The vector field
[tex]\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k[/tex]
has curl
[tex]\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath[/tex]
Parameterize [tex]S[/tex] by
[tex]\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k[/tex]
where
[tex]\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}[/tex]
with [tex]0\le u\le3[/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}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k[/tex]
Then by Stokes' theorem we have
[tex]\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S[/tex]
[tex]=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_0^{2\pi}\int_0^3u^3(2u\cos^3v\sin(u^2-9)+\cos^3v\sin v+2u\sin^3v+\cos v\sin^3v)\,\mathrm du\,\mathrm dv[/tex]
which has a value of 0, since each component integral is 0:
[tex]\displaystyle\int_0^{2\pi}\cos^3v\,\mathrm dv=0[/tex]
[tex]\displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0[/tex]
[tex]\displaystyle\int_0^{2\pi}\sin^3v\,\mathrm dv=0[/tex]
[tex]\displaystyle\int_0^{2\pi}\cos v\sin^3v\,\mathrm dv=0[/tex]
