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.


Ill answer for spherical coordinates.
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.
The integral form will be:
$$S_1 = \int_{\varphi = 0}^{2\pi}\int_{r=0}^{R} r\ \sin\theta\ d\varphi\ dr$$ 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: $$ S = S_1 + S_2 = \pi b^2 + \pi b R = \pi b(b + R)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.