Triple Integral moment of inertia

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SUMMARY

The discussion focuses on setting up a triple integral for calculating the moment of inertia Iz for a specific region defined by the sphere \(x^2+y^2+z^2=4a^2\) and the cylinder \(x^2+y^2-2ax=0\). The integral is transformed into cylindrical coordinates, resulting in limits from 0 to \(\sqrt{4a^2-r^2}\) for z, 0 to \(2a\cos\theta\) for y, and 0 to \(2\pi\) for theta. A key point of contention is the limits for theta, where the answer suggests they should range from 0 to \(\pi\), indicating that theta represents the polar angle, which cannot exceed 180 degrees.

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  • Understanding of triple integrals in calculus
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  • Basic geometry of spheres and cylinders
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Punkyc7
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set up a triple integral for the moment of inertia Iz for the region inside the sphere
x^2+y^2+z^2=4a^2 and inside the cylinder
x^2+y^2-2ax=0

so I draw my picture and convert to cylindrical coord. and i get an integral from 0 to sqrt(4a^2-r^2)
an integral from 0 to 2acostheta and an integral from 0 to 2pi

then I multiply by 2 becuase its only the top part of the volume that i set up and the integral is integrating r^3


my question is the answer says that the theta limits and goes from 0 to pi , so why is that?
 
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Is theta the polar angle? If so, the polar angle can't go above 180 degrees. (If it did, it would wrap around to the other side of the sphere, and we would just use a different phi to denote that.)
 

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