How to Stoke's Theorem?

It's not necessary to use Stokes theorem. The integral can easily be calculated directly.

First we need to parameterise C. The easiest way to do this is to view C as a graph z=f(x,y), where f(x,y)=6-x, over the circle x^2+y^2=81 in the (x,y)-plane (i.e. the plane z=0). So we can parameterise C by

r(t)=(9 cos t, 9 sin t, 6 - 9 cos t), t \in [0,2\pi]

And now use the rule

∫F · dr (over C)

= ∫F(r(t)) · dr(t)/dt dt over the parameter interval for t.

Hence

∫F · dr

= ∫(81cos t sin t, 18-27cos t, 45sin t).(-9sin t, 9cos t, 9sin t) dt

= ∫ (162 cos t -243 cos^2 t +405sin^2 t-729 cost sin^2 t )dt

= ∫ (162 cos t -243/2(1+cos 2t) +405/2(1-cos 2t) - 243 d/dt(sin^3 t) )dt

=(-243/2+405/2)pi

=162pi

(the integral is over (0,2pi) and we use the fact that integrals of cos nt over (0,2pi), where n=1,2,3,... are zero).