Given the field
[tex]\vec F(x,y,z) = \langle2x,y,3z\rangle[/tex]
Compute its divergence.
[tex]\mathrm{div}\vec F(x,y,z)= \dfrac{\partial(2x)}{\partial x} + \dfrac{\partial y}{\partial y} + \dfrac{\partial(3z)}{\partial z} = 2 + 1 + 3 = 6[/tex]
By the divergence theorem, the flux of [tex]\vec F[/tex] across [tex]\vec S[/tex] is
[tex]\displaystyle \iint_S \vec F \cdot d\vec S = \iiint_{\mathrm{int}(S)} \mathrm{div}\vec F(x,y,z)\,dV = 6 \iiint_{\mathrm{int}(S)} dV[/tex]
which is simply 6 times the volume of [tex]S[/tex]. (By [tex]\mathrm{int}(S)[/tex], I mean the "interior of [tex]S[/tex]".)
[tex]S[/tex] has radius √6, so its volume is 4/3 π (√6)³ = 8√6 π, and the flux is
[tex]\displaystyle \iint_S \vec F\cdot d\vec S = 6 \cdot 8\sqrt6\,\pi = \boxed{48\sqrt6\,\pi}[/tex]
