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?
Best Answer
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.