TITLE: Without Stokes's Theorem - Calculate $\iint_S \operatorname{curl} \mathbf{F} \cdot\; d\mathbf{S}$ for $\mathbf{F} = yz^2\mathbf{i}$ - 2013 10C
QUESTION [5 upvotes]: 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 $\color{green}{\mathbf{n} = (0, 0, 1)}$?

REPLY [2 votes]: 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 $x=r\cos\theta$, $y=r\sin \theta$,
$$I= \int_0^{2\pi} \int_0^2 8r^2 \sin^2 \theta -2r^3 \sin^2 \theta -16 r +8r^2 -r^3 drd\theta = -16\pi$$