How to evaluate the double integral over a non-closed surface? Let $\vec{F}=(x+2y)e^zi+(ye^z+x^2)j+y^2zk$ and let $S$ be the surface $x^2+y^2+z=1$, $z\geq 0$. If $\hat{n}$ is a unit normal to $S$ and $$\left|\iint_S(\nabla\times \vec{F})\cdot \hat{n}\, dS\right|=\alpha\pi.$$ Then $\alpha=?$
We can't apply Gauss divergence theorem here since the surface S is not closed. How to proceed in this question then? Please help.
 A: Notice that the boundary of the surface is the curve $\{(x,y,z):x^2+y^2=1\cap z=0\}$ By Stokes' theorem if two surfaces share the same boundary then the integral of the curl on both surfaces will be identical. I.e.
$$\iint\limits_S (\nabla \times F)\cdot dS = \iint\limits_{x^2+y^2\leq 1 \:\cap\:z=0} (\nabla \times F)\cdot dS$$
with both oriented either upwards or downwards.
Why does this make life easier? For starters, the Jacobian between the $z=0$ plane and the usual $xy$ coordinates is $1$ (the Jacobian of anything from itself to itself is $1$) and the normal vector only points in the $z$ direction, meaning we don't have to even compute the whole curl, only the $z$ component, which is
$$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x-2e^z$$
This gives us the following equality
$$\iint\limits_{x^2+y^2\leq 1 \:\cap\:z=0} (\nabla \times F)\cdot dS = \iint\limits_{x^2+y^2\leq 1}2x-2e^0\:dA = \iint\limits_{x^2+y^2\leq 1}2x\:dA - \iint\limits_{x^2+y^2\leq 1}2\:dA$$
$2x$ is an odd function, so its integral will vanish on the disk by $x$ symmetry. The only integral left is a constant, which just gives us the area of the surface times that constant:
$$\iint\limits_{x^2+y^2\leq 1 \:\cap\:z=0} (\nabla \times F)\cdot dS = -2\pi$$
Thus $\alpha =2$
