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Show that the moment of inertia of a spherical shell of radius R and mass m is [tex](2mR^2)/3[/tex] This can be done by direct integration or, more easily, by finding the increase in the moment of inertia of a solid sphere when the radius changes. To do this, first show that the moment of inertia of a solid sphere of density [tex] \rho [/tex] is [tex]I = (8/15) \pi \rho R^5[/tex] Then compute the change dI in I for a change dR, and use the fact that the mass of this shell is[tex] dm = 4 \pi R^2 \rho dr[/tex].

Obviously the question gives you much of the answer, by shoving numbers together I got

[tex] I = \frac{2}{5}mR^2 *(4/3) \pi r^3 \rho[/tex]

[tex] I = \frac{8}{15}m \pi \rho r^5[/tex]

[tex]\frac{dI}{dR} = \frac{8}{3} m \pi \rho r^4[/tex]

and then basically dividing the first equation by the dI/dR gives you the right answer, but it makes no sense how that works or why you would do that. So I tried the integration route, because I thought that might work better

[tex]z = x^2 + y^2 [/tex]

[tex]z = r^2 [/tex]

[tex]I = mr^2 [/tex]

[tex]I = \frac{m}{r}\int_{-R}^{R}r^2 dr [/tex]

Where [tex]\frac{m}{r}[\tex] is the density. This also gives the right answer, but I have a feeling I'm just seeing answer on the page and thinking up a way to get to the answer that really has nothing to do with being right, if somebody could help me conceptualize what I'm doing in both methods and whether either of these are somewhat right, I would really appreciate it.

P.S. sorry about the bad looking coding, I'm still getting used to tex