anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
There is a possibility that the rho(x,y,z) was not actually a rho sign, rather a density sign, I am unsure.
anonymous
  • anonymous
\[\int\limits_{0}^{2\pi}\int\limits_{0}^{10}\int\limits_{0}^{100-x^2-y^2}z/(100-x^2-y^2) dz dy dx\] has been converted to \[\int\limits_{0}^{2\pi}\int\limits_{0}^{10}\int\limits_{0}^{100-r^2}z/(100-r^2) dz dr d \theta\]
nikvist
  • nikvist
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anonymous
  • anonymous
so does (0,0,1125/2) sound correct?
anonymous
  • anonymous
wait, i think it is (0,0,225/2)

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