use the Divergence Theorem to calculate F Nd omega, where S is the boundary of the cylinder x^2+y^2<orequal to, 0<orequaltoZ<orequal to 3, with outward orientation and F(x,y,z)=<xy^2,yz^2,zx^2>.
The divergence theorem states that
$\displaystyle \iint_{\partial V}\vec F \cdot d\vec S = \iiint_V \nabla \vec F dV$
$\displaystyle \iiint_V (y^2+z^2+x^2)dV$
Now if we change to cylindrical coordinates we get
$\displaystyle \int_{0}^{3} \int_{0}^{2\pi} \int_{0}^{1}(r^2+z^2)rdr d\theta dz$
Since you didn't give the raduis of the cylindar I assumed it was 1.