91影视

Uniform density =1 inside a unit sphere is assumed.

In classical mechanics, the inertia tensor of a rigid body, as well as the Centreof mass (CM), is a fundamental physical parameter affecting rotational motion.

Define the region under consideration.

$S=\left\{\left(x,y,z\right)鈭�R^{3}x>0,y>0and{x}^2+y^2+z^2<1\right\}$

Fix density as 1 and obtain $dm=dV$.

Find the components of inertia.

$$\begin{gathered} I_{xx}=鈭�鈭�\left(y^2+z^2\right)dV \\ =\underset{0}{\overset{1}{鈭�}}\underset{0}{\overset{\mathrm{蟺}}{鈭�}}\underset{0}{\overset{\frac{\mathrm{蟺}}{2}}{鈭�}}\left({\sin }^2胃{\sin }^2蠒+{\cos }^2胃\right)r^4\sin 胃drd胃d蠒 \\ =\underset{0}{\overset{1}{鈭�}}{r}^{4}dr\left(\underset{0}{\overset{\frac{\mathrm{蟺}}{2}}{鈭�}}{\sin }^2蠒d蠒\underset{0}{\overset{\mathrm{蟺}}{鈭�}}{\sin }^3胃d胃+\underset{0}{\overset{\frac{\mathrm{蟺}}{2}}{鈭�}}d蠒\underset{0}{\overset{\mathrm{蟺}}{鈭�}}\sin 胃{\cos }^2胃d胃\right) \end{gathered}$$

Continue the evaluation.

$$\begin{gathered} I_{xx}=\frac{1}{5}\left(\frac{\mathrm{蟺}}{4}路\frac{4}{3}+\frac{\mathrm{蟺}}{2}路\frac{2}{3}\right) \\ =\frac{2\mathrm{蟺}}{15} \end{gathered}$$

Repeat the process.

$$\begin{gathered} I_{xx}=鈭�鈭�\left(y^2+z^2\right)dV \\ =\frac{2\mathrm{蟺}}{15} \\ {I}_{zz}=鈭�鈭�\left(x^2+y^2\right)dV \\ =\underset{0}{\overset{1}{鈭�}}\underset{0}{\overset{\mathrm{蟺}}{鈭�}}\underset{0}{\overset{\frac{3}{2}}{鈭�}}{r}^4{\sin }^3胃dr胃d蠒 \end{gathered}$$

Continue the evaluation.

$$\begin{gathered} I_{zz}=\frac{1}{5}路\frac{4}{3}路\frac{\mathrm{蟺}}{2} \\ =\frac{2\mathrm{蟺}}{15} \end{gathered}$$

Repeat the process.

$$\begin{gathered} I_{xy}=鈭�鈭�xydV \\ =\underset{0}{\overset{1}{鈭�}}{r}^{4}dr\underset{0}{\overset{\mathrm{蟺}}{鈭�}}{\sin }^3胃d胃\underset{0}{\overset{\frac{x}{2}}{鈭�}}\cos 蠒\sin 蠒d蠒 \\ =-\frac{1}{5}路\frac{4}{3}路\frac{1}{2} \\ =-\frac{2}{15} \end{gathered}$$

Calculate the same for the remaining off-diagonal components.

$$\begin{gathered} l_{xz}=-\underset{0}{\overset{1}{鈭�}}xzdV \\ =-\underset{0}{\overset{1}{鈭�}}dr\underset{0}{\overset{\mathrm{蟺}}{鈭�}}{\sin }^2胃\cos 胃d胃\underset{0}{\overset{\frac{\mathrm{蟺}}{2}}{鈭�}}\cos 蠒d蠒 \\ =0 \\ {l}_{yz}=0 \end{gathered}$$

Hence the results to obtain the final solution is $l=\frac{2}{15}\left(\begin{array}{ccc}\mathrm{蟺}& -1& 0\\ -1& \mathrm{蟺}& 0\\ 0& 0& \mathrm{蟺}\end{array}\right)$.