Q:

Force the moon exerts on the earth

Determine the force that the moon exerts on the earth. Note that since the 
diameter of the earth, 12,742 km, is not insignificant compared to the distance to 
the moon, 384,400 km, the gravitational field of the moon is not uniform over the 
volume of the earth. Therefore, to get an accurate result you will need to integrate 
the gravitational field ($\vec{g} = -GM_{moon}\frac{\hat{R}}{R^{2}}$) times the density of the earth over the 
volume of the earth (you may assume the earth is spherical and of uniform 
density). It may be expedient to provide arguments why certain components of the 
force can be ignored during the integration. 
Compare you result to the approximate solution obtained by assuming that both 
the moon and earth are point masses. 
--
My thoughts: I'm assuming the only component of the force that matters is the force directed radially towards the moon (if that makes any sense). 

A:

Indeed, by symmetry the only component is directed along the line that joins the centers of moon and earth. The force can be most easily found using the virtual work method: Compute the potential energy of the earth in the gravitational field of the moon. Let $d$ be the distance between their centers, and ${\bf r}$ a vector from the center of the earth to a point inside the eart, let $\theta$ be the angle bewteen the line joining the centers and ${\bf r}$, and $\rho$ the density if the earth. Then the potential energy is
$$
U = - \int_{Earth} \frac{GM_{moon} \rho}{\sqrt{d^2 +r^2 -2rd \cos \theta}} r^2 dr d\Omega
$$
Now expand in Legendre polynomials to obtain
$$
U = - \int_{0}^{R}\int_{S} GM_{moon}{ \rho} (1/d) \sum_{l=0}^{\infty}\left(\frac{r}{d}\right)^l P_l (\cos\theta) r^2  d\Omega dr
$$
$d\Omega$ is the differential solid angle, $S$ is the unit sphere and $R$ the radius of the earth.
In this integral only the term $l=0$ contribtes due to the orthogonality of the Legendre polynomials, the integration over $S$ gives $4\pi$. Overall you obtain 
$U= -GM_{moon}\frac{4\pi R^3}{3}\frac{\rho}{d}  = -GM_{moon}M_{earth}/d$
The force is now given by a variation of $d$, i.e. $F= -\frac{\partial U}{\partial d}$ which gives
$
F = -\frac{GM_{moon}M_{earth}}{d^2}
$
This is the same result as if they were point masses, therefore the answer doesn't change, one can see explicitely from the calculation above that it will not change even when $\rho$ depends on $r$ (to see it doesnt change for any $\rho({\bf r})$) we would need to change our origin of ${\bf r}$ to the center of mass of the earth.

