Without Stokes's Theorem - Calculate $\iint_S \operatorname{curl} \mathbf{F} \cdot\; d\mathbf{S}$ for $\mathbf{F} = yz^2\mathbf{i}$ - 2013 10C 
2013 10C. Consider the bounded surface S that is the union of $x^2 + y^2 = 4$ for $−2 \le z \le 2$
  and $(4 − z)^2 = x^2 + y^2 $ for $2 \le z \le 4.$ Sketch the surface.
  Use suitable parametrisations for the two parts of S to verify Stokes’s Theorem for 
  for $\mathbf{F} = (yz^2,0,0)$. picture

Herein, I enquire only about directly computing $ \iint_S (\nabla × F )· d\mathbf{S}$.
Denote the $2 \le z \le 4$ cone P, and the $-2 \le z \le 2$ cylinder C. I use only the first paragraph of this.
Then $\mathbf{\nabla × F} = (0, 2yz, -z^2)$, 
 $ \iint_P (\nabla × F ) · d\mathbf{S} = \iint_{x^2 + y^2 \le z^2, z = 2} (\nabla × F ) \cdot \color{darkred}{\mathbf{n}} \, dA \\ = \iint_{x^2 + y^2 \le z^2, z = 2} (♦, ♦, \underbrace{-z^2}_{=-4} ) \cdot \color{darkred}{(0, 0,-1)} \, dA = \iint_{x^2 + y^2 \le z^2, z = 2} 4 dA = 4\pi(2)^2 $. 
♦ denote objects that don't need to be computed because they're dot-producted with 0.
$\large{2.}$ To ellya especially, is it necessary to parameterise the P piece? Comparing my work to yours, I see that we differ only by a negative sign? Yet my work has far fewer steps. Does it not function? 
How would one determine that the correct normal vector is $\color{green}{\mathbf{n} = (0, 0, 1)}$?
 A: I would parameterize from the start,  so on $C$ $-2\le z\le 2 $ and $x^2+y^2=4$. This is a cylinder of radius $2$ centred at the origin with height $4$.
So let $x=2\cos\phi,y=2\sin\phi $ where $ 0\le\phi\le 2\pi $ now we parametrise our surface $C$ as $\sigma (\phi,z)=(2\cos\phi,2\sin\phi,z)$, and now $F=(2z^2\sin\phi,0,0)$.
Here we let $ z$ increase from -2 to 2 so with this orientation, the normal $n=\sigma_\phi\times\sigma_z  = \left| \begin{array}{ccc}
i & j & k \\
-2\sin\phi & 2\cos\phi & 0 \\
0 & 0 & 1 \end{array} \right|=(2\cos\phi,2\sin\phi,0)$
So we have $\int\int_C(\nabla\times F)\cdot dS=\int_0^{2\pi}\int_{-2}^2(\nabla\times F)\cdot(2\cos\phi,2\sin\phi,0)dzd\phi$
Now $ \nabla\times F=\left| \begin{array}{ccc}
i & j & k \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
yz^2 & 0 & 0 \end{array} \right|=(0,\frac{\partial}{\partial z}(yz^2),-z^2)=(0,2yz,-z^2)=(0,4z\sin\phi,-z^2)$
So $\iint_C(\nabla\times F) \cdot d\mathbf{S}=\int_0^{2\pi}\int_{-2}^2(0,4z\sin\phi,-z^2)\cdot(2\cos\phi,2\sin\phi,0)dz~d\phi       
\\ =\int_0^{2\pi}\int_{-2}^2 8z\sin^2 \phi \, dz \, d\phi
=4 \int_0^{2\pi} (1-\cos2\phi) \, d\theta \; \int_{-2}^2 z \, dz =4 \color{#009900 }{[z^2]^2_{-2}}  \;  [\phi-\frac{1}{2}\sin 2\phi]^{2\pi}_0=\color{#009900 }{0}$
I took a different route because the "$\frac{\partial z}{\partial y}\times\frac{\partial z}{\partial x}$" didn't make sense to me. They are not vectors.
To do the $P$ integral I would also parametrize, but this time you are integrating over a cone.
Parametrising $P$, here we let $\sigma(\phi,z)=((4-z)\cos\phi,(4-z)\sin\phi,z)$, here
$n=\sigma_\phi\times\sigma_z  =\left| \begin{array}{ccc}
i & j & k \\
(z-4)\sin\phi & (4-z)\cos\phi & 0 \\
-\cos\phi & -\sin\phi & 1 \end{array} \right|=((4-z)\cos\phi,(4-z)\sin\phi,4-z)$
so $\int\int_P (\nabla\times F)\cdot dS=\int_0^{2\pi}\int_2^4(0,2z(4-z)\sin\phi,-z^2)\cdot((4-z)\cos\phi,(4-z)\sin\phi,4-z)dzd\phi$
$=\int_0^{2\pi}\int_2^42z(4-z)^2\sin^2\phi-z^2(4-z)dz d\phi$
$=\int_0^{2\pi}\int_2^4z(4-z)^2(1-\cos 2\phi)-(4z^2-z^3)dzd\phi$
$=\frac{20}{3}(\phi-\frac{1}{2}\cos 2\phi|_0^{2\pi})-2\pi(\frac{44}{3})$
$=2\pi(\frac{20}{3}-\frac{44}{3})=-2\pi(\frac{24}{3})=-8(2)\pi=-16\pi$
Halelujah! To answer your new questions:


*

*Cross producting two vectors produces a vector normal to both of them. Since $z$ is oriented positively, $\sigma_\phi\times\sigma_z$ produces the outward normal which is what we want, taking the other order gives us an inward normal.

*In my belief you got lucky with your computation, since the integral you performed was on the base of the cone which is not actually part of the surface.
Addendum
I believe your main issue was with the reasoning for taking the normal to be $\sigma_\phi\times\sigma_z$ over $\sigma_z\times\sigma_\phi$. This may seem strange Please refer here.
A: Actually had a go at this a few weeks ago but kept getting the wrong answer. I'll just do the cone, since I think Ellya's calculation is correct for the cylinder. However for the cone, if we do a line integral around the base of cone we get an answer of $-16\pi$, which disagrees with his result. Here's my proof for the cone:
Let $I$ be the surface integral over the cone  with normal $\mathbf n$ taken in an upwards direction.
$$I= \iint_P (\nabla \times \mathbf F)\cdot d\mathbf S = \iint_P (0,2yz,-z^2)\cdot \mathbf n  d\mathbf S$$
Now we will use the formula for a flux integral $\iint_S \mathbf F\cdot \mathbf n \, d\mathbf S = \iint_D (F_1, F_2, F_3) \cdot \color{green}{(-\partial_x f, \partial_y f, 1)} \, dA  = \iint_D (-F_1 \partial_x f - F_2 \partial_y f +F_3 ) \, dxdy$,
where $D$ is a projection of $S$ onto the x-y plane, and $f(x,y)$ is the equation of the surface. For us, $f_x' = \frac{-x}{\sqrt{x^2 + y^2}}$ and $f_y' = \frac{-y}{\sqrt{x^2 + y^2}}$. So,
$$I = \iint_D \frac{2y^2 z}{\sqrt{x^2 + y^2}} - z^2 dxdy$$
Noting that $z = 4-\sqrt{x^2 + y^2}$ and then letting $x=r\cos\theta$, $y=r\sin \theta$,
$$I= \int_0^{2\pi} \int_0^2 8r^2 \sin^2 \theta -2r^3 \sin^2 \theta -16 r +8r^2 -r^3 drd\theta = -16\pi$$
