Let $S$ be the surface of the cone $z=\sqrt{x^2+y^2}$ bounded by the planes $z=0$ and $z=3$. Further, let $C$ be the closed curve forming the boundary of the surface $S$. A vector field $\vec{F}$ is such that it's curl is given by $\vec{T}=\langle -x,-y,0\rangle$. Then calculate the absolute value of the line integral $\int_{C}\vec{F}.\vec{dr}$
My problem here is, while applying Stokes theorem, if I choose my surface to be $z=3$ then my answer is $0$ and if I take my surface to be the cone, I am getting $18 \pi$.
Both choice of surface seem reasonable to be since both are bounded by $C$.
In order to apply Stokes theorem, you just need the surface to be piecewise continuous, not continuous,So a point at the tip of a cone is okay. It is wrong to assume otherwise.
The problem is not the cone but the curl. What you have as curl is not a curl. This can be verified by taking the divergence of [-x, -y, 0]
Divergence of a curl should be zero, ie,
$\nabla . (\nabla *(f)) = 0$
In our case it is not. Try it with a Field vector F (rather than a curl) and derive its curl manually and verify, it should match irrespective of the surface you choose