To spell it out:
Beauty knows limiting frequency (which, when known, is equal to the probability) of the coin flips that she sees right in front of her will be equal to one-half. That is, if you repeat the experiment many times (plus a little noise to determine coin flips), then you get equal numbers of the event “Beauty sees a fair coin flip and it lands Heads” and “Beauty sees a fair coin flip and it lands Tails.” Therefore Beauty assigns 50⁄50 odds to any coin flips she actually gets to see.
You can make an analogous argument from symmetry of information rather than limiting frequency, but it’s less accessible and I don’t expect people to think of it on their own. Basically, the only reason to assign thirder probabilities is if you’re treating states of the world given your information as the basic mutually-exclusive-and-exhaustive building block of probability assignment. And the states look like Mon+Heads, Mon+Tails, and Tues+Tails. If you eliminate one of the possibilities, then the remaining two are symmetrical.
If it seems paradoxical that, upon waking up, she thinks the Monday coin is more likely to have landed tails, just remember that half of the time that coin landed tails, it’s Tuesday and she never gets to see the Monday coin being flipped—as soon as she actually expects to see it flipped, that’s a new piece of information that causes her to update her probabilities.
Cool insight. We’ll just pretend constant density of 3M/4r^3.
This kind of integral shows up all the time in E and M, so I’ll give it a shot to keep in practice.
You simplify it by using the law of cosines, to turn the vector subtraction 1/|r-r’|^2 into 1/(|r|^2+|r’|^2+2|r||r’|cos(θ)). And this looks like you still have to worry about integrating two things, but actually you can just call r’ due north during the integral over r without loss of generality.
So now we need to integrate 1/(r^2+|r’|^2+2r|r’|cos(θ)) r^2 sin(θ) dr dφ dθ. First take your free 2π from φ. Cosine is the derivative of sine, so substitution makes it obvious that the θ integral gives you a log of cosine. So now we integrate 2πr (ln(r^2+|r’|^2+2r|r’|) - ln(r^2+|r’|^2-2r|r’|)) / 2|r’| dr from 0 to R. Which mathematica says is some nasty inverse-tangent-containing thing.
Okay, maybe I don’t actually want to do this integral that much :P