Integration of 3-momentum During a lecture that I missed, I was trapped when the lecturer uses the relation
$$dp_x~ dp_y ~dp_z ~=~d^3\mathbf{p} ~=~ 4\pi p^2 dp.$$
Can I know how is this relation derived please?
 A: This is not an equality, strictly speaking. Looks like your lecturer used spherical coordinates. If the integrand is spherically symmetric, i.e. it only depends on the magnitude of $\mathbf{p}$, then the integration over the angular coordinates is trivial and just gives you the solid angle subtended by a sphere, $4\pi$.
A: More mathematically, it comes from the change in volume element when making a change of variable. I will give here some intuitive arguments in 1D and 2D and give the general formula then:


*

*In 1D, if you integrate along the real line and change from a variable $x$ to $X = f(x)$, you know that the measure element $dX = f'(x) dx$ which implies that $dx = \frac{dX}{f' \circ f^{-1}(X)}$. Basically, to be consistent with the original length measure $dx$, you need to rescale the new length measure $dX$.

*In 2D, let's say you want to go from $(x,y)$ to $(X,Y)=(f_X(x,y),f_Y(x,y))$. What happens is that in general a square in coordinates $(x,y)$ with area $dxdy$ will be deformed into a parallelogram in the variables $(X,Y)$ and vice versa as depicted on the figure.

One can actually compute the area of a parallelogram by using the cross product between the two vectors from which it can be generated.
Let us look at the case of the red square. In the variables $(X,Y)$, it has an area $dXdY$. Now, we also know that they can be expressed in the space $(x,y)$ via $dX = \partial_x f_X dx + \partial_y f_X dy$ and $dY = \partial_x f_Y dx + \partial_y f_Y dy$.
I will use a notation shortcut here by introducing the wedge product $\wedge$ (so that $dx \wedge dx = dy \wedge dy = 0$ and $dx \wedge dy = - dy \wedge dx = dxdy$) that does the same geometric stuff as considering cross product for the parallelogram but without using vectors explicitely.
Basically, the area of the red parallelogram in $(x,y)$ reads $dX(x,y) \wedge dY(x,y)$ i.e.
$\left(\partial_x f_X dx + \partial_y f_X dy \right) \wedge \left(\partial_x f_Y dx + \partial_y f_Y dy \right) = \left| \partial_x f_X \partial_y f_Y - \partial_y f_X \partial_x f_Y \right| dxdy $
Now if I intoduce the Jacobian matrix with elements $J_{ij} \equiv \partial_i f_j(x,y)$, it is easy to see that $dXdY = |\det(J)|dxdy$.


*

*In 3D, we have exactly the same thing as before so that a change of variable will deform a cube when going from one representation to another. Going from a cube in the new representation $(X,Y,Z)$ with volume $dXdYdZ$ will give in a $(x,y,z)$ representation 
$dX(x,y,z) \wedge dY(x,y,z) \wedge dZ(x,y,z) = |\det(J)|dxdydz$


Application to spherical coordinates:
The spherica coordinates $(r,\theta, \phi)$ are defined via the following relations:
$x = r \sin \theta \cos \phi$
$y = r \sin \theta \sin \phi$
$z = r \cos \theta$
The corresponding Jacobian matrix reads
$J = \left( \begin{array}{ccc}
\sin \theta \cos \phi  & r \cos \theta \cos \phi & -r \sin \theta \sin \phi \\
\sin \theta \sin \phi & r \cos \theta \sin \phi & r \sin \theta \cos \phi \\
\cos \theta & -r\sin \theta & 0 \end{array} \right) $
whose determinant is simply (it's easier from the third line):
$\det(J) = \cos \theta \left(r^2 \cos^2 \phi \cos \theta \sin \theta + r^2 \sin^2 \phi \cos \theta \sin \theta \right) + r \sin \theta \left( r\sin^2 \theta \cos^2 \phi + r\sin^2 \theta \sin^2 \phi \right) = r^2 \sin \theta$
Here, I have used the jacobian of transformation from $(r,\theta, \phi)$ to the new variables $(x,y,z)$, hence a cube in $(x,y,z)$ with volume $dxdydz$ will satisfy:
$dxdydz = \det(J) dr d\theta d\phi = r^2 \sin \theta dr d \theta d\phi$
In the particular example you mention, they used spherical coordinates for the momenta because the hamiltonian is a function of the norme of the momentum and not its direction. It allows then to integrate easily over the angular degrees of freedom of the momentum (this is not the same as angular degrees of freedom of the position be careful).
