Q. 66 - 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).

Assume that the density of T is uniform throughout. Set up the integrals required to 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 $I_x=kabc\left(\frac{b^2+c^2}{60}\right)$ and the radius of gyration about the x-axis is $R_x=\sqrt{\frac{(b^2+c^2)}{10}}.$

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 moment of inertia about the x- axis.

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

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

$$\begin{gathered} I_x={\iiint }_T(y^2+z^2)\rho (x,y,z)dxdydz \\ {I}_x={\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{x}{a}\right)}{\int }_{z=0}^{c\left(1-\frac{x}{a}-\frac{y}{b}\right)}k(y^2+z^2) dxdydz \\ {I}_x=k{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{x}{a}\right)}{\overline{)\left(y^{2}z+\frac{z^3}{3}\right)}}_{z=0}^{c\left(1-\frac{x}{a}-\frac{y}{b}\right)}dxdy \\ {I}_x=k{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{x}{a}\right)}\left(y^{2}c\left(1-\frac{x}{a}-\frac{y}{b}\right)+\frac{1}{3}{c}^3{\left(1-\frac{x}{a}-\frac{y}{b}\right)}^3\right)dxdy \\ {I}_x=kc{\int }_{x=0}^a{\int }_{y=0}^{b(1-\frac{x}{a})}\left(y^2-\frac{x{y}^2}{a}-\frac{y^3}{b}\right)dxdy \\ +\left(\frac{k{c}^3}{3}\right){\int }_{x=0}^a{\int }_{y=0}^{b(1-\frac{x}{a})}{\left(1-\frac{x}{a}-\frac{y}{b}\right)}^{3}dxdy \end{gathered}$$

Step 3. Solve.

By proceeding with the calculation further,

Take the integral as $I_x=I_1+I_2.$

$$\begin{gathered} I_1=kc{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{x}{a}\right)}\left(y^2-\frac{x{y}^2}{a}-\frac{y^3}{b}\right)dxdy \\ {I}_1=={kc{\int }_{x=0}^a\overline{)\left(\frac{y^3}{3}-\frac{x{y}^3}{3a}-\frac{y^4}{4b}\right)}}_{y=0}^{y=(1-\frac{x}{a})}dx \\ {I}_1=\frac{kac{b}^3}{60} \end{gathered}$$

Now, 

$$\begin{gathered} I_2=\left(\frac{k{c}^3}{3}\right){\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{x}{a}\right)}{\left(1-\frac{x}{a}-\frac{y}{b}\right)}^{3}dxdy \\ {I}_2=\left(\frac{k{c}^3}{3}\right)\left(\frac{-b}{4}\right){\int }_{x=0}^a{\overline{)\left({\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}^4\right)}}_{y=0}^{y=b\left(1-\frac{x}{a}\right)}dx \\ {I}_2=\frac{kab{c}^3}{60} \end{gathered}$$

So,

$$\begin{gathered} I_x=I_1+I_2 \\ {I}_x=\frac{kac{b}^3}{60}+\frac{kab{c}^3}{60} \\ {I}_x=kabc\left(\frac{b^2+c^2}{60}\right) \end{gathered}$$

Step 4. 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: 

$$\begin{gathered} m={\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{x}{a}\right)}{\int }_{z=0}^{c\left(1-\frac{x}{a}-\frac{y}{b}\right)}kdxdydz \\ m=\frac{kabc}{6} \end{gathered}$$

So, the radius of gyration is:

$$\begin{gathered} R_x=\sqrt{\frac{I_x}{m}} \\ {R}_x=\sqrt{\frac{kabc(b^2+c^2)}{60}\times \frac{6}{kabc}} \\ {R}_x=\sqrt{\frac{(b^2+c^2)}{10}} \end{gathered}$$