Close off the hemisphere [tex]S[/tex] by attaching to it the disk [tex]D[/tex] of radius 3 centered at the origin in the plane [tex]z=0[/tex]. By the divergence theorem, we have
[tex]\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV[/tex]
where [tex]R[/tex] is the interior of the joined surfaces [tex]S\cup D[/tex].
Compute the divergence of [tex]\vec F[/tex]:
[tex]\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2[/tex]
Compute the integral of the divergence over [tex]R[/tex]. Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:
[tex]\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}[/tex]
So the volume integral is
[tex]\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5[/tex]
From this we need to subtract the contribution of
[tex]\displaystyle\iint_D\vec F(x,y,z)\cdot\mathrm d\vec S[/tex]
that is, the integral of [tex]\vec F[/tex] over the disk, oriented downward. Since [tex]z=0[/tex] in [tex]D[/tex], we have
[tex]\vec F(x,y,0)=\dfrac{y^3}3\,\vec\jmath+y^2\,\vec k[/tex]
Parameterize [tex]D[/tex] by
[tex]\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath[/tex]
where [tex]0\le u\le 3[/tex] and [tex]0\le v\le2\pi[/tex]. Take the normal vector to be
[tex]\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=-u\,\vec k[/tex]
Then taking the dot product of [tex]\vec F[/tex] with the normal vector gives
[tex]\vec F(x(u,v),y(u,v),0)\cdot(-u\,\vec k)=-y(u,v)^2u=-u^3\sin^2v[/tex]
So the contribution of integrating [tex]\vec F[/tex] over [tex]D[/tex] is
[tex]\displaystyle\int_0^{2\pi}\int_0^3-u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac{81\pi}4[/tex]
and the value of the integral we want is
(integral of divergence of F) - (integral over D) = integral over S
==> Â 486Ï€/5 - (-81Ï€/4) = 2349Ï€/20
