By the divergence theorem,
[tex]\displaystyle\iint_S\mathbf f\cdot\mathrm d\mathbf S=\iiint_R(\nabla\cdot\mathbf f)\,\mathrm dV[/tex]
where [tex]R[/tex] is the solid whose boundary is [tex]S[/tex]. We have
[tex]\nabla\cdot\mathbf f=\dfrac{\partial z}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial zx}{\partial z}=1+x[/tex]
so we set up the volume integral as
[tex]\displaystyle\iiint_R(\nabla\cdot\mathbf f)\,\mathrm dV=\int_{x=0}^{x=1/a}\int_{y=0}^{y=(1-ax)/b}\int_{z=0}^{z=(1-ax-by)/c}(1+x)\,\mathrm dz\,\mathrm dy\,\mathrm dx[/tex]
[tex]=\dfrac{4a+1}{24a^2bc}[/tex]
