Answer
$$2 \pi a$$
Work Step by Step
$\vec{r_u} \times \vec{r_v}=\cos u i+ \sin u j+0 k$
Now, $$I=\iint_{S} F \cdot N \ d S=\int_{0}^{2 \pi} \int_{0}^{a} (1) \ dv \ du \\= \int_{0}^{2 \pi}[v]_{0}^{a} \ du \\=a \times \int_{0}^{2 \pi} \ du \\=a \times [u]_{0}^{2\pi} \\=a \times (2 \pi-0) \\=2 \pi a$$
