Calculating Center of Mass in Tank with Muddy Suspension | Homework Help

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SUMMARY

The discussion focuses on calculating the center of mass for a tank shaped as a lower hemisphere with radius R, filled with a muddy suspension characterized by a density function ρ(x,y,z) = e^-h(x,y,z). The height function h(x,y,z) is determined to be z, where z is the vertical coordinate with z = 0 at the tank's bottom. Participants seek clarification on the limits of integration, particularly the phi limits, for solving the problem.

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  • Understanding of spherical coordinates and their application in integration.
  • Familiarity with density functions and their role in calculating center of mass.
  • Knowledge of multivariable calculus, specifically triple integrals.
  • Graphical interpretation of geometric shapes, particularly hemispheres.
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  • Study the application of spherical coordinates in triple integrals.
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  • Explore graphical methods for visualizing limits of integration in multivariable calculus.
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Homework Statement


Let Ω be a tank whose shape is that of the lower hemisphere of radius R. The tank with a muddy suspension whose density ρ is ρ(x,y,z):=e^-h(x,y,z), where h(x,y,z) is the height of (x,y,z) above the lowest point of the tank. Find the center of mass in the tank


Homework Equations





The Attempt at a Solution


First of all, how does one determine the height, h(x,y,z)? I guess it would be R but I am not able to give a reasoning to my guess. I would appreciate if someone could give me a graphical illustration on how to find the limits of integration for this problem as well (ignore this if it will cause too much hassle). Thanks
 
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If z is the vertical coordinate, and z =0 at the bottom of the tank, then h(x,y,z) = z
 
I got a feeling its z+R. Could anyone let me know what are the phi limits of integration?
 

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