Q. 68 - Calculus

Question

Let a, b, and c be positive real numbers. In Exercises 65–68, let T be the tetrahedron with vertices (0, 0, 0), (a, 0, 0), (0, b, 0), and (0, 0,c).

Assuming that the density at each point in T is proportional to the distance of the point from the yz-plane, set up the integrals required to 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 $I_x={\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{x}{a}\right)}{\int }_{z=0}^{c\left(1-\frac{x}{a}-\frac{y}{b}\right)}(y^2+z^2)kx dxdydz$ and the radius of gyration about the x-axis is $R_x=\sqrt{\frac{{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{x}{a}\right)}{\int }_{z=0}^{c\left(1-\frac{x}{a}-\frac{y}{b}\right)}\left(y^2+z^2\right)kx\hspace{0.17em}dxdydz\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{x}{a}\right)}{\int }_{z=0}^{c\left(1-\frac{x}{a}-\frac{y}{b}\right)}kx\hspace{0.17em}dxdydz}}.$

Step 1. Given Information.

The given vertices of the tetrahedron are $(0, 0, 0), (a, 0, 0), (0, b, 0), \mathrm{and} (0, 0,c).$

Step 2. Find the first moment of inertia.

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

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

$$\begin{gathered} I_x={\iiint }_T(y^2+z^2)\rho (x,y,z)dxdydz \\ {I}_x={\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{x}{a}\right)}{\int }_{z=0}^{c\left(1-\frac{x}{a}-\frac{y}{b}\right)}(y^2+z^2)kx dxdydz \end{gathered}$$

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 tetrahedron:

$$\begin{gathered} m={\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{x}{a}\right)}{\int }_{z=0}^{c\left(1-\frac{x}{a}-\frac{y}{b}\right)}\rho (x,y,z)dxdydz \\ m={\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kx(x,y,z)dxdydz \end{gathered}$$

So, the radius of the gyration about the x-axis is,

$$\begin{gathered} R_x=\sqrt{\frac{I_x}{m}} \\ {R}_x=\sqrt{\frac{{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{x}{a}\right)}{\int }_{z=0}^{c\left(1-\frac{x}{a}-\frac{y}{b}\right)}\left(y^2+z^2\right)kx\hspace{0.17em}dxdydz\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{x}{a}\right)}{\int }_{z=0}^{c\left(1-\frac{x}{a}-\frac{y}{b}\right)}kx\hspace{0.17em}dxdydz}} \end{gathered}$$