Stokes' theorem allows for any choice of [tex]S[/tex] with [tex]C[/tex] as its boundary. The simplest region would be the disk [tex]x^2+y^2\le20[/tex] in the plane [tex]z=0[/tex], which can be parameterized using polar coordinates by
[tex]\vec s(u,v)=(u\cos v,u\sin v,0)[/tex]
with [tex]0\le u\le\sqrt{20}[/tex] and [tex]0\le v\le2\pi[/tex]. Take the normal vector to [tex]S[/tex] to be
[tex]\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(0,0,u)[/tex]
Then by Stokes' theorem,
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv[/tex]
where [tex]\nabla\times\vec F[/tex] denotes the curl of [tex]\vec F[/tex]. The curl is
[tex]\nabla\times\vec F=(1,-1,-6)[/tex]
so the line integral has a value of
[tex]\displaystyle\int_0^{2\pi}\int_0^{\sqrt{20}}(1,-1,-6)\cdot(0,0,u)\,\mathrm du\,\mathrm dv=-12\pi\int_0^{\sqrt{20}}u\,\mathrm du=\boxed{-120\pi}[/tex]
