Help in choosing surface for this problem on Stoke's theorem?

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 $$C$$.

• Integrating over the surface of the cone, there is a point which is not differentiable. Stokes theorem requires the surface to be smooth. Jan 11 '19 at 1:01
• Doug M, what point is that? origin? I am not understanding this correctly. Are you saying we can never use cone as a surface in Stokes theorem(when origin is involved)? Jan 11 '19 at 1:22
• Yes, at the origin, the cone has a point. A cone is not a smooth manifold. Jan 11 '19 at 1:31
• Dough M thank you so much Jan 11 '19 at 1:53
• @Abhay What do you think about my answer. Jun 7 '20 at 5:24

$$\nabla . (\nabla *(f)) = 0$$