You can parameterize [tex]S[/tex] using spherical coordinates by
[tex]\vec s(u,v)=\langle6\cos u\sin v,6\sin u\sin v,6\cos v\rangle[/tex]
with [tex]0\le u\le2\pi[/tex] and [tex]0\le v\le\frac\pi2[/tex].
Take the normal vector to [tex]S[/tex] to be
[tex]\dfrac{\partial\vec s}{\partial\vec v}\times\dfrac{\partial\vec s}{\partial\vec u}=36\langle\cos u\sin^2v,\sin u\sin^2v,\cos v\sin v\rangle[/tex]
(I use [tex]\vec s_v\times\vec s_u[/tex] to avoid negative signs. The orientation of the normal vector doesn't matter for a scalar surface integral; you could just as easily use [tex]\vec s_u\times\vec s_v=-(\vec s_v\times\vec s_u)[/tex].)
Then
[tex]f(x,y,z)=f(6\cos u\sin v,6\sin u\sin v,6\cos v)=36\sin^2v[/tex]
and the integral of [tex]f[/tex] over [tex]S[/tex] is
[tex]\displaystyle\iint_Sf(x,y,z)\,\mathrm dS=\int_0^{\pi/2}\int_0^{2\pi}36\sin^2v\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_0^{\pi/2}\int_0^{2\pi}(36\sin^2v)(36\sin v)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle2592\pi\int_0^{\pi/2}\sin^3v\,\mathrm dv=\boxed{1728\pi}[/tex]
