TITLE: Find the mass of the unit sphere
QUESTION [6 upvotes]: I would like to find the mass of the unit sphere such that the density at any point is proportional to the distance from the surface of the sphere. I think spherical coordinates would probably be best to accomplish this. 
I know that $$ mass = \iiint_D\rho(x,y,z) dV $$ but I do not understand how to translate "proportional to the distance from the surface of the sphere" symbolically. I know that the bounds of the integral will be easy to write in spherical coordinates. I need help to understand how to find the density function. I am looking for help to set up the integrand. I can solve the integral myself. Any help would be greatly appreciated.

REPLY [11 votes]: 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:$$mass=\iiint_R\rho r^2\sin\phi dr d\phi d\theta$$
$$=k\int_0^{2\pi}\int_0^{\pi}\int_0^1(1-r)r^2\sin\phi dr d\phi d\theta$$
$$=\frac{k}{12}\int_0^{2\pi}\int_0^{\pi}\sin\phi d\phi d\theta$$
$$=\frac{k}{6}\int_0^{2\pi} d\theta$$
$$=\frac{k\pi}{3}$$