Question

As in Problem 2, complete Example 5.

Short answer

Answer

The answer is obtained.

$l=\frac{1}{15}\left(\begin{array}{ccc}\mathrm{\pi }& -1& -1\\ -1& \mathrm{\pi }& -1\\ -1& -1& \mathrm{\pi }\end{array}\right)$.

Step-by-step solution

Given information.

Physics definitions are given.

Definition of an orthogonal triad.

In a right-handed coordinate system, the unit vectors $\mathit{i}\mathbf{,}\mathit{j}\mathbf{,}\mathit{k}$are vectors of unit magnitude that point in the direction of the $\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}$axes, respectively. The orthogonal triad of unit vectors, often known as basic vectors, is a set of three unit vectors that are orthogonal to each other.

Define the region under consideration.

Define the region under consideration.

$S=\left\{\left(x,y,z\right)\in R^{3}X>0,y>>0.z>0and{x}^2+y^2+z^2\le 1\right\}$

Calculate the components of inertia.

Find the components of inertia.

$$\begin{gathered} l_{xz}=-\iint \underset{0}{\overset{1}{\int }}xzdV \\ =-\underset{0}{\overset{1}{\int }}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}r\cos \varphi \sin \theta r^2\sin \theta d\varphi d\theta dr \\ =-\underset{0}{\overset{1}{\int }}{r}^{4}dr{\int }_0^{\frac{\mathrm{\pi }}{2}}\cos \varphi d\varphi {\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^2\theta d\left(\sin \theta \right) \end{gathered}$$

Continue the evaluation.

$$\begin{gathered} l_{xz}=-\frac{1}{5}·1·\frac{1}{3} \\ =-\frac{1}{15} \end{gathered}$$

Make some analysis.

Make analysis about integrals of trigonometric ratios.

$$\begin{gathered} \underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\cos \varphi d\varphi =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\sin \varphi d\varphi \\ {l}_{yz}=l_{xz} \\ \underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\cos }^2\varphi d\varphi =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\sin }^2\varphi d\varphi \\ {l}_{yy}=l_{xx} \end{gathered}$$

Apply these equations and evaluate.

$$\begin{gathered} l_{zz}=\iint \int \left(x^2+y^2\right)dV \\ =\underset{0}{\overset{1}{\int }}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{r}^4{\sin }^3\theta d\varphi d\theta dr \\ =\underset{0}{\overset{1}{\int r^{4}dr}}\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}d\varphi \underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\sin }^3\theta d\theta \end{gathered}$$

Continue the evaluation.

Hence the final answer.

$l=\frac{1}{15}\left(\begin{array}{ccc}\mathrm{\pi }& -1& -1\\ -1& \mathrm{\pi }& -1\\ -1& -1& \mathrm{\pi }\end{array}\right)$