As near as I can tell, you're given the vector field
[tex]\vec F(x,y,z)=\langle y,-x,14\rangle[/tex]
and that [tex]S[/tex] is the part of the upper half of the sphere with equation
[tex]x^2+y^2+z^2=4[/tex]
with boundary [tex]C[/tex] the circle in the plane [tex]z=0[/tex].
Parameterize [tex]C[/tex] by
[tex]\vec r(t)=\langle2\cos t,2\sin t,0\rangle[/tex]
with [tex]0\le t\le2\pi[/tex]. Then the line integral of [tex]\vec F(x,y,z)[/tex] along [tex]C[/tex] is
[tex]\displaystyle\int_C\vec F(x,y,z)\cdot\mathrm d\vec r=\int_0^{2\pi}\langle2\sin t,-2\cos t,14\rangle\cdot\langle-2\sin t,2\cos t,0\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle-4\int_0^{2\pi}(\sin^2t+\cos^2t)\,\mathrm dt=\boxed{-8\pi}[/tex]
Parameterize [tex]S[/tex] by
[tex]\vec s(u,v)=\langle2\cos u\sin v,2\sin u\sin v,2\cos v\rangle[/tex]
with [tex]0\le u\le2\pi[/tex] and [tex]0\le v\le\dfrac\pi2[/tex]. We have
[tex]\nabla\times\vec F(x,y,z)=\langle0,0,-2\rangle[/tex]
Take the normal vector to [tex]S[/tex] to be
[tex]\vec s_v\times\vec s_u=\langle4\cos u\sin^2v,4\sin u\sin^2v,2\sin2v\rangle[/tex]
Then the surface integral of the curl of [tex]\vec F(x,y,z)[/tex] across [tex]S[/tex] is
[tex]\displaystyle\iint_S(\nabla\times\vec F(x,y,z))\cdot\mathrm d\vec S=\iint_S(\nabla\times\vec F(x(u,v),y(u,v),z(u,v)))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_0^{\pi/2}\int_0^{2\pi}\langle0,0,-2\rangle\cdot\langle4\cos u\sin^2v,4\sin u\sin^2v,2\sin2v\rangle\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle-4\int_0^{\pi/2}\int_0^{2\pi}\sin2v\,\mathrm du\,\mathrm dv=\boxed{-8\pi}[/tex]
