Answer
$$\pi $$
Work Step by Step
$F \cdot n =\dfrac{2xy^2 }{\sqrt {x^2+y^2}}+4xy +1 $
Now, $$I=\iint_{S} F \cdot n \ dS \\=\iint_{R}\dfrac{2xy^2 }{\sqrt {x^2+y^2}}+4xy +1 \ dy \ dx \\=\int_0^{2 \pi} \int_{0}^{1} \dfrac{2 }{\sqrt {r^2}} \times (r \cos \theta)(r \sin \theta)^2+4 (r \cos \theta)(r \sin \theta) +1 \\=\int_0^{2 \pi} \int_{0}^{1} (2r^3 \sin^2 \theta \cos \theta+4r^3 \sin \theta \cos \theta +r) dr d \theta \\= \int_0^{2 \pi} [\dfrac{1}{2} \sin^2 \theta \cos \theta +\sin \theta \cos \theta+\dfrac{1}{2}] \ d \theta $$
Now, we will use a calculator
So, $$I=\int_0^{2 \pi} [\dfrac{1}{2} \sin^2 \theta \cos \theta +\sin \theta \cos \theta+\dfrac{1}{2}] \ d \theta =\pi $$
