Stokes theorem to find flux Find flux of field $(z,x,y)$ on curve $x^2+y^2=1, z=1$ oriented positive looking from positive part of z axis using Stokes formula. So this should come to calculating $3\iint_S dS$ or $3\int\limits_{0}^{2\pi}\int\limits_{0}^{1}drd\phi$. Is this correct? I think we should use plus in front of integral because it is oriented positively.
 A: The Stokes' Theorem states that $\displaystyle \int_C \vec{F} \cdot d\vec{r} = \iint_{S}(\nabla \times \vec{F}) \cdot d\vec S$
The surface integral of the curl of a vector field is equal to the line integral of the vector field over the smooth closed boundary curve $C$ which is the bound of the surface.
In this case, boundary curve is $x^2+y^2 = 1$ at $z=1$. We can parametrize the curve as $\vec{r}(t) = (\cos t, \sin t, 1), 0 \leq t \leq 2\pi$.
$\vec{r'}(t) = (-\sin t, \cos t, 0)$
$\vec {F} = (z,x,y) = (1, \cos t, \sin t)$
and then the line integral.
Now if you are not applying Stokes' Theorem and are doing a double integral, you need to find a surface whose boundary curve is $x^2+y^2 = 1$ at $z=1$. It can be any simple surface, let's take a disc itself that is given by $x^2+y^2 = 1$ in plane $z=1$. The normal surface to the plane is $(0,0,1)$ based on the orientation.
$curl(\vec{F}) = \nabla \times \vec{F} = (1,1,1)$
$(\nabla \times \vec{F}) \cdot \hat{n} = 1$ which is a constant so double integral equals to nothing but the surface area of the disc which is $\pi$. Check the result to see both sides (integrals) are same.
