Cube Gravity

With the cube centered at the origin and the point mass located at $\vec{r}_c=(2,0,0)$ , consider any general volume element $dV = dx \ dy \ dz$ around the point $\vec{r}_p=(x,y,z)$ within the cube. Applying Newton's gravitational law between the elementary mass and the point mass gives:

$d\vec{F} = G\left(\frac{M \ dV}{S^3}\right)m\left( \frac{\vec{r}_p - \vec{r}_c}{\lvert \vec{r}_p - \vec{r}_c\rvert^3}\right)$

Due to symmetry of the problem, we are only interested in the X component of the force which is (after simplifications):

$dF_x = d\vec{F} \cdot \hat{i} \implies F_x = \int_{V} dFx \implies F_x= \int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} \frac{(x-2) \ dx \ dy \ dz}{8((x-2)^2 + y^2 + z^2)^{3/2}} \approx -0.235749$

Comparing this result with that of a uniform sphere suggests that the cube (just like a disk, as in the previous problem) cannot be treated as a point mass like the sphere can be, in such situations.