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 $C$.

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    $\begingroup$ Integrating over the surface of the cone, there is a point which is not differentiable. Stokes theorem requires the surface to be smooth. $\endgroup$
    – Doug M
    Jan 11 '19 at 1:01
  • $\begingroup$ 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)? $\endgroup$
    – Abhay
    Jan 11 '19 at 1:22
  • $\begingroup$ Yes, at the origin, the cone has a point. A cone is not a smooth manifold. $\endgroup$
    – Doug M
    Jan 11 '19 at 1:31
  • $\begingroup$ Dough M thank you so much $\endgroup$
    – Abhay
    Jan 11 '19 at 1:53
  • $\begingroup$ @Abhay What do you think about my answer. $\endgroup$
    – Dom Jo
    Jun 7 '20 at 5:24

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


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