State Stokes' theorem for a vector field $\mathbf{A}$.

By applying Stokes' theorem to the vector field $\mathbf{A}=\phi \mathbf{k}$, where $\mathbf{k}$ is an arbitrary constant vector in $\mathbb{R}^{3}$ and $\phi$ is a scalar field defined on a surface $S$ bounded by a curve $\partial S$, show that

$$\int_{S} d \mathbf{S} \times \nabla \phi=\int_{\partial S} \phi d \mathbf{x}$$

For the vector field $\mathbf{A}=x^{2} y^{4}(1,1,1)$ in Cartesian coordinates, evaluate the line integral

$$I=\int \mathbf{A} \cdot d \mathbf{x}$$

around the boundary of the quadrant of the unit circle lying between the $x$ - and $y$ axes, that is, along the straight line from $(0,0,0)$ to $(1,0,0)$, then the circular arc $x^{2}+y^{2}=1, z=0$ from $(1,0,0)$ to $(0,1,0)$ and finally the straight line from $(0,1,0)$ back to $(0,0,0)$.