Question

Mass of uniform density=1, bounded by the coordinate planes and the plane x +y +x=1 .

Short answer

The principal moment of inertia is$$\begin{gathered} I=\frac{1}{120}\left(4-1-1 \\ -14-1 \\ -1-14\right) \end{gathered}$$

Step-by-step solution

Given Information

Plane determined by equation $x+y+z=1$.

Definition of Principal axis and Principal moments of inertia.

An eigenvector of the mass moment of inertia tensor defined relative to a point is a principal axis of rotation. The primary moments of inertia are the eigenvalues that corresponds to them.

Specify body shape

Body is a three-sided pyramid with an equilateral base.

$$\begin{gathered} S=\left\{\left(x,y,z\right)\in {\square }^{3}0<x<1,0<y<1-x,0<z<1-x-y\right\} \\ =\left\{\left(x,y,z\right)\in {\square }^{3}0<y<1,0<z<1-y,0<x<1-y-z\right\} \\ =\left\{\left(x,y,z\right)\in {\square }^{3}0<z<1,0<x<1-z,0<z<1-x-y\right\} \end{gathered}$$

Calculate components of inertia tensor.

Find the components of inertia tensor.

$$\begin{gathered} I_x={\int }_0^1{\int }_0^{1-x}{\int }_0^{1-x-y}\left(y^2+z^2\right)dzdydx \\ \left(permutex\mapsto \mathrm{Y}\mapsto \mathrm{z}\mapsto \mathrm{x}\right) \\ ={\int }_0^1{\int }_0^{1-y}{\int }_0^{1-y-z}\left(z^2+x^2\right)dxdzdy \\ =I_{yy} \end{gathered}$$

Simplify further.

$$\begin{gathered} \left(permutex\mapsto y\mapsto z\mapsto x\right) \\ ={\int }_0^1{\int }_0^{1-z}{\int }_0^{1-z-x}\left(x^2+y^2\right)dydxdz \\ =I_{zz} \end{gathered}$$

Simply for xx .

$$\begin{gathered} I_{xx}={\int }_0^1{\int }_0^{1-x}{\overline{)\left(y^{2}z+\frac{1}{3}{z}^3\right)}}_0^{1-x-y}dydx \\ ={\int }_0^1{\int }_0^{1-x}\left(1-x\right)y^2-y^3+\frac{1}{3}{\left(1-x=y\right)}^{3}dydx \\ =\frac{1}{3}{\int }_0^1{\overline{)\left(\left(1-x\right)y^3\frac{3}{4}{y}^3-\frac{1}{4}{\left(1-x-y\right)}^4\right)}}_0^{1-x}dx \\ =\frac{1}{12}{\int }_0^1{\left(1-x\right)}^4-\left(0-{\left(1-x\right)}^4\right)dx \end{gathered}$$

Solve further.

$$\begin{gathered} =-\frac{2}{12.5}{\overline{){\left(1-x\right)}^5}}_0^1 \\ =\frac{1}{30} \end{gathered}$$

Non-diagonal elements will be same by above arguments.

$$\begin{gathered} {\mathrm{I}}_{\mathrm{xy}}={\int }_0^1{\int }_0^{1-\mathrm{x}}{\int }_0^{1-\mathrm{x}-\mathrm{y}}\mathrm{xydzdydx} \\ \left(\mathrm{permute}\mathrm{x}\mapsto \mathrm{y}\mapsto \mathrm{z}\mapsto \mathrm{x}\right) \\ =-{\int }_0^1{\int }_0^{1-\mathrm{y}}{\int }_0^{1-\mathrm{y}-\mathrm{z}}\mathrm{yzdxdzdy} \\ ={\mathrm{I}}_{\mathrm{yz}} \end{gathered}$$

Further,

$$\begin{gathered} \left(permutex\mapsto y\mapsto z\mapsto x\right) \\ =-{\int }_0^1{\int }_0^{1-z}{\int }_0^{1-y-z}zxdydxdz \\ =I_{xz} \\ =-\frac{1}{120} \end{gathered}$$

The tensor of inertia.

$$\begin{gathered} I=\frac{1}{120}\left(4-1-1 \\ -14-1 \\ -1-14\right) \end{gathered}$$.

 Step 5: Find principal axes by symmetry of body.

Find the principal axes.

$$\begin{gathered} {\omega }_1=\left(1,1,1\right) \\ {\omega }_2=\left(1,0,-1\right) \\ {\omega }_3=\left(1,-2,1\right) \end{gathered}$$

They are mutually orthogonal.

Principal axes are:

$$\begin{gathered} I{\omega }_1=\frac{2}{120}{\omega }_1 \\ I{\omega }_2=\frac{5}{120}{\omega }_2 \\ I{\omega }_3=\frac{5}{120}{\omega }^3 \end{gathered}$$

Hence, Principal moment of inertia$$\begin{gathered} I=\frac{1}{120}\left(4-1-1 \\ -14-1 \\ -1-14\right) \end{gathered}$$