Take the normal pointing outwards from the surface. Use an appropriate integral theorem $$\iint_S \textbf{F}\cdot d\textbf{S} \space \space where \space \space \textbf{F} (x,y,z)=(x^3,3yz^2,3y^2z+10) $$ and $S$ is the surface $z=-\sqrt{4-x^2-y^2}$
My attempt
I have tried using Gauss' divergence theorem which gives me $$ \iiint_V 3x^2+3y^2+3z^2 d\textbf{V}$$ and I end up with the integral $$ \int_0^2 \int_0^{2\pi} \int_{\frac{\pi}{2}}^\pi 3 r ^{4} \sin(\phi) \space \space d\theta \space d\phi \space dr$$ The extra $r^2\sin(\theta)$ is the Jacobian in spherical coordinates. Now when I integrate the above I get the incorrect answer. The correct answer is $\frac{-8\pi}{5}$. Where am i going wrong?
In the Gauss theorem one needs closed surface, so you have to subtract from the volume integral $\displaystyle I_V=\iiint_V3(x^2+y^2+z^2)d\mathbf{V}=\frac{192\pi}{5}$ the integral $\displaystyle\iint_D \mathbf{F}\cdot d\mathbf{S}$, where $D$ is the disc $x^2+y^2\leq4$ in the $xy$-plane with normal vector $\mathbf{n}=(0,0,1)$. Now since in this case $\mathbf{F}\cdot\mathbf{n}=10$, the answer will be $$I_V-10\cdot \mathrm{area}(D)=\frac{192\pi}{5}-10\cdot 4\pi=-\frac{8\pi}{5}.$$