Electric field of a spherical shell I hope this question is appropriate for this site, if not, just leave a comment and I will delete. 
I am interested in knowing how to derive the electric field due to a spherical shell by Coulomb's law without using double integrals or Gauss Law.
Relevant equations are -- Coulomb's law for electric field and the volume of a sphere: 

$\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\hat{r}$, where $Q =$
  charge, $r=$ distance. 
$V = \frac{4}{3}\pi r^3.$

From my book, I know that the spherical shell can be considered as a collection of rings piled one above the other but with each pile of rings the radius gets smaller and smaller. 
I am not interested in the final formula, just the derivation of it. Thank you! 
 A: Well you can first calculate the field of a ring centered at $z=z_0$ on the $z$ axis with radius $r$ (using CGS, multiply by ugly factors later). By symmetry, on the $z$ axis the field is only in the $z$ direction and can be shown to be:
$$E_z(z)=\frac{q(z-z_0)}{((z-z_0)^2+r^2)^{3/2}}$$
Now each ring has charge $q=Q\cos \theta d\theta$, and $z_0 = R\cos \theta$. This means you can integrate the expression $E_z(z)$ over $\theta$ to get the field at any point on the $z$ axis. By symmetry, you can choose the ring direction as you wish, so that this expression is true for points not on the $z$ axis as well, with $r$ replacing $z$. 
As I mentioned in the comments, since the field of each ring contains an integral, this is really a double integral, even if you decide to call this "two single integrals".
A: We assume that the sphere of radius $R$ centered at 0. Let us assume an observation
point $o$ above the north pole of the sphere (by symmetry this should provide a
good answer). We consider rings from the bottom up to the north polo $(0,0,R)$.
The ring at a high $z$, $-R \le z \le R$ has a radius
$\rho=\sqrt{R^2 - z^2}$. We prefer to see the problem as a function
of the polar angle from $-\pi/2$ to $\pi/2$. We have that $\rho=R \sin \theta$ with $\theta$
the polar angle.  It is well known that for a ring with uniform charge density
$\sigma$ , radius $r$ and an observation point in the axis of the ring at a
distance $d$ from the center (in the direction of the axis of the ring) produces the field
\begin{eqnarray*}
  E(d,\rho)= \frac{\sigma  \rho \, d}{2 \epsilon_0(\rho^2 + d^2)^{3/2}}.
\end{eqnarray*}
The distance between the observation point $o$ and the ring at  $z$
height is $d=o-z$, and $z=R \cos \theta$, then we find
\begin{eqnarray*}
  E(\theta)= 
  \frac{\sigma (o-R \cos \theta) 
    R \sin \theta}{2 \epsilon_0(R^2 \sin^2 \theta + (o-R \cos \theta)^2)^{3/2}}.
\end{eqnarray*}
We need to integrate along $\theta$ between $0$ and $\pi$.
Along the polar axis the element of integration is $d \ell = R \, d \theta$,
so we will need to multiply by $R \, d \theta$.
\begin{eqnarray*}
  E = \frac{\sigma}{2 \epsilon_0} \int_{-\pi/2}^{\pi/2} 
  \frac{(o-R \cos \theta) 
    R^2 \sin \theta}{(R^2 \sin^2 \theta + (o-R \cos \theta)^2)^{3/2}}
    d \theta.
\end{eqnarray*}
Let us perform the following substitution
\begin{eqnarray*}
  u= \cos \theta \quad , \quad du=- \sin \theta d \theta \\
  \theta=0 \implies u = 1\\
  \theta=\pi \implies u = -1 , 
\end{eqnarray*}
then
\begin{eqnarray*}
  E = \frac{\sigma R^2}{2 \epsilon_0} \int_{-1}^1
  \frac{o - u R}{(R^2 + o^2 - 2 o R u)^{3/2}} du
\end{eqnarray*}
We split the integrand in two fractions (forget the coefficient for now).
\begin{eqnarray*}
  \int_{-1}^1 \frac{o} {(R^2 + o^2 - 2 o R u)^{3/2}} du
  \quad  \mathrm{and} \quad
  -\int_{-1}^1 \frac{u R} {(R^2 + o^2 - 2 o R u)^{3/2}} du
\end{eqnarray*}
For the first integral, let us make $x=R^2 + o^2 - 2 o R u$,
then $dx=-2 o R du$, and  in terms of $x$,
\begin{eqnarray*}
  -\frac{1}{2 R} \int \frac{dx}{x^{3/2}} =  
  \frac{1}{ R \sqrt{x} },
\end{eqnarray*}
Then the first integral is
\begin{eqnarray*}
  \int_{-1}^1 \frac{o} {(R^2 + o^2 - 2 o R u)^{3/2}} du &=&
  \left . \frac{1}{R \sqrt{R^2 + o^2 - 2 o R u}} \right |_0^1 \\
\end{eqnarray*}
Let us do the second integral usig integration by parts.
We write
\begin{eqnarray*}
  -\int_{-1}^1 \frac{u R} {(R^2 + o^2 - 2 o R u)^{3/2}} du &=&
    -\frac{u }{o  \sqrt{R^2 + o^2 - 2 o R u}} \\
    && + \int
  \frac{1}{o  \sqrt{R^2 + o^2 - 2 o R u}} du 
\end{eqnarray*}
Now,
\begin{eqnarray*}
\int \frac{1}{o  \sqrt{R^2 + o^2 - 2 o R u}} du =
-\frac{1}{o^2 R} \sqrt{R^2 + o^2 - 2 o R u},
\end{eqnarray*}
then
\begin{eqnarray*}
  \int_{-1}^1 \frac{u R} {(R^2 + o^2 - 2 o R u)^{3/2}} du &=&
    \frac{u }{o  \sqrt{R^2 + o^2 - 2 o R u}} 
    + \frac{ \sqrt{R^2 + o^2 - 2 o R u}}{o^2 R} \\
    &=& \frac{ R^2 + o^2 -  o R u}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}}
\end{eqnarray*}
Putting the first and the second integrals back together we get
\begin{eqnarray*}
  \frac{1}{R \sqrt{R^2 + o^2 - 2 o R u}} 
  - \frac{ R^2 + o^2 -  o R u}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}}
  =
  \frac{-R^2 + o Ru}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}}
\end{eqnarray*}
Hence we found that
\begin{eqnarray*}
  \int \frac{o - u R}{(R^2 + o^2 - 2 o R u)^{3/2}} du
  = \frac{o u - R}{o^2  \sqrt{o^2 - 2 o R u + R^2}},
\end{eqnarray*}
and so
\begin{eqnarray*}
  \left .
  \frac{o u - R}{o^2  \sqrt{o^2 - 2 o R u + R^2}} \right |_{-1}^1
  &=&  \frac{o  - R}{o^2  \sqrt{o^2 - 2 o R  + R^2}} 
  + \frac{o + R}{o^2  \sqrt{o^2 + 2 o R  + R^2}} \\
  &=&
  \frac{o  - R}{o^2  |o - R|}
  + \frac{o + R}{o^2 |o + R|}
\end{eqnarray*}
\begin{eqnarray*}
E= \frac{\sigma R^2}{2 \epsilon_0} 
\left [
  \frac{o  - R}{o^2  |o - R|}
  + \frac{o + R}{o^2 |o + R|}
\right ].
\end{eqnarray*}
That is
\begin{eqnarray*}
E = \left \{
  \begin{array}{cc}
    \frac{\sigma R^2}{o^2 \epsilon_0} &  o > R \\
    \\
      0 &  o < R 
\end{array}
  \right .
\end{eqnarray*}
but
\begin{eqnarray*}
    \frac{\sigma R^2}{o^2 \epsilon_0} 
    = \frac{4 \pi \sigma R^2}{4 \pi o^2 \epsilon_0} 
    = \frac{Q}{4 \pi o^2 \epsilon_0} 
\end{eqnarray*}
where $4 \pi R^2 \sigma$ is the total charge in the sphere.
Then
\begin{eqnarray*}
E = \left \{
  \begin{array}{cc}
  \frac{Q}{4 \pi o^2 \epsilon_0} &  o > R \\
    \\
      0 &  o < R 
\end{array}
  \right .
\end{eqnarray*}
What if $o=R$?
A: You can derive the electric field without using double integrals explicitely, using Gauss law:
$$ \Phi = \epsilon_0 Q $$
Where $\Phi$ is the flow of the electric field across the Gaussian surface. By symmetry you can choose a sphere of radius $R$ bigger than the radius of the charged sphere and the field will be normal and constant on all the surface, so $\Phi = 4\pi R^2 E$, from here you find 
$${\bf E} = \frac{Q}{4\pi \epsilon_0 r^2} \hat{\bf r} $$
If one insists in dividing the shpere in rings I see no way to avoid integration.
