By Stokes' theorem,
[tex]\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S[/tex]
where [tex]\mathcal C[/tex] is the circular boundary of the hemisphere [tex]\mathcal M[/tex] in the [tex]y[/tex]-[tex]z[/tex] plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting
[tex]\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle[/tex]
where [tex]0\le t\le2\pi[/tex]. Then the line integral is
[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi[/tex]
We can check this result by evaluating the equivalent surface integral. We have
[tex]\nabla\times\mathbf f=\langle1,0,0\rangle[/tex]
and we can parameterize [tex]\mathcal M[/tex] by
[tex]\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle[/tex]
so that
[tex]\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv[/tex]
where [tex]0\le v\le\dfrac\pi2[/tex] and [tex]0\le u\le2\pi[/tex]. Then,
[tex]\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi[/tex]
as expected.
