Without Stokes's Theorem - Calculate $\iint_S \operatorname{curl} \mathbf{F} \cdot\; d\mathbf{S}$ for $\mathbf{F} = (-y^3,x^3,z^3)$ - 2012 9C Question: 2012 9C. Consider the (cutoff) paraboloid defined by $z= x^2 + y^2 , \frac{1}{9} \le z \le 1$. Sketch the surface. 
Verify Stokes’s Theorem for for $\mathbf{F} = (-y^3,x^3,z^3)$. 
Herein, I enquire only about directly computing $ \iint_S (\nabla × F )· d\mathbf{S}$, which I separate from the computation with Stokes's Theorem. 
My effort: Parameterise with $\mathbf{z} (x,y) = ( x, y, x^2 + y^2 )$
Then $\mathbf{\nabla × F} = (0, 0, 3(x^2 + y^2))$.
I think that S is supposed to be the projection of $ x^2 + y^2 \le 1/9 $ onto z = 0 plane.

$1.$ This guess proved correct, but why is this S? Why isn't S the projection of $ x^2 + y^2 \le 1 $ or something else ? Why must it be projected onto the z = 0 plane? Why not somewhere else?

♦ denote objects that don't need to be computed because they're dot-producted with 0.
$ \iint_{S} (\nabla × F ) · {\mathbf{n}} \, dS = \iint_{ x^2 + y^2 \le z, 1/9 \le z \le 1  } (\nabla × F ) \cdot {\mathbf{n}} \, dA 
\\ = \iint_{ x^2 + y^2 \le z, 1/9 \le z \le 1 } (0, 0, 3(x^2 + y^2)) \cdot (-\partial_x \mathbf{z}, -\partial_y \mathbf{z},1) \, dA
\\ = \iint_{ x^2 + y^2 \le z, 1/9 \le z \le 1 } (0, 0, 3\color{orangered}{(x^2 + y^2)}) \cdot (♦, ♦,1) \, dA.$

$2.$ How would one determine that it's $\color{green}{(\partial_{x} \mathbf{z} \times \partial_{y} \mathbf{z}} )$ that 'produces the outward normal which is what we want', and not $\color{darkred}{(\partial_{y} \mathbf{z}  \times \partial_{x} \mathbf{z} )}$?
  I know that everything must be positively oriented by convention. If the answer is geometric or visual, would you please provide a picture? 

Conver tot polar coordinates, where $ z = radius^2 = 1/9  \implies radius = 1/3 $.
Then above = $ 3 \int^{2\pi}_{0} \, d\theta  \int^{1}_{1/3} \color{orangered}{r^2} \, r \, dr  \, d\theta  = 3 \cdot 2\pi \cdot \frac{1}{4} (1^4 - (1/3)^4) = 40\pi/27.   $ 

For the normal you have $\frac{\partial \mathbf{z}}{\partial x}=(1,0,2x), \quad\frac{\partial \mathbf{z}}{\partial y}=(0,1,2y)$ so $$\frac{\partial \mathbf{z}}{\partial x}\times \frac{\partial \mathbf{z}}{\partial y}=(-2x,-2y,1)$$ So you were correct that the normal is $(-\partial_x\mathbf{z},-\partial_y\mathbf{z},1)=(-2x,-2y,1)$.
So the direct computation gives $$\iint_S \nabla\times\mathbf{F}\cdot\mathrm{d}\mathbf{S}=\iint_S 3(x^2+y^2)\,\mathrm{d}A=\iint_S3r^3\,\mathrm{d}r\,\mathrm{d}\theta$$ where I have switched to polar as you did. Your error here was with the region of integration $S$. We have $1/9\le z\le 1$ so $$1/9\le x^2+y^2=r^2\le 1\implies 1/3\le r\le 1$$ Therefore we get for the integral $$3\cdot{1\over 4}(2\pi)(1^4-(1/3)^4)=\frac{3\pi}{2}\frac{80}{81}=\frac{40\pi}{27}$$

EDIT: $1.$ To determine the region $S$, you need to look at what your parameters range between to span out $S$. In this case, the parameters are the two variables $x$ and $y$. When $z$ goes from $1/9$ to $1$, this means that $x,y$ cover the region between two circles of radius $1/3$ and $1$. If you switch to polar coords, which is equivalent to making the substitutions $$x=r\cos\theta \\ y=r\sin\theta$$ then the region can be expressed nicely as $S=\{(r,\theta): 1/3\le r\le 1,\;0\le\theta\le 2\pi\}$.
$2.$ Determining which direction the normal should take is, as you surmised, entirely a geometric matter. In this case, the problem only says to "verify Stokes' Theorem" so it does not specify the direction of the normal. We get to choose. I took the normal as $(-2x,-2y,1)$. Note that the $z$-component is positive, so the normal faces up. If you imaging (or graph) a paraboloid, you will see that the inward normal points up, and the outward normal faces down. Therefore this vector, which points up, must be the inward normal. If points to the inside the paraboloid instead of to the outside.
To apply Stokes' Theorem by using a line integral(s) over the boundary circles (when $z=1/9$ and when $z=1$) you have to orient them consistently with this choice of normal (remember, I chose arbitrarily - but in this step we don't have any freedom to choose). I will explain this in the other post.

Second part was successfully moved, apparently there was a period when nobody could answer questions which is why I was getting an error message.
