Hydrostatic equlibrium and Centre of pressure

[Taylor_1989]

Homework Statement

A uniform star has a mass of M and radius of R,use the hydrostatic equilibrium equation to find its centre pressure. (note: not the minimum
centre pressure)

Homework Equations

$$\frac{dp}{dr}=-\frac{Gm\left(r\right)\rho \left(r\right)}{r^2}[1]$$

The Attempt at a Solution

Have I followed the right assumptions in my working? The reason I followed this is because it states uniform star.

I assume density of the star dose not change w/r to radius

$$M(r)=4/3 \pi r^3<\rho> [2]$$

subbing [2] into [1]

$$\frac{dp}{dr}=\frac{-G4\pi \:r^3<\rho >\:}{3r^2}=\frac{-G4\pi \:r<\rho \:>\:}{3}[3]$$

$$\int _{P_c}^0\:dp=\int _0^{R_s}\:\frac{-G4\pi \:r<\rho \:>\:}{3}[4]$$

$$P_c=\:\frac{-G2\pi \:R_s^2<\rho \:>\:}{3}[5]$$

So my $<\rho>$ is the average density of the star and as the pressure at the centre of the star is much greater than the surface I have assumed the pressure at the surface is 0.

I am a bit ifify on this as it the uniform mass which is tripping me up, so I just thought how I would do it, if I asked the question my self, have I missed something in the question should I assume mean density of the star?

 

[haruspex]

What happened to the ρ(r) term in eqn 1?

 

[Taylor_1989]

What happened to the ρ(r) term in eqn 1?

Sorry my corrections as follows:
$$\frac{dp}{dr}=-\frac{4}{3}G\pi r<\rho >^2$$

carrying through my edit

$$P_c=\frac{2}{3}G\pi R^2<\rho >^2$$

 

[haruspex]

Sorry my corrections as follows:
$$\frac{dp}{dr}=-\frac{4}{3}G\pi r<\rho >^2$$

carrying through my edit

$$P_c=\frac{2}{3}G\pi R^2<\rho >^2$$

Looks right to me.