I guess you're supposed to evaluate the integral,
[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S[/tex]
where [tex]\vec F(x,y,z)=\langle xz+x^2,-2xy,x^2+z^2\rangle[/tex] and [tex]S[/tex] is the hemisphere [tex]z=\sqrt{4-x^2-y^2}[/tex].
[tex]\vec F[/tex] has divergence
[tex]\mathrm{div}\vec F(x,y,z)=(z+2x)+(-2x)+(2z)=3z[/tex]
By the divergence theorem,
[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\iiint_V\mathrm{div}\vec F\,\mathrm dV[/tex]
where [tex]V[/tex] is the space with [tex]S[/tex] as its boundary.
Convert to spherical coordinates, taking
[tex]\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]
Then the integral is
[tex]\displaystyle\int_0^{\pi/2}\int_0^{2\pi}\int_0^2(3\rho\cos\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]
[tex]\displaystyle6\pi\int_0^{\pi/2}\int_0^2\rho^3\sin\varphi\cos\varphi\,\mathrm d\rho\,\mathrm d\varphi[/tex]
[tex]\displaystyle3\pi\int_0^{\pi/2}\int_0^2\rho^3\sin2\varphi\,\mathrm d\rho\,\mathrm d\varphi[/tex]
[tex]\displaystyle3\pi\left(\int_0^{\pi/2}\sin2\varphi\,\mathrm d\varphi\right)\left(\int_0^2\rho^3\,\mathrm d\rho\right)=\boxed{12\pi}[/tex]
