Answer :
If you're familiar with surface integrals, start by parameterizing the surface by the vector-valued function,
r(u, v) = 3 cos(u) sin(v) i + 3 sin(u) sin(v) j + 3 cos(v) k
with 0 ≤ u ≤ 2π and 0 ≤ v ≤ arccos(1/√8).
Then the area of the surface (I denote it by S) is
[tex]\displaystyle\iint_S\mathrm dA = \int_0^{2\pi}\int_0^{\arccos\left(1/\sqrt8\right)}\left\|\dfrac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm dv\,\mathrm du \\\\ = \int_0^{2\pi}\int_0^{\arccos\left(1/\sqrt8\right)}9\sin(v)\,\mathrm dv\,\mathrm du \\\\ =18\pi \int_0^{\arccos\left(1/\sqrt8\right)}\sin(v)\,\mathrm dv = \boxed{\frac{9(4-\sqrt2)\pi}2}[/tex]