I am trying to prove Newton's Shell Theorem, and the natural integral I think of to use does not seem to give the right answer, so I am trying to find my mistake.
For simplicity, I will work in one dimension. Assume there is a uniform "sphere" of one dimension (ie, a segment) of mass $M$ and radius $R$ centered at $x = 0$, which exerts a gravitational force on a particle of mass $m$ located at $x = r$, where $r > R$.
According to Newton's Theorem, the gravitational force exerted by the sphere on the particle should be $G\dfrac{Mm}{r^2}$.
It seems to me that the force exerted on the particle by an infinitesimal slice of the sphere at location $x$ with width $dx$ should be $\dfrac{G(M\dfrac{dx}{2R})m}{(x – r)^2}$. Integrating over $[-R,R]$ should then give the total force. But
$$\int_{-R}^R \dfrac{GMm}{2R(x-r)^2}dx = \dfrac{GMm}{r^2-R^2}.$$
Not only is this different from $G\dfrac{Mm}{r^2}$, but it gives the absurd result that the force increases to infinity as $r$ approaches $R$. Can anyone explain the flaw in my setup?
[Note: In 3 dimensions my integral becomes
$$\int\int\int \dfrac{GMm}{\frac{4}{3}\pi R^3[(x-r)^2+y^2+z^2]}dV,$$
where the integration is over the sphere of radius $R$ centered at the origin and the particle is located at $(r,0,0)$. This too is incorrect, but the problem already arises in one dimension where the integral is far easier to evaluate.]
Best Answer
What you've proven is that the gravitational field along the axis of a uniform rod is in fact not proportional to $1/r^2$, and diverges as you approach the end of the rod. Both of these are true statements, so well done! But you seem confused about these answers, so I should probably elucidate a bit more.
The shell theorem (for spherically symmetric objects) can be viewed as a consequence of Gauss's Law for gravity. If you're familiar with the version from electrostatics, this works pretty much the same way: the flux integral of the gravitational acceleration field $\vec{g}$ over any surface is proportional to the amount of mass enclosed within that surface. In the case of a spherically symmetric mass distribution, one can draw an imaginary spherical surface surrounding it. By symmetry, $\vec{g}$ must be purely radial: $\vec{g} = - g(r) \hat{r}$. This means that we have $$ - 4 \pi r^2 g(r) = - 4 \pi G M $$ and so $g(r) = GM/r^2$, which is what you expect. But if you have a situation like a uniform rod of finite length, there's no way to do this: there's no analogue to the statement above that starts "By symmetry...". You simply can't get any further using Gauss's Law. Instead, what you find is that the gravitational acceleration is no longer uniform over the surface of a sphere; in fact, it can even diverge so long as its flux integral over the enclosing surface stays finite.
In fact, it can be shown (using infinitesimal Gaussian surfaces) that we should expect $\vec{g}$ to be discontinuous at points where there is an infinitesimally thin sheet of mass; it should diverge proportionally to $1/d$ near an infinitesimally thin rod, where $d$ is the distance to the rod; and it should diverge proportionally to $1/d^2$ near a point charge. The second case is, in fact, precisely what you found.