# Calculate surface area of a cone using spherical coordinate double integral?

I can understand that to calculate the surface area of the cone, one can write down the Cartesian equation $z^2=x^2+y^2$ and use double integral in Cartesian coordinate to calculate the surface area.

But my question is

How could I calculate the area using spherical or polar coordinates? (Namely the one ending with $drd\theta$)?

I think this would be more convenient than Cartesian coordinate.

Lets say the maximum radius of the cone(in spherical coordinates!) is $R$. If you dont have it then:
$$R=\sqrt{h^2+b^2}$$ Where $h$ is the cone height, and $b$ is the base radius.
To clarify the coordinates: $\varphi$ is the angle on the xy plane(azimuth, I believe?), $\theta$ the height angle, and $r$ the radius in spherical coordinates.
$$S_1 = \int_{\varphi = 0}^{2\pi}\int_{r=0}^{R} r\ \sin\theta\ d\varphi\ dr$$ Solving the integral we get: $$S_1 = 2\pi [{{r^2}\over {2}}]|^{R}_{0}\sin\theta = \pi R^2\sin\theta$$ Now we'll notice that geometrically: $\sin\theta = \frac bR$
So finally we get: $$S_1 = \pi bR$$ All that remains is adding the surface area of the base, which is a circle: $$S_2 = \pi b^2$$ And finally our result is: $$S = S_1 + S_2 = \pi b^2 + \pi b R = \pi b(b + R)$$