Find the mass of the unit sphere

Question

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.

Answer 1

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

Answer 2

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

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

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

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

Answer 3

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

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