Electric field of cone I can easily do it with triple integration, but my brother asked me to do it with the help of single integration, I cant work it out. Can you please help me? Thanks
 A: With single integration, it's doable for points on the axis of the cone. 
Using symmetry, we show that the electric field is directed along the axis of the cone. We can start from a formula for the electric field of a charged ring, subdivide the cone into "very thin" rings and integrate. 
We are given the vertex angle $2\theta$, slant height $L$, surface charge density $\sigma$, and Coulomb's constant (electrostatic constant) $k$.
Let's place the origin of $(x,y)$ coordinates at the apex, so that the cone is symmetric about the $x$ axis.  We compute the electric field $E$ at a point on the cone axis,  at the distance $b$ from the apex.  We already know $dE$ (the  contribution to the field from a thin ring):
$$
dE = {k(x+b)dQ \over ((x + b)^2 + (x \tan\theta)^2)^{3/2}}
$$
where the charge $dQ$ of the ring is
$$
dQ = \sigma\cdot dA = {2\pi\sigma y\, dx\over\cos\theta}
={2\pi\sigma (x\tan\theta)\, dx\over\cos\theta}.
$$
Combining the above we find
$$
E = \int dE= {2\pi\sigma k \sin\theta\over\cos^2\theta}
\int_0^{L\cos\theta}{(x + b) x\over((x + b)^2 + (x \tan\theta)^2)^{3/2}} dx
$$
Here is a WolframAlpha command to evaluate the indefinite integral:
integrate (x+b)x/((x+b)^2+(xt)^2)^(3/2) dx

It is not pretty, but it's single integration nonetheless. (We denoted $t=\tan\theta$.)
