Parameterize the surface [tex]S[/tex] by
[tex]\mathbf r(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+u^2\,\mathbf k[/tex]
where we use the polar coordinate transformation [tex]x(u,v)=u\cos v[/tex] and [tex]y(u,v)=u\sin v[/tex], from which it follows that [tex]x(u,v)^2+y(u,v)^2=r^2=z(u,v)[/tex]. For this region, we have [tex]0\le u\le1[/tex] and [tex]0\le v\le2\pi[/tex].
The surface integral is then
[tex]\displaystyle\iint_S\mathbf F\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf F(\mathbf r(u,v))\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dv\,\mathrm du[/tex]
where [tex]\mathbf r_u[/tex] and [tex]\mathbf r_v[/tex] are the respective partial derivatives of [tex]\mathbf r(u,v)[/tex]. We have
[tex]\mathbf F(\mathbf r(u,v))=u\cos v\,\mathbf i+u\sin v\,\mathbf j+u(\cos v+\sin v)\,\mathbf k[/tex]
[tex]\mathbf r_u=\cos v\,\mathbf i+\sin v\,\mathbf j+2u\,\mathbf k[/tex]
[tex]\mathbf r_v=-u\sin v\,\mathbf i+u\cos v\,\mathbf j[/tex]
[tex]\implies\mathbf r_u\times\mathbf r_v=-2u^2\cos v\,\mathbf i-2u^2\sin v\,\mathbf j+u\,\mathbf k[/tex]
Now, presumably you know what a paraboloid looks like. An outward-pointing normal vector to the surface would probably refer to any normal that points downward, so check what happens to the cross product above at, let's say, the point [tex](u,v)=(0.1,0)[/tex] (something close to the origin; the origin itself won't work because [tex]\mathbf r(0,0)=\mathbf 0[/tex]):
[tex]\mathbf r_u\times\mathbf r_v\bigg|_{(u,v)=(0.1,0)}=-0.02\,\mathbf i+0.01\,\mathbf k[/tex]
Now in Cartesian space, the point [tex](-0.02,0,0.01)[/tex] would lie above the paraboloid, since
[tex](-0.02)^2+0^2=0.0004>0[/tex]
which means this normal vector has an inward orientation. To rectify this, swap the sign and replace [tex]\mathbf r_u\times\mathbf r_v[/tex] with [tex]\mathbf r_v\times\mathbf r_u[/tex] (which differs by a sign because the cross product is anticommutative).
[tex]\implies\mathbf F(\mathbf r(u,v))\cdot(\mathbf r_v\times\mathbf r_u)=u^2(\cos v+\sin v)-2u^3[/tex]
So the value of the surface integral is
[tex]\displaystyle\iint_S\mathbf F\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}(u^2(\cos v+\sin v)-2u^3)\,\mathrm dv\,\mathrm du=\pi[/tex]
