[tex]\mathbf f(x,y,z)=\langle z,y,x\rangle\implies\nabla\cdot\mathbf f=\dfrac{\partial z}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial x}{\partial z}=0+1+0=1[/tex]
Converting to spherical coordinates, we have
[tex]\displaystyle\iiint_E\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\int_{\varphi=0}^{\varphi=\pi}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=6}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=288\pi[/tex]
On the other hand, we can parameterize the boundary of [tex]E[/tex] by
[tex]\mathbf s(u,v)=\langle6\cos u\sin v,6\sin u\sin v,6\cos v\rangle[/tex]
with [tex]0\le u\le2\pi[/tex] and [tex]0\le v\le\pi[/tex]. Now, consider the surface element
[tex]\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\dfrac{\mathbf s_v\times\mathbf s_u}{\|\mathbf s_v\times\mathbf s_u\|}\|\mathbf s_v\times\mathbf s_u\|\,\mathrm du\,\mathrm dv[/tex]
[tex]\mathrm d\mathbf S=\mathbf s_v\times\mathbf s_u\,\mathrm du\,\mathrm dv[/tex]
[tex]\mathrm d\mathbf S=36\langle\cos u\sin^2v,\sin u\sin^2v,\sin v\cos v\rangle\,\mathrm du\,\mathrm dv[/tex]
So we have the surface integral - which the divergence theorem says the above triple integral is equal to -
[tex]\displaystyle\iint_{\partial E}\mathbf f\cdot\mathrm d\mathbf S=36\int_{v=0}^{v=\pi}\int_{u=0}^{u=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle36\int_{v=0}^{v=\pi}\int_{u=0}^{u=2\pi}(12\cos u\cos v\sin^2v+6\sin^2u\sin^3v)\,\mathrm du\,\mathrm dv=288\pi[/tex]
as required.
