Question

Find the center of mass of the region of density rho(x,y,z)=1/(100-x^2-y^2) bounded by the paraboloid z=100-x^2-y^2 and the xy-plane.

I have already figured out that I need to use the center of mass problems and have attempted to switch to cylindrical coordinates by replacing x^2+y^2 with r^2, making the limits 0<=z<=100-x^2-y^2, 0<=r<=10, 0<=theta<=2pi. Please help, I keep getting ridiculously high numbers for the z value (11250pi or 20000pi) and have not even gotten the x and y values yet.

Answer 1

There is a possibility that the rho(x,y,z) was not actually a rho sign, rather a density sign, I am unsure.

Answer 2

\[\int\limits_{0}^{2\pi}\int\limits_{0}^{10}\int\limits_{0}^{100-x^2-y^2}z/(100-x^2-y^2) dz dy dx\]
has been converted to \[\int\limits_{0}^{2\pi}\int\limits_{0}^{10}\int\limits_{0}^{100-r^2}z/(100-r^2) dz dr d \theta\]

Answer 3

so does (0,0,1125/2) sound correct?

Answer 4

wait, i think it is (0,0,225/2)