I can see why mathematically, but why conceptually?
[Physics] Why is the moment of inertia of a solid hemisphere the same as that of a full sphere revolving around the same axis
moment of inertiarotational-dynamics
Related Solutions
The meaning of moment of inertia tensor comes from $$\vec{L}=I\vec{\omega}$$
So for example if you consider an object rotating about the $y$ axis with angular velocity $\omega$, you have
$$\left(\begin{array}{c} L_x\\L_y\\L_z \end{array}\right)=\left(\begin{array}{ccc} I_{xx}&I_{xy}&I_{xz}\\I_{yx}&I_{yy}&I_{yz}\\I_{zx}&I_{zy}&I_{zz}\end{array}\right)\left(\begin{array}{c}0\\ \omega\\0\end{array}\right)$$ $$=\left(\begin{array}{c}I_{xy}\\I_{yy}\\I_{zy}\end{array}\right)\omega$$
So $I_{xy}\omega$ is the $x$-component of angular momentum when the object rotates about the $y$ axis with angular velocity $\omega$, and similarly for the meaning of the other components.
In general $I_{ij}\omega$ is the $i$-th component of the angular momentum of the object rotating with angular velocity $\omega$ about the $j$-th axis.
For axes along which the rotation of an object will give angular momentum along the same axes as well, then the off-diagonal elements are zero if these axes are chosen as basis. Because in that case you have, e.g., for rotation about the $x$-axis $$\vec{L}=I_{xx}\omega \hat{x}$$ for rotation about $y$-axis $$\vec{L}=I_{yy}\omega \hat{y}$$ for rotation about $z$-axis $$\vec{L}=I_{zz}\omega \hat{z}$$ and hence in general for $$\vec{\omega}=\omega_x\hat{x}+\omega_y\hat{y}+\omega_z\hat{z}$$ $$\vec{L}=\left(\begin{array}{ccc}I_{xx}&0&0\\0&I_{yy}&0\\0&0&I_{zz}\end{array}\right)\left(\begin{array}{c}\omega_x\\\omega_y\\\omega_z\end{array}\right)$$
- But I don't know what is the axis of rotation of Ixy, Iyz ,Izx I mean about which axis they are rotating?
$I_{ij}$ is talking about the rotation about the $j$-th axis.
The moment of inertia is defined relative to an axis, not a point. Therefore the distance you need is from a mass element in the sphere to the $z$-axis, not to the origin. This is why the disk method is used -- it takes advantage of the symmetry. Your expression uses the distance $r$ to the origin, whereas the distance to the $z$ axis in spherical coordinates is $r\sin \theta$ (where $\theta$ is the angle between the $z$ axis and the point of interest).
The correct expression for a spherical object with arbitrarily varying density in spherical coordinates is
\begin{eqnarray} I &=& \int_0^R {\rm d} r r^2 \int_0^{\pi} {\rm d\theta} \sin \theta \int_0^{2\pi}{\rm d}\phi \rho(r,\theta,\phi)\Delta(r,\theta,\phi)^2, \end{eqnarray} where $\rho(r,\theta,\phi)$ is the density and $\Delta(r,\theta,\phi)=r\sin\theta$ is the distance from a mass element at $r,\theta,\phi$ to the $z$ axis.
For completeness we can work this out for a constant density sphere, $\rho(r,\theta,\phi)=3 M/(4\pi R^3)$ (note that the integrand doesn't depend on $\phi$, so the $\phi$ integral is just $2\pi$) \begin{eqnarray} I &=& 2\pi \frac{3M}{4\pi R^3} \int_0^R {\rm d} r r^2 \int_0^{\pi} {\rm d\theta} \sin \theta\ \left(r \sin\theta\right)^2 \\ &=& \frac{3 M}{2 R^3} \int_0^R {\rm d} r r^4 \int_0^\pi {\rm d}\theta \sin^3 \theta \\ &=& \frac{3 M}{2 R^3} \times \frac{R^5}{5} \times \frac{4}{3} \\ &=& \frac{2}{5} MR^2 \end{eqnarray}
Best Answer
Because when you think about what fraction of the mass is at what distance from the axis (say, between $r$ and $r+dr$), they’re the same.