By the divergence theorem, the surface integral over [tex]S_2[/tex] is
[tex]\displaystyle\iint_{S_2}\mathbf F\cdot\mathrm dS=\iiint_R\nabla\cdot\mathbf F\,\mathrm dR[/tex]
where [tex]R[/tex] denotes the space bounded by [tex]S_2[/tex]. Assuming the vector field is given to be
[tex]\mathbf F(x,y,z)=z^2x\,\mathbf i+(y^3+\tan z)\,\mathbf j+(x^2z+y^2)\,\mathbf k[/tex]
then
[tex]\nabla\cdot\mathbf F=\dfrac\partial{\partial x}[z^2x]+\dfrac\partial{\partial y}[y^3+\tan z]+\dfrac\partial{\partial z}[x^2z+y^2]=z^2+3y^2+x^2[/tex]
Converting to spherical coordinates, we take
[tex]\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}[/tex]
so that the triple integral becomes
[tex]\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=3}(\rho^2\cos^2\varphi+3\rho^2\sin^2\theta\sin^2\varphi+\rho^2\cos^2\theta\sin^2\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]
[tex]=\displaystyle\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4(\cos^2\varphi+2\sin^2\theta\sin^2\varphi+\sin^2\varphi)\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]
[tex]=162\pi[/tex]
Now the integral over [tex]S[/tex] alone will be the difference of the integral over [tex]S_2[/tex] and the integral over [tex]S_1[/tex], i.e.
[tex]\displaystyle\iint_{S_2}=\iint_{S_1\cup S}=\iint_{S_1}+\iint_S\implies\iint_S=\iint_{S_2}-\iint_{S_1}[/tex]
We can parameterize the points in [tex]S_1[/tex] by
[tex]\mathbf s(r,\theta)=\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=0\end{cases}[/tex]
so that the integral over [tex]S_1[/tex] is
[tex]\displaystyle\iint_{S_1}\mathbf F\cdot\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\mathbf F(x(r,\theta),y(r,\theta),z(r,\theta))\cdot\left(\dfrac{\partial\mathbf s}{\partial r}\times\dfrac{\partial\mathbf s}{\partial\theta}\right)\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle\int_0^{2\pi}\int_0^3(r^3\sin^3\theta\,\mathbf j+r^2\sin^2\theta\,\mathbf k)\cdot(r\,\mathbf k)\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle\int_0^{2\pi}\int_0^3r^3\sin^2\theta\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\dfrac{81\pi}4[/tex]
So, the integral over [tex]S[/tex] alone evaluates to
[tex]\displaystyle\iint_S=\iint_{S_2}-\iint_{S_1}=162\pi-\dfrac{81\pi}4=\dfrac{567\pi}4[/tex]
