Deriving Moment of inertia for a hollow sphere

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SUMMARY

The moment of inertia for a hollow sphere can be derived using a double integral approach. The correct formula for the moment of inertia is (8πr^5)/9. The integration bounds utilized are from 0 to 4πr² for one integral and from 0 to r for the other, with the integrand being r² + z² dz dA. The initial attempt yielded (16πr^5)/3, which does not align with the expected result, indicating a need for careful evaluation of the integration process.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with the concept of moment of inertia
  • Knowledge of spherical coordinates
  • Basic proficiency in integrating functions involving multiple variables
NEXT STEPS
  • Study the derivation of moment of inertia for different geometric shapes
  • Learn about spherical coordinates and their applications in integration
  • Explore advanced integration techniques, particularly in three dimensions
  • Review examples of calculating moments of inertia using double and triple integrals
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Students studying physics or engineering, particularly those focusing on mechanics and dynamics, as well as educators teaching calculus and its applications in real-world scenarios.

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Homework Statement



How do you derive the moment of inertia for a hollow sphere?
I am not ending up with what i need to get which is: (8[tex]pi[/tex]r^5)/9

Homework Equations



I am not sure if the bounds are correct or if we need to factor something else in.

The Attempt at a Solution



A double integral of the area multiplied by the height.
for our integrals we do from 0 to 4[tex]pi[/tex]r^2 and the other integral is from 0 to r. The equation in the integral is r^2 + z^2 dzdA. (we get r^2 because x^2+y^2=r^2)

solving that out and get (16[tex]pi[/tex]r^5)/3.

plugging that in doesn't give the 2/3 we need for the moment of inertia.
 
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