How to calculate the surface area of a cone using cylindrical coordinates? In cylindrical coordinates, the infinitesimal surface area is $dA=sd\theta dz$. 
In order to find the surface area of the curved portion of a cone,with radius R and height h, I compute the integral:
$$A  = \int_{\theta=0}^{2\pi}\int_{z=0}^{h}dA = \int_{\theta=0}^{2\pi}\int_{z=0}^{h} sd\theta dz$$
Using the straight line equation which gives $s = \frac{R}{h}(h-z)$, I obtain $A = \pi R h$. 
This however does not give the literature solution, $ A = \pi R (R + \sqrt{R^2 + h^2})$. Have I gone wrong somewhere in my calculations?
 A: What you tried to calculate is the surface of the lateral area, not including the bottom of the cone. So you did 2 mistakes, you didn't take the $S_{circle} =πR^2 $ area of the circle in mind and also you made a mistake calculating the lateral area.
Now for the lateral surface there is an older question already so you can go check it out: Setting Up an Integral to Find A Cone's Surface Area
which gives you $S_{lateral} = πR \sqrt{ R^2 + h^2 } $ so the result is what you expect.
EDIT: the reason you are wrong is because the infinitesimal surface you used is that of a surface of constant radius (so you can use that in a cylinder for example). But in a cone the radius, the height and the azimuth all change.
A: Parameterize your surface.
$S = (r\cos\theta, r\sin\theta, r\frac {H}{R})$
Find $dS$ and $\|dS\|$
$dS = (\cos\theta, \sin\theta, \frac {H}{R})\times(-r\sin\theta, r\cos\theta, 0)\ dr\ d\theta\\
dS = (-\frac {H}{R}r\cos\theta, -\frac {H}{R}r\sin\theta, r)\\
\|dS\| = r\sqrt {\frac {H^2}{R^2} + 1}$
Giving us the integral.
$\int_0^{2\pi}\int_0^R r\sqrt {\frac {H^2}{R^2} + 1}\ dr\ d\theta$
And resolve the integral:
$2\pi (\frac 12R^2)(\sqrt {\frac {H^2}{R^2} + 1})\\
\pi R\sqrt{H^2 + R^2}$
