{\int }_0^{3}dy\right]dx \\ {I}_x={\int }_0^{4}26dx \\ {I}_x=104 \end{gathered}$$

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:

$$\begin{gathered} M={\iiint }_R\rho (x,y,z)dxdydz \\ M={\int }_0^4{\int }_0^3{\int }_0^{2}dxdydz \\ M==(4-0)(3-0)(2-0) \\ M=24 \end{gathered}$$

So, the radius of gyration is:

$$\begin{gathered} R_x=\sqrt{\frac{I_x}{m}} \\ {R}_x=\sqrt{\frac{104}{24}} \\ {R}_x\approx 2.0816 \end{gathered}$$