Answer
$8 \pi$
Work Step by Step
Stoke's Theorem: Let us consider that $C$ be the closed curve which encloses surface $S$ for a vector field $F$. This also states that the line integral and the surface integral must be equal.
That is, $\oint_C F \ dx=\iint_S (\nabla \times F) \cdot n \ dS$
Here, $\nabla \times F=\begin{vmatrix} i&j&k \\ \cos \phi & -1 & \sin \phi \\ -\cos \phi & -1 & -\sin \phi \end{vmatrix}=\lt \sin \phi, 0, -\cos \phi\gt$
Thus, the surface integral can be computed as:
$\iint_S 1 \cdot ds=\int_0^{2 \pi} |n| \ dA\\= \int_0^{2\pi} 4( \sin^2 \phi +\cos^2 \phi) \ d \phi \\=(4) \times 2 \pi\\=8 \pi$
