Finding Mass and Center of Mass in a Solid Hemisphere

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SUMMARY

The discussion focuses on calculating the mass and center of mass of a solid hemisphere of radius 'a' with a density proportional to the distance from the center of the base. The mass is derived using spherical coordinates, resulting in the equation M = ∫∫∫ δ dV, leading to a mass of M = (a^4 π)/2. The center of mass is determined to be at (0, 0, 2a/5). The calculations confirm the correct application of spherical coordinates and integration techniques.

PREREQUISITES
  • Understanding of spherical coordinates in calculus
  • Familiarity with triple integrals and volume integrals
  • Knowledge of density functions and their implications in physics
  • Ability to compute center of mass using integrals
NEXT STEPS
  • Study the derivation of mass using triple integrals in spherical coordinates
  • Learn about density functions and their role in calculating mass
  • Explore applications of center of mass in physics and engineering
  • Investigate variations in density and their effects on mass calculations
USEFUL FOR

Students in physics and engineering, particularly those studying mechanics and calculus, will benefit from this discussion. It is especially relevant for anyone tackling problems involving mass distribution and center of mass in three-dimensional objects.

squeeky
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Homework Statement


Use Spherical Coordinates.
Let H be a solid hemisphere of radius a whose density at any point is proportional to its distance from the center of the base.
a) Find the mass of H.
b) Find the center of mass of H.


Homework Equations


M=\int\int_D\int\delta dV
M_{yz}\int\int_D\int x \delta dV;M_{xz}\int\int_D\int y \delta dV;M_{xy}\int\int_D\int z \delta dV
C.O.M.=(\bar{x},\bar{y},\bar{z})
\bar{x}=\frac{M_{yz}}{M};\bar{y}\frac{M_{xz}}{M};\bar{z}\frac{M_{xy}}{M}


The Attempt at a Solution


I think that if we place the hemisphere's center at (0,0,0), then the limit of theta is from 0 to 2pi, phi is from 0 to pi/2, and rho is from 0 to a, while the density is equal to rho. This gives me the equation:
M=\int^{2\pi}_0\int^{\pi/2}_0\int^a_0 \delta \rho^2 sin \phi d \rho d \phi d \theta
Solving this, I get a mass of \frac{a^4 \pi}{2}, M_{xy}=\frac{a^5\pi}{5}, Mxz=Myz=0. Then the center of mass is (0,0,2a/5).
Is this right?
 
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You are told only that the density is proportional to the distance from the center. How can you possibly get a specific number as the mass? What happened to the "proportionality"? Other than that, I think you are correct.
 

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