Use Stokes’ Theorem to calculate $\iint_{S} \nabla \times V· dS$ on the given paraboloid 

On the boundary of the surface we have $z=4$, and hence $4=\frac{x^2+y2}{4}$. Hence the boundary is 
\[C:       z=4, \ \ \ \ x^2+y^2=16,\]
which can be parametrized as $(x,y,z)=(4\cos t, 4\sin t, 4)$, $ 0\leq t \leq 2\pi $. Using Stokes' Theorem and noting that $\frac{dz}{dt}=0$ we have 
\[\iint_{S} \nabla \times V \cdot dS=\int_{C}V \cdot dr=\int_C xy^2zdx- 4x^2y dy+\frac{z-1}{x^2+2y^2+1}dz\]
\[\int_0^{2\pi}(xy^2z\frac{dx}{dt}- 4x^2y \frac{dy}{dt}+\frac{z-1}{x^2+2y^2+1}\frac{dz}{dt}) dt\]
\[=\int_0^{2\pi}4\cos t (4\sin t)^2(4)(-4\sin t)-4(4\cos t)^2(4\sin t)(4\cos t )+0  dt\]
\[=\int_0^{2\pi}-4^5(\cos t \sin^3 t+\cos^3 t \sin t) dt=-4^5\int_0^{2\pi} \sin t \cos t (\sin^2 t+\cos^2 t)dt\]
\[=-4^{5}\int_0^{2\pi}\sin t \cos t dt=-\frac{4^5}{2}\int_0^{2 \pi} \sin 2t dt=-\frac{4^5}{2} (\frac{-\cos 2t}{2})|_0^{2\pi}=0.\]