Is the distance involved in calculating angular momentum to an axis or a point? I'm a high school student.I still don't really understand angular momentum and moment of inertia. I know the moment of inertia of a point mass is defined as $mr^2$. For any other shape, we integrate this to produce a formula. For instance, the formula for a sphere is $\frac{2}{5}MR^2$. We integrate $r^2$ with respect to dm. My question is how do you measure the $r$ in a sphere? Some books say it's the distance from the point mass to its axis of rotation, but there are so many points on the axis of rotation. Which distance is it really inferring to? The center of the sphere?
Also, imagine the sphere as earth. Does a point mass on the equator have the same angular momentum as a mass point at a higher latitude? I always heard people say that when a point moving to higher latitude, its tangential speed decreases because it gets closer to the axis. Here is the question again. If I see the earth as infinite amount of disks chopped horizontally, then as the point goes up, it goes onto another "disk", which has its own origin. But if call the distance to the axis as the distance from a mass point to the center of the earth, then the distance isn't changing. I think I really messed up something in my understanding of this concept.
 A: Caution: Calculus ahead
The $r$ is the position of the mass particle ${\rm d}m$. It works best in vector form as $$\vec{r} = \pmatrix{x \\ y \\ z}$$
The mass moment of inertia ${\rm I}$ is a 3×3 matrtix (a tensor) defined as follows (first the shape mass and then the shape MMOI):
$$ m = \int {\rm d}m = \int \rho {\rm d}V $$
and
$$ \mathrm{I} = \int \begin{vmatrix} y^2+z^2 & -x y & -x z \\ -x y & x^2+z^2 & -y z \\ -x z & -y z & x^2+y^2  \end{vmatrix} \rho {\rm d}V $$
Example of Sphere
Imagine a solid sphere of radius $R$, where any interior point can be expressed as a function of two angles $\varphi$ and $\psi$ as well as the distance from the center $r$. The position of each ${\rm d}m$ is described using spherical coordinates as:
$$ \vec{r} = \pmatrix{x \\ y \\ z} = \pmatrix{r \cos\varphi \cos\psi \\ r \sin \psi \\ r \sin\varphi \cos\psi} $$
with some looking up online, you will find that in spherical coordinates $${\rm d}V = r^2 \cos \psi\, {\rm d}\varphi\, {\rm d}\psi\, {\rm d}r$$
First, the density is found from
$$ m = \int \rho {\rm d}V = \int_0^R \int_{-\pi/2}^{\pi/2} \int_{0}^{2\pi} \rho r^2 \cos \psi \,{\rm d}\varphi\, {\rm d}\psi\, {\rm d}r = \rho \frac{4}{3} \pi R^3 = \rho V \; \checkmark $$
Now with a lot of math, you can find that
$$\mathrm{I} = \int_0^R \int_{-\pi/2}^{\pi/2} \rho \pi r^2 \begin{vmatrix} 1+\sin^2\psi & & \\ & 2 \cos^2\psi & \\ & & 1+\sin^2 \psi \end{vmatrix} r^2 \cos \psi \, {\rm d}\psi\, {\rm d}r $$
$$ \Rightarrow \;\; \mathrm{I} = \rho \begin{vmatrix} \frac{8 \pi R^5}{15} & & \\ & \frac{8 \pi R^5}{15} & \\ & & \frac{8 \pi R^5}{15} \end{vmatrix} $$
and since $\rho = \frac{m}{\tfrac{4}{3} \pi R^3}$ the above becomes
$$  \mathrm{I} = \begin{vmatrix} \tfrac{2}{5} m R^2 & & \\ & \tfrac{2}{5} m R^2 & \\ & & \tfrac{2}{5} m R^2 \end{vmatrix}  \;\; \checkmark $$
So we have derived the mass moment of inertia for a symmetrical sphere
Angular Momentum
Mass moment of inertia is important because it allows us to calculate angular momentum from a rotational velocity vector $\vec{\omega}$
$$ \vec{L}_{\rm com} = \mathrm{I}_{\rm com} \vec{\omega} $$
where ${\rm com}$ is the center of mass.
Additionally, the angular momentum about a different point is
$$ \vec{L} = \mathrm{I}_{\rm com} \vec{\omega} + \vec{r}_{\rm com} \times \vec{p} $$
where $\vec{p} =m \vec{v}_{\rm com}$ is the linear momentum.
So the complex integral around a shape to get the mass moment of inertia only needs to happen once, and then you just use the 3×3 matrix is the equations above to get the angular momentum vector.
Well almost. In reality the mass moment of inertia is defined in a coordinate system aligned with the body geometry (such as x-axis is along the long dimension for example) and to use it in dynamics it has to be transformed into an inertial world coordinate frame. See this post on how this is done.

PS. To find the infinitesimal volume in any coordinate system, describe the position of the volume in terms of thre parameters, like $\vec{\rm pos}(t,s,e)$ and define the infinitesimal volume as
$$ {\rm d}V = \frac{ \partial \vec{\rm pos}}{\partial t} \cdot \left( \frac{ \partial \vec{\rm pos}}{\partial s} \times \frac{ \partial \vec{\rm pos}}{\partial e} \right) {\rm d}t\, {\rm d}s\, {\rm d}e $$
where $\cdot$ is the vector dot prodcut, and $\times$ is the vector cross product.
A: The distance from a point to a line (such as your axis of rotation) is the shortest such distance.  If you drew it as a line segment, it would always be perpendicular to the axis.

Does a point mass on the equator have the same angular momentum as a mass point at a higher latitude?

What's one formula for angular momentum?  $L = I \omega$.  Because the earth's surface is rigid, the angular speed is the same everywhere.  But the moment of inertia of that point changes as it gets closer or further from the axis.  The distance is equal to the radius of the earth at the equator, and zero at the poles.  So the angular momentum is maximal at the equator and zero at the poles (on the axis).
Since you mention integration and dm, I assume you are taking a calculus based physics course.  In that case, hyperphysics has a treatment for calculating the moment of inertia of a sphere that you should be able to follow.
