Find the center of mass in 3D

Question

Vector Calculus:

[Using Integration] Find the center of mass of the "snow cone" of uniform density bounded above by the sphere $x^2+y^2+z^2=a^2$ and below by the cone $z=\sqrt{x^2+y^2}$.

Answer 1

Note that your snow cone is symmetric with respect to the $z$ axis, thus all you need is $\bar z$

The formula is $$\bar z = \frac { \int zdv }{\int dv}$$

The spherical coordinates seems the natural choice.

$$ \int zdv = \int _0 ^{2\pi} \int _0^{\pi /4}\int _0^a (\rho \cos \phi )\rho ^2 \sin \phi d\rho d\phi d \theta $$

And $$ \int dv = \int _0 ^{2\pi} \int _0^{\pi /4}\int _0^a \rho ^2 \sin \phi d\rho d\phi d \theta $$