Surface Integral How to find the surface integral of such term
$\int \int_{F_+} (y-z)dydz + (z-x)dzdx +(x-y)dxdy$
where $F_+$ is the surface $x^2+y^2 = z^2$ $(0 \leq z \leq h )$ oriented outward.
There were other problems, where I could just parametrize the $F_+$ and then compute $\vec{n}$ and simply put in the surface integral formula, but here I'm confused on how to carry on. Hints please.
Thanks.
 A: First, parametrize the surface $S$:
$$
x=x, \quad y=y, \quad z=\sqrt{x^2+y^2},
$$
with $x,y \in D := \{(x,y)\;|\; x^2+y^2\le h^2\} $.
Second, compute $\vec{r}_x \times \vec{r}_y= (\frac{-x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},1)$.
Third, note that you can rewrite your integral as follows:
$$
\iint_S \vec{F}\cdot d\vec{S},
$$
with $\vec{F}=(y-z,z-x,x-y)$.
You now have everything you need to finish:
$$
\iint_S \vec{F}\cdot d\vec{S} = \iint_D \vec{F}(x,y)\cdot (\vec{r}_x \times \vec{r}_y) dA =
 \iint_{D}(y-\sqrt{x^2+y^2})\frac{-x}{\sqrt{x^2+y^2}}+(\sqrt{x^2+y^2}-x)\frac{y}{\sqrt{x^2+y^2}}+(x-y)dA,
$$
i.e.:
$$
\iint_{D}\frac{-xy}{\sqrt{x^2+y^2}}-1+1-\frac{xy}{\sqrt{x^2+y^2}}+(x-y)dA=\iint_{D}\frac{-2xy}{\sqrt{x^2+y^2}}+x-y dA
$$
Using polar coordinates to simplify the integral yields:
$$
\int_{0}^{2\pi}\int_0^h\left(\frac{-2r\cos{\theta}r\sin{\theta}}{r}+r\cos{\theta}-r\sin{\theta}\right)r drd\theta=0.
$$
Note. Alternatively, with the Divergence theorem:
$$
\iint_S \vec{F}\cdot d\vec{S} = \iiint_V div \vec{F} \;dV - \iint_{S_1}\vec{F} \cdot d\vec{S},
$$
where $V$ is the closed region above the cone and below the plane $z=h$, and $S_1$ is the surface of the plane $z=h$ above $D$.
It follows that 
$$
\iint_S \vec{F}\cdot d\vec{S} =0 - \int_{0}^{2\pi}\int_0^h (r\cos{\theta}-r\sin{\theta})r drd\theta =0.
$$
