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 $z=\sqrt{x^2+y^2}$.

  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ – José Carlos Santos Apr 25 '18 at 11:23
  • $\begingroup$ okay, thank you and much appreciation for the advice! I am new to MSE and still learning how it all works. Is there a way to upload a photo of my work? I tried to do so originally and it wasn't working for me. $\endgroup$ – Emily J Bapst Apr 25 '18 at 12:38
  • $\begingroup$ Yes, you can upload images, but it's much better if you type what you did. $\endgroup$ – José Carlos Santos Apr 25 '18 at 12:39

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

  • $\begingroup$ That actually makes a lot of sense about the symmetry, thank you for pointing that out! $\endgroup$ – Emily J Bapst Apr 25 '18 at 12:56

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