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Clearly, the R^2 we use is the R at which it is sitting.
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But what's the mass I should use?
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Well, for the mass I should use,
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I argue that if the total mass is that
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M / (π·R^2) is the mass per unit area.
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Then, I need the area of this shaded region.
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That's the tricky part.
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Have you got that? Yes?
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Right, so let's ask why it's 2·π·R·dR.
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If you take this annulus,
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and you take a pair of scissors and you cut it out,
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and you open it out,