Let S to be the closed surface of the cone given by x^2+y^2<=z^2, where 0<=z<=h
Let n be a unit normal to S, pointing outward from the cone.
I have to evaluate the surface integral (r dot product n dS) where r is the position vector.
I was thinking that I have to consider the curved part of the cone and its flat part.
Now to do the surface interval over the curved area I was intending to integrate using cylindrical polars. If I set r=(Rcosf, Rsinf, z) and ndS = partial derivative of r w.r.t f cross product partial derivative of r w.r.t z. Here is the point I am not sure. Since with cylindrical polars the cone is described using three parameters (R, f, z) so which two parameters do I use for findind ndS?
I may be in a totally wrong path, any help for this would be appreciated!!!
Or maybe integrate the curved surface using cartesian coordiantes by saying z=sqrt(x^2+y^2)?
And then I will just have to integrate over the circle x^2+y^2=h^2, since 0<=z<=h?