Center of Mass of a Solid Enclosed by a Spherical Coordinate Surface?

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SUMMARY

The discussion focuses on finding the center of mass of a solid with a uniform density of \(\delta = 1\) enclosed by the spherical coordinate surface defined by \(\rho = 1 - \cos\phi\). The correct approach involves setting up a triple integral in spherical coordinates, specifically \(\int_{0}^{2\pi}\int_{\pi/4}^{\pi}\int_{0}^{1-\cos\phi} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta\). Participants emphasize the importance of correctly determining the limits of integration for \(\phi\) and advise against converting to Cartesian coordinates, as this complicates the calculation. The integration should maintain spherical coordinates to accurately compute the center of mass.

PREREQUISITES
  • Understanding of spherical coordinates and their applications in calculus.
  • Familiarity with triple integrals and volume integrals.
  • Knowledge of density functions and their role in calculating mass.
  • Ability to compute moments in different coordinate systems.
NEXT STEPS
  • Study the derivation of limits of integration for spherical coordinates.
  • Learn how to compute the center of mass using spherical coordinates.
  • Explore the conversion between spherical and Cartesian coordinates for integrals.
  • Investigate examples of mass calculations for solids defined by various coordinate surfaces.
USEFUL FOR

Students and educators in physics and engineering, particularly those studying mechanics and calculus, will benefit from this discussion. It is especially relevant for anyone tackling problems involving center of mass calculations in spherical coordinates.

forestmine
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Homework Statement



Find the center of mass of a solid of density \delta = 1 enclosed by the spherical coordinate surface \rho = 1-cos\phi.

Homework Equations


The Attempt at a Solution



I'm a bit confused about how to start here, mainly because the surface is defined by spherical coordinates. But here's what I've done so far..

I figured I should start by finding the mass, so, setting up a triple integral over the density, and in this case, it would be a triple integral in spherical coordinates.

$$\int_{0}^{2\pi}\int_{\pi/4}^{\pi}\int_{0}^{1-cos\phi} \rho^2sin\phi d\rho d\phi dθ$$

I'm not sure about the phi limits of integration...the reason I'm thinking beginning at \pi/4 is really just an assumption made by looking at the image. I'm not sure how to mathematically reach that conclusion, however.

I understand how to calculate the center of mass in cartesian coordinates, but the whole spherical surface part is throwing me off. Can I just convert my above integral into a cartesian coordinate system and then just go about finding x-bar, y-bar, and z-bar?

Ah, thank you!
 
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forestmine said:

Homework Statement



Find the center of mass of a solid of density \delta = 1 enclosed by the spherical coordinate surface \rho = 1-cos\phi.

What are the ranges of the variables ##\theta,\phi\, ##?

I'm a bit confused about how to start here, mainly because the surface is defined by spherical coordinates. But here's what I've done so far..

I figured I should start by finding the mass, so, setting up a triple integral over the density, and in this case, it would be a triple integral in spherical coordinates.

$$\int_{0}^{2\pi}\int_{\pi/4}^{\pi}\int_{0}^{1-cos\phi} \rho^2sin\phi d\rho d\phi dθ$$

I'm not sure about the phi limits of integration...the reason I'm thinking beginning at \pi/4 is really just an assumption made by looking at the image. I'm not sure how to mathematically reach that conclusion, however.

Those limits don't define a closed surface for any kind of volume integral.

I understand how to calculate the center of mass in cartesian coordinates, but the whole spherical surface part is throwing me off. Can I just convert my above integral into a cartesian coordinate system and then just go about finding x-bar, y-bar, and z-bar?

No, that is the last thing you would want to do. Once you figure out the correct limits of the solid you will know the limits for the triple integral. Then if you want the ##x## moment, you just put the ##x## moment arm in the integrand like you would do in rectangular. It's just that you do it in spherical coordinates ##x=\rho\sin\phi\cos\theta##. Same idea for the other moments.
 

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