Electric field in space created by intersection of spheres of charge I am trying to calculate the electric field in space created by a body assembled by the intersection of 2 spheres.
The upper sphere, its center is at $$\frac{d}{2}\mathbf{\hat{z}}$$ with radius $R$ and the second is centered at $$-\frac{d}{2}\mathbf{\hat{z}}$$ with radius $R$.
The region of intersection is neutral, whilst the upper region is charged with constant density $\rho$ and the lower region is charged with density $-\rho$.
I calculated the volume of the region of intersection, $$V=\frac{\pi}{12}(16R^3 -4R^2d-d^3)$$, which makes sense because if $$d\rightarrow 0$$ (the spheres coincide) then $$V\approx \frac{\pi}{12}\cdot 16R^3=\frac{4\pi R^3}{3}$$ which is the volume of a single sphere.
Then I treated each sphere's charge as superposition of a whole sphere charged with $\rho$ (or $-\rho$ respectively) and the region of intersection with the opposite charge.
As expected, the total charge of the lower region is the opposite of the total charge of the upper region, i.e $$q_1 = \rho\cdot \frac{4}{3}\pi(R^2d+\frac{d^3}{4}), q_2 =-\rho\cdot \frac{4}{3}\pi(R^2d+\frac{d^3}{4})$$.
Then, again, I used superposition to calculate the field everywhere in space:


*

*Inside the region of intersection: I used the formula of a field created inside a charged sphere, and found out that $$\vec{E}_{in}(P)=\frac{kq_1(\vec{r}_p-\frac{d}{2}\mathbf{\hat{z}})}{R^3}-\frac{kq_1(\vec{r}_p+\frac{d}{2}\mathbf{\hat{z}})}{R^3}=-\frac{kq_1d}{R^3}\mathbf{\hat{z}}=\frac{-d\rho}{12\varepsilon_0}(\frac{d}{R}+\frac{d^3}{4R^3})\mathbf{\hat{z}}$$

*Inside the upper charged region: I used the same formula for a field inside a charged sphere, plus the fact that the field that a sphere inducts outside its region is like a point charge. After the calculation I got that $$\vec{E}_{in+}(P)=\frac{d\rho}{12\varepsilon_0}(\frac{d}{R}+\frac{d^3}{4R^3})(\vec{r}_P-\frac{d}{2}\mathbf{\hat{z}}-R^3\frac{\vec{r}_P+\frac{d}{2}\mathbf{\hat{z}}}{|\vec{r}_P+\frac{d}{2}\mathbf{\hat{z}}|^3})$$.

*Inside the lower charged region: Same reasons, I got: $$\vec{E}_{in-}(P)=\frac{d\rho}{12\varepsilon_0}(\frac{d}{R}+\frac{d^3}{4R^3})(\vec{r}_P+\frac{d}{2}\mathbf{\hat{z}}-R^3\frac{\vec{r}_P-\frac{d}{2}\mathbf{\hat{z}}}{|\vec{r}_P-\frac{d}{2}\mathbf{\hat{z}}|^3})$$

*Outside: I use the fact that both spheres act like a point charge, and got:$$\frac{\rho}{12\varepsilon_0}(R^2d+\frac{d^3}{4})(\frac{\vec{r}_P-\frac{d}{2}\mathbf{\hat{z}}}{|\vec{r}_P-\frac{d}{2}\mathbf{\hat{z}}|^3}-\frac{\vec{r}_P+\frac{d}{2}\mathbf{\hat{z}}}{|\vec{r}_P+\frac{d}{2}\mathbf{\hat{z}}|^3})$$
My question is, is this OK?
And also, I am suppose to write an expression for the field outside of the body at the limit that $R>>d$, but I can't seem to get the intuition for the result I am suppose to get.
 A: First, when you found your volume it looks like you missed a sign. In my last integration over $\phi$, I had 
$$ \frac{4\pi}{3}\int^{\phi}_{0}d\phi R^3\sin\phi-\frac{d^3\sin\phi}{8\cos^3\phi}\Rightarrow$$
$$ \frac{4\pi}{3}\left[ -R^3\cos\phi - \frac{d^3}{16\cos^2\phi}\right]_0^{\cos\phi=\frac{d}{2R}}.$$
If you switch the sign on the second term, it looks like out pops the answer you got. 
Secondly, you've actually made the problem very difficult for yourself by slightly misusing the superposition principle. If you superimpose two opposite charge densities on top of each other, the electric fields created by them naturally cancel and it creates a situation equivalent to there being no charge in the region. 
The problem in your derivation is that you mix up the superimposed states and the final state. For an example, look at the charge you calculate outside the two spheres. You use the formula for the electric field outside a perfectly uniform sphere, $\vec{E}=\frac{q}{\epsilon_0r^2}$. But then you take $q=\rho V$ and for $V$ plug in the volume of the charge in the final configuration with a chunk missing. However, the electric field of a sphere with a chunk missing is not simply $\vec{E}=\frac{q}{\epsilon_0r^2}$. With the chunk missing you've completely lost the spherical symmetry that allowed you to derive the simple field. 
To calculate the field outside of the two spheres you directly use the full charge on a spherical distribution $q=\frac{\ 4\pi R^3\rho}{3}$ instead of $q=\rho\frac{4\pi}{3}(R^2d + \frac{d^3}{4})$. The rest of your term looks fine. If $d$ is small enough and the two spheres do happen to overlap, then the superposition principle guarantees that wherever the spheres overlap, the electric field from those overlapping parts will completely cancel because they carry equal and opposite charge. It takes care of itself without you having to worry about manually taking out the overlapping charge by calculating volumes. 
The same principle holds for calculating the electric field inside of the spheres. It's actually a lot simpler than what you've written out.
