Evaluate integral of $\langle xz^2,x^2y-z^3,2xy+y^2z\rangle$ with divergence theorem

Question

Q. Evaluate by Gauss divergence theorem $$\iint_S xz^2\,dy\,dz + \left(x^2y-z^3\right)\,dz\,dx + \left(2xy+y^2z\right)\,dx\,dy$$ where $S$ is the surface bounded by $z=0$ and $z=\sqrt{a^2-x^2-y^2}$.

I was not able to understand this question properly and when I solved this question I got the answer (2/3)π(a^5). Is this answer correct? In this question, I solve like this

∫∫∫ divF dv
div F =(x^2+y^2+z^2)
∫∫∫(x^2+y^2+z^2) dzdydx
a^2∫∫∫1 dzdydx from limit z=0 to z=(a^2-x^2-y^2)
               from limit y=-(a^2-x^2) to y=(a^2-x^2)        
                    limit x=-a to x=a

In this way, I got (2/3)π(a^5) as the answer.

Answer 1

The integral in your post is over a closed volume bounded by half a sphere of radius $a$. By divergence theorem, $$\begin{aligned} \oint_{S}\langle xz^2,x^2y-z^3,2xy+y^2z\rangle\cdot\mathrm d\mathbf S&=\iiint_V\text{div}\langle xz^2,x^2y-z^3,2xy+y^2z\rangle \mathrm dV\\ &=\iiint_V(z^2+ x^2+y^2)\mathrm dV\\ &=\iiint_V r^2dV\\ &=\int_0^{2\pi}\int_0^{\pi/2}\int_0^a r^2r^2\sin\theta\mathrm dr\mathrm d\theta\mathrm d\Phi\\ &=2\pi\int_0^{\pi/2}\int_0^a r^4\sin\theta\mathrm dr\mathrm d\theta\\ &=2\pi\int_0^{\pi/2}\sin\theta\mathrm d\theta\int_0^a r^4\mathrm dr\\ &=2\pi\left[-\cos\theta\right]_0^{\pi/2}\left[\frac{r^5}{5}\right]_0^a=\frac{2\pi}{5}a^5 \end{aligned} $$ Notice how the divergence is $x^2+y^2+z^2=r^2\neq a^2$ because you need to integrate over $r\in[0,a]$. If not what you'd be doing is integrating over a surface and not a volume, and thus, divergence theorem wouldn't hold.