The density of a black hole is a constant

the volume formula V$=$(4/3)$\pi $R$^{\mathrm{3}}$, the density of the stars $\rho =$M/V. From the density formula, we can see: the density and radius of the stars in the process of star change According to the law of conservation of angular momentum, MVR$=$MCRS, so V$^{\mathrm{3}}$R$^{\mathrm{3}}=$C$^{\mathrm{3}}$R$^{\mathrm{3}}_{\mathrm{S}}$. Conclusion: The radius cube is inversely proportional to the cube of the spin speed, and the density is inversely proportional to the cube of the radius. Then the density must be proportional to the cube of the velocity. Proportional: $\rho =$KV$^{\mathrm{3}}$. The speed of the black hole's rotation is the speed of light, so the density of the black hole $\rho =$KC$^{\mathrm{3}}$ where K is the constant of proportionality and C is the speed of light, so the density of the black hole is a constant. The density of the sun is: 1.408\texttimes 10\textthreesuperior kg/m3, and the speed of the sun's rotation is 2\texttimes 10$^{\mathrm{3}}$ m/s, substitute $\rho =$KV$^{\mathrm{3}}$ to calculate K$=$1.76\texttimes 10$^{\mathrm{-7}}$, substitute $\rho =$KC$^{\mathrm{3}}==$1.76\texttimes 10$^{\mathrm{-7}}$\texttimes (3\texttimes 10$^{\mathrm{8}})$3$=$4.752\texttimes 10$^{\mathrm{18}}$(kg/m$^{\mathrm{3}})$