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I was attempting to calculate the moment of inertia for a ball of radius R about the z axis when a mistake still led me to the correct answer and the seemingly correct method leads to the wrong answer.

Using the result of the moment of inertia for a solid hoop of radius R we have $I=\frac{1}{2}mR^2$ where $m$ is the mass of one solid hoop gives $dI=\frac{1}{2}r^2dm$, now for a sphere of radius $R$ and mass $M$ the density is $\rho=\frac{3M}{4\pi R^3}$ therefore $dm=\rho\cdot\pi r^2\, dz$ $$I=\frac{1}{2}\int_{-R}^{R} r^2(\rho\pi r^2)dz=\frac{\rho\pi}{2}\int_{-R}^{R}r^4\, dz$$ now if I choose to integrate this with by changing to the variable $\theta$ then $dz$ should become $Rd\theta$ then the integral becomes ($r=R\cos\theta$) $$I=\frac{\rho\pi}{2}\int_{-\pi /2}^{\pi /2}R^4\cos^4\theta Rd\theta=\rho\pi R^5\int_{0}^{\pi /2} \cos^4 \theta d\theta=\rho\pi R^5(\frac{3\pi}{16})=(\frac{3M}{4\pi R^3})*\pi R^5*\frac{3\pi}{16}=\frac{9\pi M R^2}{64}$$ which is clearly the wrong result. However if instead the substitution $dz=rd\theta$ is made $$I=\rho\pi\int_{0}^{\pi /2}r^5d\theta=\rho\pi R^5\int_0^{\pi /2}cos^5\theta d\theta=(\frac{3M}{4\pi R^3})\pi R^5(\frac{8}{15})=\frac{2MR^2}{5}$$ which is the correct result however I can't see why $dz$ would be equal to $rd\theta$ and not $Rd\theta$.

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  • $\begingroup$ Why not use spherical coordinates? It is much easier that way. $\endgroup$ – Henricus V. Mar 31 '16 at 2:05
  • $\begingroup$ I don't see how $dm = \pi r^2dz$. If we're using cylindrical coordinates then wouldn't $dm = (4\pi r z dr + 2\pi r^2 dz)\rho$ (assuming it's symmetrical, i.e. $z$ extends in both directions--maybe I don't understand your coordinate system)? And I agree with HenryW, it doesn't seem to make since to try and parameterize a sphere using cylindrical coordinates. $\endgroup$ – Jared Mar 31 '16 at 2:08
  • $\begingroup$ Also doesn't a "hoop" have $I = mr^2$? (If you're trying to derive it using a "hoop" rotating along the "wrong" axis then you seem to be trying to do this is almost the most possibly difficult way). $\endgroup$ – Jared Mar 31 '16 at 2:13
  • $\begingroup$ @"Henry W." I've already completed the integral in spherical coordinates but was curious as to this result. Here's a link to a site where they are using the same coordinate system just instead of switching to theta they rewrite r in terms of z. miniphysics.com/… And sorry I'm not sure why I wrote cylindrical coordinates up there, also I mentioned it is a solid hoop, perhaps that's also the wrong wording. $\endgroup$ – Craig Mar 31 '16 at 2:18

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