Denote the cylindrical surface by [tex]S[/tex], and its interior by [tex]R[/tex]. By the divergence theorem, the integral of [tex]\vec v[/tex] across [tex]S[/tex] (the outward flow of the fluid) is equal to the integral of the divergence of [tex]\vec v[/tex] over the space it contains, [tex]R[/tex]:
[tex]\displaystyle\iint_S\vec v\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec v)\,\mathrm dV[/tex]
The given velocity vector has divergence
[tex]\vec v=z\,\vec\imath+y^2\,\vec\jmath+x^2\,\vec k\implies\nabla\cdot\vec v=\dfrac{\partial(z)}{\partial x}+\dfrac{\partial(y^2)}{\partial y}+\dfrac{\partial(x^2)}{\partial z}=2y[/tex]
Then the total outward flow is
[tex]\displaystyle\iiint_R2y\,\mathrm dV[/tex]
Converting to cylindrical coordinates gives the integral
[tex]\displaystyle2\int_0^{2\pi}\int_0^3\int_0^4r^2\sin\theta\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\boxed{0}[/tex]
