In the book: Integral Geometry and Geometric Probability, (p16-17), the author proved that the measure of randomly throwing three points P1, P2, and P3 on the plane such that the circumdisk and the triangle formed by the points P1,P2,P3 both are in a given circle with fixed radius $\rho$ is $\frac{\pi^3\rho^6}{15}.$ The measure here is defined by the integral over the kinematic density.
In the content (p17) of the book, he used some technique of polar coordinates. Does anyone know how to derive the equation (2.19) from the equation (2.16) ? In particular, it seems that $R^3dR=\frac{ (\rho-r)^4}{4}$ ?! Does anyone know how to get this?
Thanks.
Finally, I got the answer. Since we assume that the circumdisk formed by $P_1P_2P_3$ has to be in the given circle $C_\rho$ with the radius $\rho$, the circumradius must be constrained; namely, for every fixed circumcenter of triangle $P_1P_2P_3$, , say $(r, \phi)$ in the polar coordinates, the range of the circumradius goes from zero to $(\rho-r)$.
Therefore, when we integrate the differential $3$-form in equation $(2.16)$ in page 16, the term related to the variable $R$ should be $$\int_0^{\rho-r}R^3dR=\frac{R^4}{4}\Bigg|^{\rho-r}_0=\frac{(\rho-r)^4}{4}.$$
Thanks for Paul's inspiration.