Find the mass of the unit sphere

The nice people in the comments have told me that the distance from any point to the surface of the unit sphere is given by: $(1-r)$ , where $r^2=x^2+y^2+z^2$. Since the density is said to be proportional to the distance from any point to the surface of the unit sphere, the density must differ by a constant and is given by: $\rho=k(1-r)$ , where $k$ is a constant. Converting $mass=\iiint_D\rho(x,y,z) dV$ into spherical coordinates by change of variable we have:$$mass=\iiint_R\rho r^2\sin\phi dr d\phi d\theta$$ $$=k\int_0^{2\pi}\int_0^{\pi}\int_0^1(1-r)r^2\sin\phi dr d\phi d\theta$$ $$=\frac{k}{12}\int_0^{2\pi}\int_0^{\pi}\sin\phi d\phi d\theta$$ $$=\frac{k}{6}\int_0^{2\pi} d\theta$$ $$=\frac{k\pi}{3}$$

The surface area of a sphere is $4\pi r^2$, so that the mass is

$$\int_0^1\rho(1-r)4\pi r^2\,dr.$$

Put $$x=r\cos(t)\sin (f) $$ $$y=r\sin(t)\sin(f) $$ $$z=r\cos(f) $$

if $\rho(x,y,z)=k (1-r ) $, then

$$mass=$$ $$k\int_0^1\int_0^{2\pi}\int_0^\pi(1-r)r^2\sin(f)drdtdf$$

$$=k\int_0^{2\pi}dt\int_0^\pi\sin(f)df\int_0^1 (r^2-r^3)dr $$

$$=4k\pi (\frac 1 3-\frac 1 4) $$